Chapter 10 Vector Algebra

Given:

Vector $\vec{a} = 2\hat{i}− \hat{j}+ 2\hat{k}$

Vector $\vec{b}= −\hat{i}+ \hat{j}+ 3\hat{k}$

To Find:

The unit vector in the direction of the sum of $\vec{a}$ and $\vec{b}$.

Solution:

Let $\vec{c}$ be the sum of the vectors $\vec{a}$ and $\vec{b}$.

$\vec{c} = \vec{a} + \vec{b}$

Substitute the given vectors:

$\vec{c} = (2\hat{i}− \hat{j}+ 2\hat{k}) + (−\hat{i}+ \hat{j}+ 3\hat{k})$

Combine the corresponding components:

$\vec{c} = (2 - 1)\hat{i} + (-1 + 1)\hat{j} + (2 + 3)\hat{k}$

$\vec{c} = 1\hat{i} + 0\hat{j} + 5\hat{k}$

$\vec{c} = \hat{i} + 5\hat{k}$

Next, we need to find the magnitude of the vector $\vec{c}$. The magnitude of a vector $x\hat{i} + y\hat{j} + z\hat{k}$ is given by $\sqrt{x^2 + y^2 + z^2}$.

$|\vec{c}| = \sqrt{(1)^2 + (0)^2 + (5)^2}$

$|\vec{c}| = \sqrt{1 + 0 + 25}$

$|\vec{c}| = \sqrt{26}$

The unit vector in the direction of $\vec{c}$, denoted by $\hat{c}$, is given by the formula $\hat{c} = \frac{\vec{c}}{|\vec{c}|}$.

Substitute the values of $\vec{c}$ and $|\vec{c}|$:

$\hat{c} = \frac{\hat{i} + 5\hat{k}}{\sqrt{26}}$

This can also be written by distributing the denominator:

$\hat{c} = \frac{1}{\sqrt{26}}\hat{i} + \frac{0}{\sqrt{26}}\hat{j} + \frac{5}{\sqrt{26}}\hat{k}$

$\hat{c} = \frac{1}{\sqrt{26}}\hat{i} + \frac{5}{\sqrt{26}}\hat{k}$

Final Answer:

The unit vector in the direction of the sum of the given vectors is $\frac{1}{\sqrt{26}}\hat{i} + \frac{5}{\sqrt{26}}\hat{k}$.

Given:

Point P with coordinates (1, 3, 2).

Point Q with coordinates (–1, 0, 8).

Desired magnitude of the vector is 11.

To Find:

A vector of magnitude 11 in the direction opposite to that of $\overrightarrow{PQ}$.

Solution:

First, we find the vector $\overrightarrow{PQ}$. The position vector of point P is $\vec{p} = 1\hat{i} + 3\hat{j} + 2\hat{k}$, and the position vector of point Q is $\vec{q} = -1\hat{i} + 0\hat{j} + 8\hat{k}$.

The vector $\overrightarrow{PQ}$ is given by the difference of the position vectors of Q and P:

$\overrightarrow{PQ} = \vec{q} - \vec{p}$

$\overrightarrow{PQ} = (-\hat{i} + 0\hat{j} + 8\hat{k}) - (\hat{i} + 3\hat{j} + 2\hat{k})$

$\overrightarrow{PQ} = (-1 - 1)\hat{i} + (0 - 3)\hat{j} + (8 - 2)\hat{k}$

$\overrightarrow{PQ} = -2\hat{i} - 3\hat{j} + 6\hat{k}$

Next, we find the magnitude of $\overrightarrow{PQ}$. The magnitude of a vector $x\hat{i} + y\hat{j} + z\hat{k}$ is $|\vec{v}| = \sqrt{x^2 + y^2 + z^2}$.

$|\overrightarrow{PQ}| = \sqrt{(-2)^2 + (-3)^2 + (6)^2}$

$|\overrightarrow{PQ}| = \sqrt{4 + 9 + 36}$

$|\overrightarrow{PQ}| = \sqrt{49}$

$|\overrightarrow{PQ}| = 7$

The unit vector in the direction of $\overrightarrow{PQ}$ is $\widehat{PQ} = \frac{\overrightarrow{PQ}}{|\overrightarrow{PQ}|}$.

$\widehat{PQ} = \frac{-2\hat{i} - 3\hat{j} + 6\hat{k}}{7}$

$\widehat{PQ} = -\frac{2}{7}\hat{i} - \frac{3}{7}\hat{j} + \frac{6}{7}\hat{k}$

The unit vector in the direction opposite to $\overrightarrow{PQ}$ is $-\widehat{PQ}$.

$-\widehat{PQ} = - \left( -\frac{2}{7}\hat{i} - \frac{3}{7}\hat{j} + \frac{6}{7}\hat{k} \right)$

$-\widehat{PQ} = \frac{2}{7}\hat{i} + \frac{3}{7}\hat{j} - \frac{6}{7}\hat{k}$

Finally, to find a vector of magnitude 11 in the direction opposite to $\overrightarrow{PQ}$, we multiply the unit vector $-\widehat{PQ}$ by the desired magnitude, 11.

Required vector = $11 \times (-\widehat{PQ})$

Required vector = $11 \left( \frac{2}{7}\hat{i} + \frac{3}{7}\hat{j} - \frac{6}{7}\hat{k} \right)$

Required vector = $\frac{11 \times 2}{7}\hat{i} + \frac{11 \times 3}{7}\hat{j} + \frac{11 \times (-6)}{7}\hat{k}$

Required vector = $\frac{22}{7}\hat{i} + \frac{33}{7}\hat{j} - \frac{66}{7}\hat{k}$

Final Answer:

The vector of magnitude 11 in the direction opposite to that of $\overrightarrow{PQ}$ is $\frac{22}{7}\hat{i} + \frac{33}{7}\hat{j} - \frac{66}{7}\hat{k}$.

Given:

Position vector of point P, $\overrightarrow{OP} = \vec{p} = 2\vec{a} + \vec{b}$.

Position vector of point Q, $\overrightarrow{OQ} = \vec{q} = \vec{a} - 2\vec{b}$.

The ratio of division is $m:n = 1:2$.

To Find:

The position vector of point R ($\overrightarrow{OR}$ or $\vec{r}$) which divides the line segment PQ in the ratio 1:2, both internally and externally.

Solution:

Let R be the point that divides the line segment PQ in the ratio $m:n$. According to the section formula, the position vector $\vec{r}$ of point R is given by:

(i) Internal Division:

When R divides PQ internally in the ratio $m:n$, the position vector $\vec{r}$ is given by:

$\vec{r}_{\text{internal}} = \frac{n\vec{p} + m\vec{q}}{m+n}$

Here, $m=1$, $n=2$, $\vec{p} = 2\vec{a} + \vec{b}$, and $\vec{q} = \vec{a} - 2\vec{b}$.

Substitute the values into the formula:

$\vec{r}_{\text{internal}} = \frac{2(2\vec{a} + \vec{b}) + 1(\vec{a} - 2\vec{b})}{1+2}$

$\vec{r}_{\text{internal}} = \frac{4\vec{a} + 2\vec{b} + \vec{a} - 2\vec{b}}{3}$

Combine like terms:

$\vec{r}_{\text{internal}} = \frac{(4\vec{a} + \vec{a}) + (2\vec{b} - 2\vec{b})}{3}$

$\vec{r}_{\text{internal}} = \frac{5\vec{a} + 0\vec{b}}{3}$

$\vec{r}_{\text{internal}} = \frac{5\vec{a}}{3}$

(ii) External Division:

When R divides PQ externally in the ratio $m:n$, the position vector $\vec{r}$ is given by:

$\vec{r}_{\text{external}} = \frac{n\vec{p} - m\vec{q}}{n-m}$

Here, $m=1$, $n=2$, $\vec{p} = 2\vec{a} + \vec{b}$, and $\vec{q} = \vec{a} - 2\vec{b}$. Note that for external division using the ratio $m:n$, the point R lies outside the segment PQ on the line passing through P and Q. The formula $\frac{n\vec{p} - m\vec{q}}{n-m}$ is used when R divides PQ externally in the ratio $m:n$ (i.e., PR:QR = m:n). It can also be written as $\frac{m\vec{q} - n\vec{p}}{m-n}$. We will use the first form.

Substitute the values into the formula:

$\vec{r}_{\text{external}} = \frac{2(2\vec{a} + \vec{b}) - 1(\vec{a} - 2\vec{b})}{2-1}$

$\vec{r}_{\text{external}} = \frac{4\vec{a} + 2\vec{b} - \vec{a} + 2\vec{b}}{1}$

Combine like terms:

$\vec{r}_{\text{external}} = (4\vec{a} - \vec{a}) + (2\vec{b} + 2\vec{b})$

$\vec{r}_{\text{external}} = 3\vec{a} + 4\vec{b}$

Final Answer:

(i) The position vector of R when it divides PQ internally in the ratio 1:2 is $\frac{5\vec{a}}{3}$.

(ii) The position vector of R when it divides PQ externally in the ratio 1:2 is $3\vec{a} + 4\vec{b}$.

Given:

Point A = (–1, –1, 2)

Point B = (2, m, 5)

Point C = (3, 11, 6)

The points A, B, and C are collinear.

To Find:

The value of m.

Solution:

If three points A, B, and C are collinear, then the vector $\overrightarrow{AB}$ must be parallel to the vector $\overrightarrow{BC}$. This means that $\overrightarrow{AB}$ is a scalar multiple of $\overrightarrow{BC}$, i.e., $\overrightarrow{AB} = k \overrightarrow{BC}$ for some scalar $k$.

First, we find the vector $\overrightarrow{AB}$. The coordinates of A are $(-1, -1, 2)$ and B are $(2, m, 5)$.

$\overrightarrow{AB} = (2 - (-1))\hat{i} + (m - (-1))\hat{j} + (5 - 2)\hat{k}$

$\overrightarrow{AB} = (2 + 1)\hat{i} + (m + 1)\hat{j} + 3\hat{k}$

$\overrightarrow{AB} = 3\hat{i} + (m + 1)\hat{j} + 3\hat{k}$

Next, we find the vector $\overrightarrow{BC}$. The coordinates of B are $(2, m, 5)$ and C are $(3, 11, 6)$.

$\overrightarrow{BC} = (3 - 2)\hat{i} + (11 - m)\hat{j} + (6 - 5)\hat{k}$

$\overrightarrow{BC} = 1\hat{i} + (11 - m)\hat{j} + 1\hat{k}$

Since A, B, C are collinear, $\overrightarrow{AB} = k \overrightarrow{BC}$.

$3\hat{i} + (m + 1)\hat{j} + 3\hat{k} = k(1\hat{i} + (11 - m)\hat{j} + 1\hat{k})$

$3\hat{i} + (m + 1)\hat{j} + 3\hat{k} = k\hat{i} + k(11 - m)\hat{j} + k\hat{k}$

Equating the coefficients of $\hat{i}$, $\hat{j}$, and $\hat{k}$:

Coefficient of $\hat{i}$: $3 = k$

Coefficient of $\hat{k}$: $3 = k$

From both equations, we find that $k = 3$.

Now, equate the coefficients of $\hat{j}$ and substitute the value of $k$:

Coefficient of $\hat{j}$: $m + 1 = k(11 - m)$

$m + 1 = 3(11 - m)$

$m + 1 = 33 - 3m$

Add $3m$ to both sides:

$m + 3m + 1 = 33$

$4m + 1 = 33$

Subtract 1 from both sides:

$4m = 33 - 1$

$4m = 32$

Divide by 4:

$m = \frac{32}{4}$

$m = 8$

Final Answer:

The value of m is 8.

Given:

Magnitude of the vector $\vec{r}$, $|\vec{r}| = 3\sqrt{2}$ units.

Angle with the y-axis, $\beta = \frac{\pi}{4}$.

Angle with the z-axis, $\gamma = \frac{\pi}{2}$.

To Find:

The vector $\vec{r}$.

Solution:

Let the vector $\vec{r}$ make an angle $\alpha$ with the x-axis, $\beta$ with the y-axis, and $\gamma$ with the z-axis.

The direction cosines of the vector $\vec{r}$ are $\cos\alpha$, $\cos\beta$, and $\cos\gamma$.

We are given $\beta = \frac{\pi}{4}$ and $\gamma = \frac{\pi}{2}$.

Let's calculate the cosine of these angles:

$\cos\beta = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$

$\cos\gamma = \cos\left(\frac{\pi}{2}\right) = 0$

We know that the sum of the squares of the direction cosines of any vector is equal to 1:

$\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$

Substitute the known values:

$\cos^2\alpha + \left(\frac{1}{\sqrt{2}}\right)^2 + (0)^2 = 1$

$\cos^2\alpha + \frac{1}{2} + 0 = 1$

$\cos^2\alpha = 1 - \frac{1}{2}$

$\cos^2\alpha = \frac{1}{2}$

Taking the square root of both sides:

$\cos\alpha = \pm\sqrt{\frac{1}{2}}$

$\cos\alpha = \pm\frac{1}{\sqrt{2}}$

Thus, there are two possible values for $\cos\alpha$, which means there are two possible vectors that satisfy the given conditions.

Let the vector $\vec{r}$ be represented as $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.

The components of the vector are related to its magnitude and direction cosines by:

$x = |\vec{r}|\cos\alpha$

$y = |\vec{r}|\cos\beta$

$z = |\vec{r}|\cos\gamma$

We are given $|\vec{r}| = 3\sqrt{2}$, $\cos\beta = \frac{1}{\sqrt{2}}$, and $\cos\gamma = 0$.

For the y-component:

$y = (3\sqrt{2})\left(\frac{1}{\sqrt{2}}\right) = 3$

For the z-component:

$z = (3\sqrt{2})(0) = 0$

For the x-component, we have two cases based on the values of $\cos\alpha$:

Case 1: $\cos\alpha = \frac{1}{\sqrt{2}}$

$x = (3\sqrt{2})\left(\frac{1}{\sqrt{2}}\right) = 3$

In this case, the vector is $\vec{r}_1 = 3\hat{i} + 3\hat{j} + 0\hat{k} = 3\hat{i} + 3\hat{j}$.

Case 2: $\cos\alpha = -\frac{1}{\sqrt{2}}$

$x = (3\sqrt{2})\left(-\frac{1}{\sqrt{2}}\right) = -3$

In this case, the vector is $\vec{r}_2 = -3\hat{i} + 3\hat{j} + 0\hat{k} = -3\hat{i} + 3\hat{j}$.

Final Answer:

There are two possible vectors that satisfy the given conditions:

$\vec{r}_1 = 3\hat{i} + 3\hat{j}$

$\vec{r}_2 = -3\hat{i} + 3\hat{j}$

Given:

Vector $\vec{a} = 2\hat{i} − \hat{j} + \hat{k}$

Vector $\vec{b} = \hat{i} + \hat{j}$

Vector $\vec{c} = \hat{i} + 3\hat{j} − \hat{k}$

The vector $\vec{a}$ is perpendicular to the vector $\lambda\vec{b} + \vec{c}$.

To Find:

The value of the scalar $\lambda$.

Solution:

Two vectors are perpendicular if their dot product is zero.

The given condition states that $\vec{a} \perp (\lambda\vec{b} + \vec{c})$.

Therefore, their dot product must be zero:

$\vec{a} \cdot (\lambda\vec{b} + \vec{c}) = 0$

First, let's find the vector $\lambda\vec{b} + \vec{c}$:

$\lambda\vec{b} = \lambda(\hat{i} + \hat{j}) = \lambda\hat{i} + \lambda\hat{j}$

$\lambda\vec{b} + \vec{c} = (\lambda\hat{i} + \lambda\hat{j}) + (\hat{i} + 3\hat{j} − \hat{k})$

Combine the corresponding components:

$\lambda\vec{b} + \vec{c} = (\lambda + 1)\hat{i} + (\lambda + 3)\hat{j} + (0 - 1)\hat{k}$

$\lambda\vec{b} + \vec{c} = (\lambda + 1)\hat{i} + (\lambda + 3)\hat{j} - \hat{k}$

Now, calculate the dot product of $\vec{a}$ and $(\lambda\vec{b} + \vec{c})$:

$\vec{a} = 2\hat{i} − \hat{j} + \hat{k}$

$(\lambda\vec{b} + \vec{c}) = (\lambda + 1)\hat{i} + (\lambda + 3)\hat{j} - \hat{k}$

$\vec{a} \cdot (\lambda\vec{b} + \vec{c}) = (2)(\lambda + 1) + (-1)(\lambda + 3) + (1)(-1)$

$\vec{a} \cdot (\lambda\vec{b} + \vec{c}) = 2\lambda + 2 - \lambda - 3 - 1$

$\vec{a} \cdot (\lambda\vec{b} + \vec{c}) = (2\lambda - \lambda) + (2 - 3 - 1)$

$\vec{a} \cdot (\lambda\vec{b} + \vec{c}) = \lambda - 2$

According to the condition of perpendicularity, the dot product must be zero:

$\lambda - 2 = 0$

Solve for $\lambda$:

$\lambda = 2$

Final Answer:

The value of $\lambda$ is 2.

Given:

Vector $\vec{u} = \hat{i} + 2\hat{j} + \hat{k}$

Vector $\vec{v} = −\hat{i} + 3\hat{j} + 4\hat{k}$

Required magnitude of the vector is $10\sqrt{3}$.

To Find:

All vectors of magnitude $10\sqrt{3}$ that are perpendicular to the plane containing vectors $\vec{u}$ and $\vec{v}$.

Solution:

A vector perpendicular to the plane containing $\vec{u}$ and $\vec{v}$ can be found by calculating their cross product $\vec{u} \times \vec{v}$.

$\vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ -1 & 3 & 4 \end{vmatrix}$

$= \hat{i}((2)(4) - (1)(3)) - \hat{j}((1)(4) - (1)(-1)) + \hat{k}((1)(3) - (2)(-1))$

$= \hat{i}(8 - 3) - \hat{j}(4 + 1) + \hat{k}(3 + 2)$

$= 5\hat{i} - 5\hat{j} + 5\hat{k}$

Let $\vec{w} = 5\hat{i} - 5\hat{j} + 5\hat{k}$. This vector is perpendicular to the given plane.

Now, find the magnitude of $\vec{w}$:

$|\vec{w}| = |5\hat{i} - 5\hat{j} + 5\hat{k}|$

$|\vec{w}| = \sqrt{(5)^2 + (-5)^2 + (5)^2}$

$|\vec{w}| = \sqrt{25 + 25 + 25}$

$|\vec{w}| = \sqrt{75}$

$|\vec{w}| = \sqrt{25 \times 3} = 5\sqrt{3}$

The unit vector in the direction of $\vec{w}$ is $\hat{w} = \frac{\vec{w}}{|\vec{w}|}$.

$\hat{w} = \frac{5\hat{i} - 5\hat{j} + 5\hat{k}}{5\sqrt{3}}$

$\hat{w} = \frac{5(\hat{i} - \hat{j} + \hat{k})}{5\sqrt{3}}$

$\hat{w} = \frac{\hat{i} - \hat{j} + \hat{k}}{\sqrt{3}}$

Vectors perpendicular to the plane can be in the direction of $\hat{w}$ or $-\hat{w}$. We need vectors with magnitude $10\sqrt{3}$.

The required vectors are given by $\pm (\text{desired magnitude}) \times \hat{w}$.

Required vectors = $\pm (10\sqrt{3}) \times \left( \frac{\hat{i} - \hat{j} + \hat{k}}{\sqrt{3}} \right)$

Required vectors = $\pm 10 (\hat{i} - \hat{j} + \hat{k})$

Required vectors = $\pm (10\hat{i} - 10\hat{j} + 10\hat{k})$

The two such vectors are $10\hat{i} - 10\hat{j} + 10\hat{k}$ and $-10\hat{i} + 10\hat{j} - 10\hat{k}$.

Final Answer:

The vectors of magnitude $10 \sqrt{3}$ perpendicular to the plane of the given vectors are $10\hat{i} - 10\hat{j} + 10\hat{k}$ and $-10\hat{i} + 10\hat{j} - 10\hat{k}$.

Given:

Angles A and B.

To Prove:

$\cos (A – B) = \cos A \cos B + \sin A \sin B$

Proof:

Consider two unit vectors $\mathbf{a}$ and $\mathbf{b}$ in the xy-plane. Let $\mathbf{a}$ make an angle A with the positive x-axis, and $\mathbf{b}$ make an angle B with the positive x-axis.

The coordinates of the vectors can be written in terms of their components:

$\mathbf{a} = \cos A \hat{\mathbf{i}} + \sin A \hat{\mathbf{j}}$

$\mathbf{b} = \cos B \hat{\mathbf{i}} + \sin B \hat{\mathbf{j}}$

The angle between the vectors $\mathbf{a}$ and $\mathbf{b}$ is the absolute difference between their angles with the x-axis, i.e., $|A - B|$.

Using the definition of the dot product of two vectors, $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos\theta$, where $\theta$ is the angle between $\mathbf{a}$ and $\mathbf{b}$.

Since $\mathbf{a}$ and $\mathbf{b}$ are unit vectors, their magnitudes are $|\mathbf{a}| = 1$ and $|\mathbf{b}| = 1$. The angle between them is $|A - B|$.

Therefore, the dot product is:

Alternatively, we can compute the dot product using the component form:

$\mathbf{a} \cdot \mathbf{b} = (\cos A \hat{\mathbf{i}} + \sin A \hat{\mathbf{j}}) \cdot (\cos B \hat{\mathbf{i}} + \sin B \hat{\mathbf{j}})$

Using the property $\hat{\mathbf{i}} \cdot \hat{\mathbf{i}} = 1$, $\hat{\mathbf{j}} \cdot \hat{\mathbf{j}} = 1$, and $\hat{\mathbf{i}} \cdot \hat{\mathbf{j}} = 0$, we get:

Equating the two expressions for the dot product from equations (1) and (2), we get:

Thus, the identity is proved using vectors.

Hence Proved.

Given:

A triangle ABC with side lengths opposite to vertices A, B, and C being a, b, and c respectively.

To Prove:

$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$

Proof:

Let the position vectors of the vertices A, B, and C be $\vec{A}$, $\vec{B}$, and $\vec{C}$ respectively. We can represent the sides of the triangle using vectors:

$\vec{AB} = \vec{B} - \vec{A}$ (Magnitude $|\vec{AB}| = c$)

$\vec{BC} = \vec{C} - \vec{B}$ (Magnitude $|\vec{BC}| = a$)

$\vec{CA} = \vec{A} - \vec{C}$ (Magnitude $|\vec{CA}| = b$)

The area of triangle ABC can be calculated using the magnitude of the cross product of two vectors representing two sides originating from the same vertex.

Area of $\triangle ABC = \frac{1}{2} |\vec{AB} \times \vec{AC}|$

The vector $\vec{AC} = \vec{C} - \vec{A} = -(\vec{A} - \vec{C}) = -\vec{CA}$. The angle between $\vec{AB}$ and $\vec{AC}$ is angle A.

Area $= \frac{1}{2} |\vec{AB}| |\vec{AC}| \sin A = \frac{1}{2} c b \sin A$

Similarly, using vertex B, the area is $\frac{1}{2} |\vec{BA} \times \vec{BC}|$. The angle between $\vec{BA}$ and $\vec{BC}$ is angle B. Note that $\vec{BA} = -\vec{AB}$.

Area $= \frac{1}{2} |\vec{BA}| |\vec{BC}| \sin B = \frac{1}{2} c a \sin B$

Using vertex C, the area is $\frac{1}{2} |\vec{CB} \times \vec{CA}|$. The angle between $\vec{CB}$ and $\vec{CA}$ is angle C. Note that $\vec{CB} = -\vec{BC}$.

Area $= \frac{1}{2} |\vec{CB}| |\vec{CA}| \sin C = \frac{1}{2} a b \sin C$

Since the area of the triangle is unique, we can equate the expressions from (1), (2), and (3):

$\frac{1}{2} bc \sin A = \frac{1}{2} ca \sin B = \frac{1}{2} ab \sin C$

Multiply the entire equation by 2 to remove the fraction:

$bc \sin A = ca \sin B = ab \sin C$

Now, divide the entire equation by $abc$ (since a, b, c are side lengths of a triangle, they are non-zero):

$\frac{bc \sin A}{abc} = \frac{ca \sin B}{abc} = \frac{ab \sin C}{abc}$

Cancel out the common terms in the numerator and the denominator in each fraction:

$\frac{\cancel{b}\cancel{c} \sin A}{\cancel{a}\cancel{b}\cancel{c}} = \frac{\cancel{c}\cancel{a} \sin B}{\cancel{a}\cancel{b}\cancel{c}} = \frac{\cancel{a}\cancel{b} \sin C}{\cancel{a}\cancel{b}\cancel{c}}$

This simplifies to:

$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$

Thus, the Law of Sines is proved using vectors.

Hence Proved.

Solution:

Let the given vector be $\mathbf{v}$.

$\mathbf{v} = 6\hat{i} + 2\hat{j} + 3\hat{k}$

The magnitude of a vector $\mathbf{v} = a\hat{i} + b\hat{j} + c\hat{k}$ is given by the formula $|\mathbf{v}| = \sqrt{a^2 + b^2 + c^2}$.

In this case, $a = 6$, $b = 2$, and $c = 3$.

So, the magnitude of the vector is:

$|\mathbf{v}| = \sqrt{(6)^2 + (2)^2 + (3)^2}$

$|\mathbf{v}| = \sqrt{36 + 4 + 9}$

$|\mathbf{v}| = \sqrt{49}$

$|\mathbf{v}| = 7$

The magnitude of the vector $6\hat{i} + 2\hat{j} + 3\hat{k}$ is 7.

The correct option is (B).

Solution:

Let the position vectors of the two points be $\vec{p}_1$ and $\vec{p}_2$.

Given, $\vec{p}_1 = \vec{a} + \vec{b}$

Given, $\vec{p}_2 = 2\vec{a} - \vec{b}$

The ratio of division is m : n = 1 : 2.

Using the section formula for internal division, the position vector $\vec{r}$ of the point dividing the line segment joining points with position vectors $\vec{p}_1$ and $\vec{p}_2$ in the ratio m : n is given by:

$\vec{r} = \frac{n\vec{p}_1 + m\vec{p}_2}{m+n}$

Substitute the given values $m=1$, $n=2$, $\vec{p}_1 = \vec{a} + \vec{b}$, and $\vec{p}_2 = 2\vec{a} - \vec{b}$ into the formula:

$\vec{r} = \frac{2(\vec{a} + \vec{b}) + 1(2\vec{a} - \vec{b})}{1+2}$

$\vec{r} = \frac{2\vec{a} + 2\vec{b} + 2\vec{a} - \vec{b}}{3}$

Combine like terms:

$\vec{r} = \frac{(2\vec{a} + 2\vec{a}) + (2\vec{b} - \vec{b})}{3}$

$\vec{r} = \frac{4\vec{a} + \vec{b}}{3}$

The position vector of the point is $\frac{4\vec{a} + \vec{b}}{3}$.

The correct option is (D).

Given:

Initial point P with coordinates $(2, -3, 5)$.

Terminal point Q with coordinates $(3, -4, 7)$.

To Find:

The vector $\vec{PQ}$.

Solution:

Let the position vector of point P be $\vec{p}$ and the position vector of point Q be $\vec{q}$.

$\vec{p} = 2\hat{i} - 3\hat{j} + 5\hat{k}$

$\vec{q} = 3\hat{i} - 4\hat{j} + 7\hat{k}$

The vector from initial point P to terminal point Q is given by the difference between the position vector of the terminal point and the position vector of the initial point:

$\vec{PQ} = \vec{q} - \vec{p}$

Substitute the position vectors of P and Q:

$\vec{PQ} = (3\hat{i} - 4\hat{j} + 7\hat{k}) - (2\hat{i} - 3\hat{j} + 5\hat{k})$

$\vec{PQ} = (3 - 2)\hat{i} + (-4 - (-3))\hat{j} + (7 - 5)\hat{k}$

$\vec{PQ} = (3 - 2)\hat{i} + (-4 + 3)\hat{j} + (7 - 5)\hat{k}$

$\vec{PQ} = 1\hat{i} - 1\hat{j} + 2\hat{k}$

$\vec{PQ} = \hat{i} - \hat{j} + 2\hat{k}$

The vector with initial point P(2, -3, 5) and terminal point Q(3, -4, 7) is $\hat{i} - \hat{j} + 2\hat{k}$.

This matches option (A).

The correct option is (A).

Solution:

Let $\mathbf{a} = \hat{i} - \hat{j}$ and $\mathbf{b} = \hat{j} - \hat{k}$ be the two given vectors.

The angle $\theta$ between two vectors $\mathbf{a}$ and $\mathbf{b}$ is given by the formula:

$\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|}$

First, calculate the dot product $\mathbf{a} \cdot \mathbf{b}$:

$\mathbf{a} \cdot \mathbf{b} = (\hat{i} - \hat{j} + 0\hat{k}) \cdot (0\hat{i} + \hat{j} - \hat{k})$

$\mathbf{a} \cdot \mathbf{b} = (1)(0) + (-1)(1) + (0)(-1)$

$\mathbf{a} \cdot \mathbf{b} = 0 - 1 + 0$

Next, calculate the magnitudes of the vectors $\mathbf{a}$ and $\mathbf{b}$.

$|\mathbf{a}| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{1 + 1 + 0} = \sqrt{2}$

$|\mathbf{b}| = \sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{0 + 1 + 1} = \sqrt{2}$

Now, substitute the values from (1), (2), and (3) into the formula for $\cos \theta$:

$\cos \theta = \frac{-1}{(\sqrt{2})(\sqrt{2})}$

$\cos \theta = \frac{-1}{2}$

To find the angle $\theta$, we take the inverse cosine:

$\theta = \text{cos}^{-1}\left(-\frac{1}{2}\right)$

The principal value for $\text{cos}^{-1}\left(-\frac{1}{2}\right)$ is $\frac{2\pi}{3}$.

$\theta = \frac{2\pi}{3}$

The angle between the vectors $\hat{i}−\hat{j}$ and $\hat{j}−\hat{k}$ is $\frac{2\pi}{3}$.

This matches option (B).

The correct option is (B).

Given:

Two vectors: $\mathbf{a} = 2\hat{i} - \hat{j} + 2\hat{k}$ and $\mathbf{b} = 3\hat{i} + \lambda\hat{j} + \hat{k}$.

The vectors are perpendicular.

To Find:

The value of $\lambda$ for which the vectors are perpendicular.

Solution:

Two vectors $\mathbf{a}$ and $\mathbf{b}$ are perpendicular if and only if their dot product is zero.

So, we must have $\mathbf{a} \cdot \mathbf{b} = 0$.

Calculate the dot product of the given vectors:

$\mathbf{a} \cdot \mathbf{b} = (2\hat{i} - \hat{j} + 2\hat{k}) \cdot (3\hat{i} + \lambda\hat{j} + \hat{k})$

Using the property that $\hat{i} \cdot \hat{i} = 1$, $\hat{j} \cdot \hat{j} = 1$, $\hat{k} \cdot \hat{k} = 1$, and the dot product of orthogonal unit vectors is 0 ($\hat{i} \cdot \hat{j} = 0$, etc.), we get:

$\mathbf{a} \cdot \mathbf{b} = (2)(3) + (-1)(\lambda) + (2)(1)$

$\mathbf{a} \cdot \mathbf{b} = 6 - \lambda + 2$

$\mathbf{a} \cdot \mathbf{b} = 8 - \lambda$

Since the vectors are perpendicular, their dot product must be zero:

Now, solve for $\lambda$ from equation (1):

$8 = \lambda$

$\lambda = 8$

The value of $\lambda$ for which the two vectors are perpendicular is 8.

This matches option (D).

The correct option is (D).

Solution:

Let the adjacent sides of the parallelogram be represented by the vectors $\mathbf{a}$ and $\mathbf{b}$.

Given, $\mathbf{a} = \hat{i} + \hat{k} = \hat{i} + 0\hat{j} + \hat{k}$

Given, $\mathbf{b} = 2\hat{i} + \hat{j} + \hat{k}$

The area of a parallelogram whose adjacent sides are vectors $\mathbf{a}$ and $\mathbf{b}$ is given by the magnitude of their cross product, $|\mathbf{a} \times \mathbf{b}|$.

First, calculate the cross product $\mathbf{a} \times \mathbf{b}$:

$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 1 \\ 2 & 1 & 1 \end{vmatrix}$

$\mathbf{a} \times \mathbf{b} = \hat{i} \begin{vmatrix} 0 & 1 \\ 1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 0 \\ 2 & 1 \end{vmatrix}$

$\mathbf{a} \times \mathbf{b} = \hat{i} ((0)(1) - (1)(1)) - \hat{j} ((1)(1) - (1)(2)) + \hat{k} ((1)(1) - (0)(2))$

$\mathbf{a} \times \mathbf{b} = \hat{i} (0 - 1) - \hat{j} (1 - 2) + \hat{k} (1 - 0)$

$\mathbf{a} \times \mathbf{b} = -1\hat{i} - (-1)\hat{j} + 1\hat{k}$

Now, calculate the magnitude of the cross product vector from equation (1):

$|\mathbf{a} \times \mathbf{b}| = \sqrt{(-1)^2 + (1)^2 + (1)^2}$

$|\mathbf{a} \times \mathbf{b}| = \sqrt{1 + 1 + 1}$

$|\mathbf{a} \times \mathbf{b}| = \sqrt{3}$

The area of the parallelogram is $\sqrt{3}$ square units.

This matches option (B).

The correct option is (B).

Given:

Magnitude of vector $\vec{a}$, $|\vec{a}| = 8$.

Magnitude of vector $\vec{b}$, $|\vec{b}| = 3$.

Magnitude of the cross product of $\vec{a}$ and $\vec{b}$, $|\vec{a} \times \vec{b}| = 12$.

To Find:

The value of the dot product $\vec{a} \cdot \vec{b}$.

Solution:

We know the vector identity that relates the magnitudes of the dot product and the cross product to the magnitudes of the individual vectors:

$|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$

Substitute the given values into this identity:

$(12)^2 + (\vec{a} \cdot \vec{b})^2 = (8)^2 (3)^2$

$144 + (\vec{a} \cdot \vec{b})^2 = 64 \times 9$

$144 + (\vec{a} \cdot \vec{b})^2 = 576$

Now, isolate $(\vec{a} \cdot \vec{b})^2$ by subtracting 144 from both sides of the equation:

$(\vec{a} \cdot \vec{b})^2 = 576 - 144$

To find the value of $\vec{a} \cdot \vec{b}$, take the square root of both sides of equation (1):

$\vec{a} \cdot \vec{b} = \pm \sqrt{432}$

Let's simplify the square root of 432:

$432 = 2 \times 216$

$216 = 6^3 = 216 = 36 \times 6$

So, $432 = 144 \times 3$ (since $144 = 12 \times 12$ and $12^2 = 144$)

$\sqrt{432} = \sqrt{144 \times 3} = \sqrt{144} \times \sqrt{3} = 12\sqrt{3}$

Therefore, the possible values for $\vec{a} \cdot \vec{b}$ are $12\sqrt{3}$ and $-12\sqrt{3}$.

Looking at the given options, $12\sqrt{3}$ is one of the options.

The correct option is (C).

Given:

Vectors representing two adjacent sides of $\triangle ABC$ originating from vertex A.

$\vec{AB} = \hat{j} + \hat{k}$

$\vec{AC} = 3\hat{i} - \hat{j} + 4\hat{k}$

To Find:

The length of the median through A.

Solution:

Let $\vec{AB} = \mathbf{u} = \hat{j} + \hat{k}$ and $\vec{AC} = \mathbf{v} = 3\hat{i} - \hat{j} + 4\hat{k}$.

Let M be the midpoint of the side BC.

The median through A is the line segment AM, represented by the vector $\vec{AM}$.

The vector representing the median from a vertex A to the midpoint of the opposite side BC is given by half the sum of the vectors representing the sides originating from A, i.e., $\vec{AB}$ and $\vec{AC}$.

So, the vector representing the median through A is:

$\vec{AM} = \frac{1}{2}(\vec{AB} + \vec{AC})$

First, calculate the sum of the vectors $\vec{AB}$ and $\vec{AC}$:

$\vec{AB} + \vec{AC} = (\hat{j} + \hat{k}) + (3\hat{i} - \hat{j} + 4\hat{k})$

$\vec{AB} + \vec{AC} = 3\hat{i} + (1 - 1)\hat{j} + (1 + 4)\hat{k}$

$\vec{AB} + \vec{AC} = 3\hat{i} + 0\hat{j} + 5\hat{k}$

Now, find the vector $\vec{AM}$:

$\vec{AM} = \frac{1}{2}(3\hat{i} + 5\hat{k})$

$\vec{AM} = \frac{3}{2}\hat{i} + \frac{5}{2}\hat{k}$

The length of the median through A is the magnitude of the vector $\vec{AM}$.

$|\vec{AM}| = \left|\frac{3}{2}\hat{i} + 0\hat{j} + \frac{5}{2}\hat{k}\right|$

The magnitude of a vector $a\hat{i} + b\hat{j} + c\hat{k}$ is $\sqrt{a^2 + b^2 + c^2}$.

$|\vec{AM}| = \sqrt{\left(\frac{3}{2}\right)^2 + (0)^2 + \left(\frac{5}{2}\right)^2}$

$|\vec{AM}| = \sqrt{\frac{9}{4} + 0 + \frac{25}{4}}$

$|\vec{AM}| = \sqrt{\frac{9 + 25}{4}}$

$|\vec{AM}| = \sqrt{\frac{34}{4}}$

$|\vec{AM}| = \frac{\sqrt{34}}{\sqrt{4}}$

$|\vec{AM}| = \frac{\sqrt{34}}{2}$

The length of the median through A is $\frac{\sqrt{34}}{2}$.

This matches option (A).

The correct option is (A).

Given:

Vector $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$.

Vector $\vec{b} = \hat{i} + 2\hat{j} + 2\hat{k}$.

To Find:

The projection of vector $\vec{a}$ along vector $\vec{b}$.

Solution:

The scalar projection of vector $\vec{a}$ along vector $\vec{b}$ is given by the formula:

Projection $= \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$

First, we calculate the dot product $\vec{a} \cdot \vec{b}$:

$\vec{a} \cdot \vec{b} = (2\hat{i} - \hat{j} + \hat{k}) \cdot (\hat{i} + 2\hat{j} + 2\hat{k})$

$\vec{a} \cdot \vec{b} = (2)(1) + (-1)(2) + (1)(2)$

$\vec{a} \cdot \vec{b} = 2 - 2 + 2$

$\vec{a} \cdot \vec{b} = 2$

Next, we calculate the magnitude of vector $\vec{b}$:

$|\vec{b}| = \sqrt{(1)^2 + (2)^2 + (2)^2}$

$|\vec{b}| = \sqrt{1 + 4 + 4}$

$|\vec{b}| = \sqrt{9}$

$|\vec{b}| = 3$

Now, substitute the values into the projection formula:

Projection $= \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{2}{3}$

The projection of vector $\vec{a}$ along vector $\vec{b}$ is $\frac{2}{3}$.

This matches option (A).

The correct option is (A).

Given:

$\vec{a}$ and $\vec{b}$ are unit vectors, so $|\vec{a}| = 1$ and $|\vec{b}| = 1$.

The vector $\sqrt{3}\vec{a} - \vec{b}$ is a unit vector, so $|\sqrt{3}\vec{a} - \vec{b}| = 1$.

To Find:

The angle between $\vec{a}$ and $\vec{b}$.

Solution:

Let $\theta$ be the angle between vectors $\vec{a}$ and $\vec{b}$.

We are given that $|\sqrt{3}\vec{a} - \vec{b}| = 1$. Squaring both sides, we get:

$|\sqrt{3}\vec{a} - \vec{b}|^2 = 1^2$

$(\sqrt{3}\vec{a} - \vec{b}) \cdot (\sqrt{3}\vec{a} - \vec{b}) = 1$

Using the distributive property of the dot product, expand the left side:

$(\sqrt{3}\vec{a}) \cdot (\sqrt{3}\vec{a}) - (\sqrt{3}\vec{a}) \cdot \vec{b} - \vec{b} \cdot (\sqrt{3}\vec{a}) + \vec{b} \cdot \vec{b} = 1$

$(\sqrt{3})^2 (\vec{a} \cdot \vec{a}) - \sqrt{3}(\vec{a} \cdot \vec{b}) - \sqrt{3}(\vec{a} \cdot \vec{b}) + (\vec{b} \cdot \vec{b}) = 1$

Recall that $\vec{x} \cdot \vec{x} = |\vec{x}|^2$ and $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$.

$3 |\vec{a}|^2 - 2\sqrt{3} (\vec{a} \cdot \vec{b}) + |\vec{b}|^2 = 1$

Substitute the given magnitudes $|\vec{a}| = 1$ and $|\vec{b}| = 1$:

$3 (1)^2 - 2\sqrt{3} (\vec{a} \cdot \vec{b}) + (1)^2 = 1$

$3 - 2\sqrt{3} (\vec{a} \cdot \vec{b}) + 1 = 1$

$4 - 2\sqrt{3} (\vec{a} \cdot \vec{b}) = 1$

Rearrange the equation to solve for $\vec{a} \cdot \vec{b}$:

$2\sqrt{3} (\vec{a} \cdot \vec{b}) = 4 - 1$

$2\sqrt{3} (\vec{a} \cdot \vec{b}) = 3$

Rationalize the denominator of equation (1):

$\vec{a} \cdot \vec{b} = \frac{3}{2\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{3\sqrt{3}}{2 \times 3} = \frac{\sqrt{3}}{2}$

Now, use the definition of the dot product in terms of magnitudes and the angle between the vectors:

Substitute $|\vec{a}| = 1$, $|\vec{b}| = 1$, and $\vec{a} \cdot \vec{b} = \frac{\sqrt{3}}{2}$ into equation (2):

$\frac{\sqrt{3}}{2} = (1)(1)\cos\theta$

$\cos\theta = \frac{\sqrt{3}}{2}$

The angle $\theta$ in the range $[0^\circ, 180^\circ]$ (or $[0, \pi]$ radians) for which $\cos\theta = \frac{\sqrt{3}}{2}$ is $30^\circ$ (or $\frac{\pi}{6}$ radians).

$\theta = 30^\circ$

The angle between $\vec{a}$ and $\vec{b}$ is $30^\circ$.

This matches option (A).

The correct option is (A).

Solution:

Let the two given vectors be $\mathbf{a} = \hat{i} - \hat{j}$ and $\mathbf{b} = \hat{i} + \hat{j}$.

A vector perpendicular to both $\mathbf{a}$ and $\mathbf{b}$ is given by their cross product $\mathbf{a} \times \mathbf{b}$.

We calculate the cross product:

$\mathbf{a} \times \mathbf{b} = (\hat{i} - \hat{j}) \times (\hat{i} + \hat{j})$

Using the determinant form (adding a zero $\hat{k}$ component for clarity):

$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 0 \\ 1 & 1 & 0 \end{vmatrix}$

$\mathbf{a} \times \mathbf{b} = \hat{i}((-1)(0) - (0)(1)) - \hat{j}((1)(0) - (0)(1)) + \hat{k}((1)(1) - (-1)(1))$

$\mathbf{a} \times \mathbf{b} = \hat{i}(0 - 0) - \hat{j}(0 - 0) + \hat{k}(1 - (-1))$

$\mathbf{a} \times \mathbf{b} = 0\hat{i} - 0\hat{j} + \hat{k}(1 + 1)$

The vector $2\hat{k}$ is perpendicular to both $\hat{i} - \hat{j}$ and $\hat{i} + \hat{j}$.

For the vectors $\hat{i} - \hat{j}$, $\hat{i} + \hat{j}$, and the required unit vector to form a right-handed system, the required unit vector must be in the direction of $\mathbf{a} \times \mathbf{b}$.

The direction of $\mathbf{a} \times \mathbf{b}$ is the direction of $2\hat{k}$, which is the direction of $\hat{k}$.

The unit vector in the direction of $2\hat{k}$ is found by dividing the vector by its magnitude.

The magnitude of $2\hat{k}$ is:

The unit vector perpendicular to the given vectors and forming a right-handed system is:

$\hat{u} = \frac{2\hat{k}}{|2\hat{k}|} = \frac{2\hat{k}}{2} = \hat{k}$

The unit vector is $\hat{k}$.

This matches option (A).

The correct option is (A).

Given:

Magnitude of vector $\vec{a}$, $|\vec{a}| = 3$.

The range of scalar k, $-1 \le k \le 2$.

To Find:

The interval in which $|k\vec{a}|$ lies.

Solution:

We use the property of vectors that the magnitude of a scalar multiple of a vector is the absolute value of the scalar times the magnitude of the vector:

$|k\vec{a}| = |k| |\vec{a}|$

Substitute the given value of $|\vec{a}| = 3$:

$|k\vec{a}| = |k| \times 3$

We are given the range for k as $-1 \le k \le 2$. We need to find the range of $|k|$ for this interval of k.

If $-1 \le k \le 2$, the possible values for k include negative values, zero, and positive values.

The absolute value $|k|$ is the distance of k from zero on the number line.

For the interval $[-1, 2]$, the minimum value of $|k|$ occurs at $k=0$, which is $|0| = 0$.

The maximum value of $|k|$ occurs at the endpoint furthest from zero. Comparing $|-1| = 1$ and $|2| = 2$, the maximum value is 2.

Therefore, for $-1 \le k \le 2$, the range of $|k|$ is $0 \le |k| \le 2$, which can be written as $|k| \in [0, 2]$.

Now, substitute this range of $|k|$ into the expression for $|k\vec{a}| = 3|k|$:

Multiply the inequality $0 \le |k| \le 2$ by 3:

$3 \times 0 \le 3|k| \le 3 \times 2$

$0 \le |k\vec{a}| \le 6$

So, $|k\vec{a}|$ lies in the interval $[0, 6]$.

This matches option (A).

The correct option is (A).

Given:

Vector $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$.

Vector $\vec{b} = 2\hat{j} + \hat{k}$.

To Find:

The unit vector in the direction of the sum of $\vec{a}$ and $\vec{b}$.

Solution:

First, find the sum of the vectors $\vec{a}$ and $\vec{b}$. Let the sum be $\vec{c}$.

$\vec{c} = \vec{a} + \vec{b}$

$\vec{c} = (2\hat{i} - \hat{j} + \hat{k}) + (0\hat{i} + 2\hat{j} + \hat{k})$

Combine the components:

$\vec{c} = (2 + 0)\hat{i} + (-1 + 2)\hat{j} + (1 + 1)\hat{k}$

Next, find the magnitude of the sum vector $\vec{c}$. The magnitude of a vector $a\hat{i} + b\hat{j} + c\hat{k}$ is $\sqrt{a^2 + b^2 + c^2}$.

$|\vec{c}| = \sqrt{(2)^2 + (1)^2 + (2)^2}$

$|\vec{c}| = \sqrt{4 + 1 + 4}$

$|\vec{c}| = \sqrt{9}$

The unit vector in the direction of $\vec{c}$ is given by $\hat{c} = \frac{\vec{c}}{|\vec{c}|}$.

Substitute the vector $\vec{c}$ from equation (1) and its magnitude $|\vec{c}|$ from equation (2):

$\hat{c} = \frac{2\hat{i} + \hat{j} + 2\hat{k}}{3}$

The unit vector can be written as:

$\hat{c} = \frac{2}{3}\hat{i} + \frac{1}{3}\hat{j} + \frac{2}{3}\hat{k}$

The unit vector in the direction of the sum of the given vectors is $\frac{2}{3}\hat{i} + \frac{1}{3}\hat{j} + \frac{2}{3}\hat{k}$.

Given:

Vector $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$.

Vector $\vec{b} = 2\hat{i} + \hat{j} - 2\hat{k}$.

To Find:

(i) The unit vector in the direction of $6\vec{b}$.

(ii) The unit vector in the direction of $2\vec{a} - \vec{b}$.

Solution:

(i) Unit vector in the direction of $6\vec{b}$:

First, calculate the vector $6\vec{b}$:

$6\vec{b} = 6(2\hat{i} + \hat{j} - 2\hat{k})$

Next, find the magnitude of the vector $6\vec{b}$:

$|6\vec{b}| = \sqrt{(12)^2 + (6)^2 + (-12)^2}$

$|6\vec{b}| = \sqrt{144 + 36 + 144}$

$|6\vec{b}| = \sqrt{324}$

The unit vector in the direction of $6\vec{b}$ is obtained by dividing the vector $6\vec{b}$ by its magnitude $|6\vec{b}|$.

Unit vector $= \frac{6\vec{b}}{|6\vec{b}|}$

Substitute the results from equation (1) and (2):

Unit vector $= \frac{12\hat{i} + 6\hat{j} - 12\hat{k}}{18}$

Unit vector $= \frac{12}{18}\hat{i} + \frac{6}{18}\hat{j} - \frac{12}{18}\hat{k}$

Simplify the fractions:

Unit vector $= \frac{2}{3}\hat{i} + \frac{1}{3}\hat{j} - \frac{2}{3}\hat{k}$

(ii) Unit vector in the direction of $2\vec{a} - \vec{b}$:

First, calculate the vector $2\vec{a}$:

$2\vec{a} = 2(\hat{i} + \hat{j} + 2\hat{k})$

$2\vec{a} = 2\hat{i} + 2\hat{j} + 4\hat{k}$

Next, calculate the vector $2\vec{a} - \vec{b}$:

$2\vec{a} - \vec{b} = (2\hat{i} + 2\hat{j} + 4\hat{k}) - (2\hat{i} + \hat{j} - 2\hat{k})$

$2\vec{a} - \vec{b} = (2 - 2)\hat{i} + (2 - 1)\hat{j} + (4 - (-2))\hat{k}$

$2\vec{a} - \vec{b} = 0\hat{i} + 1\hat{j} + (4 + 2)\hat{k}$

Now, find the magnitude of the vector $2\vec{a} - \vec{b}$:

$|2\vec{a} - \vec{b}| = \sqrt{(0)^2 + (1)^2 + (6)^2}$

$|2\vec{a} - \vec{b}| = \sqrt{0 + 1 + 36}$

The unit vector in the direction of $2\vec{a} - \vec{b}$ is obtained by dividing the vector $2\vec{a} - \vec{b}$ by its magnitude $|2\vec{a} - \vec{b}|$.

Unit vector $= \frac{2\vec{a} - \vec{b}}{|2\vec{a} - \vec{b}|}$

Substitute the results from equation (3) and (4):

Unit vector $= \frac{\hat{j} + 6\hat{k}}{\sqrt{37}}$

The unit vector can be written as:

Unit vector $= \frac{1}{\sqrt{37}}\hat{j} + \frac{6}{\sqrt{37}}\hat{k}$

Given:

Coordinates of point P are $(5, 0, 8)$.

Coordinates of point Q are $(3, 3, 2)$.

To Find:

A unit vector in the direction of $\overrightarrow{PQ}$.

Solution:

Let the position vector of point P be $\vec{p}$ and the position vector of point Q be $\vec{q}$.

From the given coordinates, the position vectors are:

$\vec{p} = 5\hat{i} + 0\hat{j} + 8\hat{k} = 5\hat{i} + 8\hat{k}$

$\vec{q} = 3\hat{i} + 3\hat{j} + 2\hat{k}$

The vector $\overrightarrow{PQ}$ is the vector from the initial point P to the terminal point Q. It is given by the difference of the position vectors:

$\overrightarrow{PQ} = \vec{q} - \vec{p}$

$\overrightarrow{PQ} = (3\hat{i} + 3\hat{j} + 2\hat{k}) - (5\hat{i} + 0\hat{j} + 8\hat{k})$

$\overrightarrow{PQ} = (3 - 5)\hat{i} + (3 - 0)\hat{j} + (2 - 8)\hat{k}$

Next, find the magnitude of the vector $\overrightarrow{PQ}$. The magnitude of a vector $a\hat{i} + b\hat{j} + c\hat{k}$ is $\sqrt{a^2 + b^2 + c^2}$.

$|\overrightarrow{PQ}| = \sqrt{(-2)^2 + (3)^2 + (-6)^2}$

$|\overrightarrow{PQ}| = \sqrt{4 + 9 + 36}$

$|\overrightarrow{PQ}| = \sqrt{49}$

The unit vector in the direction of $\overrightarrow{PQ}$ is obtained by dividing the vector $\overrightarrow{PQ}$ by its magnitude $|\overrightarrow{PQ}|$.

Unit vector $\widehat{PQ} = \frac{\overrightarrow{PQ}}{|\overrightarrow{PQ}|}$

Substitute the results from equation (1) and (2):

Unit vector $\widehat{PQ} = \frac{-2\hat{i} + 3\hat{j} - 6\hat{k}}{7}$

The unit vector can be written as:

Unit vector $\widehat{PQ} = -\frac{2}{7}\hat{i} + \frac{3}{7}\hat{j} - \frac{6}{7}\hat{k}$

The unit vector in the direction of $\overrightarrow{PQ}$ is $-\frac{2}{7}\hat{i} + \frac{3}{7}\hat{j} - \frac{6}{7}\hat{k}$.

Given:

Position vector of point A is $\vec{a}$.

Position vector of point B is $\vec{b}$.

Point C lies in BA produced such that BC = 1.5 BA.

To Find:

The position vector of point C, denoted as $\vec{c}$.

Solution:

The statement "point C in BA produced" means that C lies on the line passing through B and A, and A is located between B and C. The points are collinear in the order B, A, C.

The vector $\overrightarrow{BA}$ is given by $\vec{a} - \vec{b}$.

The vector $\overrightarrow{BC}$ is given by $\vec{c} - \vec{b}$.

Since C is in BA produced, the vectors $\overrightarrow{BC}$ and $\overrightarrow{BA}$ are in the same direction.

The given condition is BC = 1.5 BA. This refers to the magnitudes of the line segments. Since the vectors are in the same direction, the vector equation is:

Substitute the position vector forms into the equation:

$\vec{c} - \vec{b} = 1.5 (\vec{a} - \vec{b})$

$\vec{c} - \vec{b} = \frac{3}{2} (\vec{a} - \vec{b})$

Now, solve for $\vec{c}$:

$\vec{c} = \vec{b} + \frac{3}{2} (\vec{a} - \vec{b})$

$\vec{c} = \vec{b} + \frac{3}{2}\vec{a} - \frac{3}{2}\vec{b}$

Combine the terms involving $\vec{b}$:

$\vec{c} = \frac{3}{2}\vec{a} + \left(1 - \frac{3}{2}\right)\vec{b}$

$\vec{c} = \frac{3}{2}\vec{a} + \left(\frac{2}{2} - \frac{3}{2}\right)\vec{b}$

$\vec{c} = \frac{3}{2}\vec{a} - \frac{1}{2}\vec{b}$

We can also write this as:

Alternate Solution (Using Section Formula):

Since C is in BA produced such that BC = 1.5 BA, it means that A lies between B and C, and BC = 1.5 BA.

The distance AC = BC - BA = 1.5 BA - BA = 0.5 BA.

This means that point A divides the line segment BC internally in the ratio BA : AC = BA : 0.5 BA = 1 : 0.5 = 2 : 1.

Using the section formula for internal division, the position vector of A ($\vec{a}$) dividing the segment BC with position vectors $\vec{b}$ and $\vec{c}$ in the ratio $m:n = 2:1$ is:

$\vec{a} = \frac{n\vec{b} + m\vec{c}}{n+m}$

Substitute $m=2$, $n=1$, the position vector of B ($\vec{b}$), and the position vector of C ($\vec{c}$):

$\vec{a} = \frac{1\vec{b} + 2\vec{c}}{1+2}$

$\vec{a} = \frac{\vec{b} + 2\vec{c}}{3}$

Now, solve for $\vec{c}$:

$3\vec{a} = \vec{b} + 2\vec{c}$

$2\vec{c} = 3\vec{a} - \vec{b}$

Both methods yield the same result.

The position vector of point C is $\frac{3\vec{a} - \vec{b}}{2}$.

Given:

Three points P(k, -10, 3), Q(1, -1, 3), and R(3, 5, 3).

To Find:

The value of k for which the points P, Q, and R are collinear.

Solution:

For three points P, Q, and R to be collinear, the vector $\overrightarrow{PQ}$ must be parallel to the vector $\overrightarrow{QR}$. This means that $\overrightarrow{PQ} = \lambda \overrightarrow{QR}$ for some scalar $\lambda$.

First, find the vectors $\overrightarrow{PQ}$ and $\overrightarrow{QR}$.

The position vector of point P is $\vec{p} = k\hat{i} - 10\hat{j} + 3\hat{k}$.

The position vector of point Q is $\vec{q} = \hat{i} - \hat{j} + 3\hat{k}$.

The position vector of point R is $\vec{r} = 3\hat{i} + 5\hat{j} + 3\hat{k}$.

Calculate $\overrightarrow{PQ} = \vec{q} - \vec{p}$:

$\overrightarrow{PQ} = (1\hat{i} - \hat{j} + 3\hat{k}) - (k\hat{i} - 10\hat{j} + 3\hat{k})$

$\overrightarrow{PQ} = (1 - k)\hat{i} + (-1 - (-10))\hat{j} + (3 - 3)\hat{k}$

Calculate $\overrightarrow{QR} = \vec{r} - \vec{q}$:

$\overrightarrow{QR} = (3\hat{i} + 5\hat{j} + 3\hat{k}) - (1\hat{i} - \hat{j} + 3\hat{k})$

$\overrightarrow{QR} = (3 - 1)\hat{i} + (5 - (-1))\hat{j} + (3 - 3)\hat{k}$

For collinearity, $\overrightarrow{PQ} = \lambda \overrightarrow{QR}$ for some scalar $\lambda$.

From equations (1) and (2):

$(1 - k)\hat{i} + 9\hat{j} + 0\hat{k} = \lambda (2\hat{i} + 6\hat{j} + 0\hat{k})$

$(1 - k)\hat{i} + 9\hat{j} = 2\lambda \hat{i} + 6\lambda \hat{j}$

Equating the coefficients of $\hat{i}$ and $\hat{j}$ on both sides:

From equation (4), solve for $\lambda$:

$6\lambda = 9$

$\lambda = \frac{9}{6} = \frac{3}{2}$

Now substitute the value of $\lambda$ into equation (3):

$1 - k = 2 \times \left(\frac{3}{2}\right)$

$1 - k = 3$

Solve for k:

$-k = 3 - 1$

$-k = 2$

$k = -2$

The value of k for which the points are collinear is -2.

Given:

A vector $\vec{r}$ is inclined at equal angles to the positive x, y, and z axes.

The magnitude of $\vec{r}$ is $|\vec{r}| = 2\sqrt{3}$.

To Find:

The vector $\vec{r}$.

Solution:

Let the angles that the vector $\vec{r}$ makes with the positive x, y, and z axes be $\alpha$, $\beta$, and $\gamma$, respectively.

We are given that the vector is inclined at equal angles to the three axes, so $\alpha = \beta = \gamma$. Let this equal angle be $\theta$.

The direction cosines of the vector $\vec{r}$ are $\cos\alpha$, $\cos\beta$, and $\cos\gamma$. Since the angles are equal, the direction cosines are also equal: $\cos\alpha = \cos\beta = \cos\gamma = \cos\theta$.

We know that the sum of the squares of the direction cosines of any vector is equal to 1:

$\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$

Substituting $\cos\alpha = \cos\beta = \cos\gamma = \cos\theta$:

$\cos^2\theta + \cos^2\theta + \cos^2\theta = 1$

$3\cos^2\theta = 1$

$\cos^2\theta = \frac{1}{3}$

Taking the square root of both sides:

This means the direction cosines are either all positive ($\frac{1}{\sqrt{3}}$) or all negative ($-\frac{1}{\sqrt{3}}$).

A vector $\vec{r}$ can be expressed in terms of its magnitude and direction cosines as:

$\vec{r} = |\vec{r}| (\cos\alpha \hat{i} + \cos\beta \hat{j} + \cos\gamma \hat{k})$

We are given $|\vec{r}| = 2\sqrt{3}$ and we found $\cos\alpha = \cos\beta = \cos\gamma = \cos\theta = \pm \frac{1}{\sqrt{3}}$.

Case 1: Direction cosines are all positive.

$\cos\alpha = \cos\beta = \cos\gamma = \frac{1}{\sqrt{3}}$

$\vec{r} = 2\sqrt{3} \left(\frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k}\right)$

$\vec{r} = 2\sqrt{3} \times \frac{1}{\sqrt{3}} (\hat{i} + \hat{j} + \hat{k})$

$\vec{r} = 2(\hat{i} + \hat{j} + \hat{k})$

Case 2: Direction cosines are all negative.

$\cos\alpha = \cos\beta = \cos\gamma = -\frac{1}{\sqrt{3}}$

$\vec{r} = 2\sqrt{3} \left(-\frac{1}{\sqrt{3}}\hat{i} - \frac{1}{\sqrt{3}}\hat{j} - \frac{1}{\sqrt{3}}\hat{k}\right)$

$\vec{r} = 2\sqrt{3} \times \left(-\frac{1}{\sqrt{3}}\right) (\hat{i} + \hat{j} + \hat{k})$

$\vec{r} = -2(\hat{i} + \hat{j} + \hat{k})$

Both vectors $2\hat{i} + 2\hat{j} + 2\hat{k}$ and $-2\hat{i} - 2\hat{j} - 2\hat{k}$ are inclined at equal angles to the axes and have a magnitude of $2\sqrt{3}$.

Let's verify the magnitude for $\vec{r} = 2\hat{i} + 2\hat{j} + 2\hat{k}$:

$|\vec{r}| = \sqrt{2^2 + 2^2 + 2^2} = \sqrt{4 + 4 + 4} = \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$.

Let's verify the magnitude for $\vec{r} = -2\hat{i} - 2\hat{j} - 2\hat{k}$:

$|\vec{r}| = \sqrt{(-2)^2 + (-2)^2 + (-2)^2} = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3}$.

Thus, there are two possible vectors that satisfy the given conditions.

The vector $\vec{r}$ can be $2\hat{i} + 2\hat{j} + 2\hat{k}$ or $-2\hat{i} - 2\hat{j} - 2\hat{k}$.

Given:

Magnitude of vector $\vec{r}$, $|\vec{r}| = 14$.

Direction ratios of $\vec{r}$ are 2, 3, -6.

Vector $\vec{r}$ makes an acute angle with the x-axis.

To Find:

The direction cosines of $\vec{r}$.

The components of $\vec{r}$.

Solution:

Let the direction ratios of the vector $\vec{r}$ be $a=2$, $b=3$, and $c=-6$.

To find the direction cosines, we first calculate the magnitude of the vector formed by these direction ratios. Let this magnitude be $d$.

$d = \sqrt{a^2 + b^2 + c^2}$

$d = \sqrt{(2)^2 + (3)^2 + (-6)^2}$

$d = \sqrt{4 + 9 + 36}$

$d = \sqrt{49}$

The direction cosines (l, m, n) of the vector with direction ratios (a, b, c) can be $\left(\frac{a}{d}, \frac{b}{d}, \frac{c}{d}\right)$ or $\left(\frac{-a}{d}, \frac{-b}{d}, \frac{-c}{d}\right)$.

In our case, the possible direction cosines are:

$\left(\frac{2}{7}, \frac{3}{7}, \frac{-6}{7}\right)$ or $\left(\frac{-2}{7}, \frac{-3}{7}, \frac{6}{7}\right)$.

We are given that $\vec{r}$ makes an acute angle with the x-axis. The direction cosine with the x-axis is $\cos\alpha = l$. For an acute angle ($0^\circ \le \alpha < 90^\circ$), $\cos\alpha$ must be positive.

Comparing the two sets of direction cosines, the first set $\left(\frac{2}{7}, \frac{3}{7}, \frac{-6}{7}\right)$ has a positive x-component ($\frac{2}{7}$). The second set $\left(\frac{-2}{7}, \frac{-3}{7}, \frac{6}{7}\right)$ has a negative x-component ($-\frac{2}{7}$).

Therefore, the direction cosines of $\vec{r}$ must be:

Now, find the components of the vector $\vec{r}$. A vector can be written as $\vec{r} = |\vec{r}| (l\hat{i} + m\hat{j} + n\hat{k})$.

Substitute the magnitude $|\vec{r}| = 14$ and the direction cosines from equation (2):

$\vec{r} = 14 \left(\frac{2}{7}\hat{i} + \frac{3}{7}\hat{j} - \frac{6}{7}\hat{k}\right)$

Distribute the magnitude to each component:

$\vec{r} = \left(14 \times \frac{2}{7}\right)\hat{i} + \left(14 \times \frac{3}{7}\right)\hat{j} + \left(14 \times -\frac{6}{7}\right)\hat{k}$

$\vec{r} = (2 \times 2)\hat{i} + (2 \times 3)\hat{j} + (2 \times -6)\hat{k}$

The components of $\vec{r}$ are the coefficients of $\hat{i}$, $\hat{j}$, and $\hat{k}$, which are 4, 6, and -12.

The direction cosines of $\vec{r}$ are $\left(\frac{2}{7}, \frac{3}{7}, -\frac{6}{7}\right)$.

The components of $\vec{r}$ are (4, 6, -12), so the vector is $4\hat{i} + 6\hat{j} - 12\hat{k}$.

Given:

Vector $\vec{a} = 2\hat{i} - \hat{j} + 2\hat{k}$.

Vector $\vec{b} = 4\hat{i} - \hat{j} + 3\hat{k}$.

Required magnitude of the resultant vector is 6.

To Find:

A vector of magnitude 6 that is perpendicular to both $\vec{a}$ and $\vec{b}$.

Solution:

A vector that is perpendicular to both vectors $\vec{a}$ and $\vec{b}$ is given by their cross product, $\vec{a} \times \vec{b}$.

Calculate the cross product $\vec{a} \times \vec{b}$:

$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 2 \\ 4 & -1 & 3 \end{vmatrix}$

$\vec{a} \times \vec{b} = \hat{i}((-1)(3) - (2)(-1)) - \hat{j}((2)(3) - (2)(4)) + \hat{k}((2)(-1) - (-1)(4))$

$\vec{a} \times \vec{b} = \hat{i}(-3 + 2) - \hat{j}(6 - 8) + \hat{k}(-2 + 4)$

$\vec{a} \times \vec{b} = -1\hat{i} - (-2)\hat{j} + 2\hat{k}$

Let $\vec{c} = \vec{a} \times \vec{b} = -\hat{i} + 2\hat{j} + 2\hat{k}$. The required vector is in the direction of $\vec{c}$ or $-\vec{c}$.

Next, find the magnitude of the vector $\vec{c}$.

$|\vec{c}| = \sqrt{(-1)^2 + (2)^2 + (2)^2}$

$|\vec{c}| = \sqrt{1 + 4 + 4}$

$|\vec{c}| = \sqrt{9}$

The unit vector in the direction perpendicular to both $\vec{a}$ and $\vec{b}$ is the unit vector in the direction of $\vec{c}$, which is $\hat{c} = \frac{\vec{c}}{|\vec{c}|}$, or the unit vector in the opposite direction, $-\hat{c}$.

$\hat{c} = \frac{-\hat{i} + 2\hat{j} + 2\hat{k}}{3} = -\frac{1}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}$

The unit vectors perpendicular to both $\vec{a}$ and $\vec{b}$ are $\pm \left(-\frac{1}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}\right)$.

We need a vector of magnitude 6 in this direction. Let the required vector be $\vec{v}$.

$\vec{v} = 6 \times (\pm \hat{c})$

$\vec{v} = \pm 6 \left(-\frac{1}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}\right)$

$\vec{v} = \pm \left(6 \times -\frac{1}{3}\hat{i} + 6 \times \frac{2}{3}\hat{j} + 6 \times \frac{2}{3}\hat{k}\right)$

$\vec{v} = \pm (-2\hat{i} + 4\hat{j} + 4\hat{k})$

Thus, there are two vectors of magnitude 6 that are perpendicular to both $\vec{a}$ and $\vec{b}$: $-2\hat{i} + 4\hat{j} + 4\hat{k}$ and $2\hat{i} - 4\hat{j} - 4\hat{k}$. The question asks for "a vector", so either answer is valid.

One such vector is $-2\hat{i} + 4\hat{j} + 4\hat{k}$.

Given:

Vector $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$.

Vector $\vec{b} = 3\hat{i} + 4\hat{j} - \hat{k}$.

To Find:

The angle between vectors $\vec{a}$ and $\vec{b}$.

Solution:

Let $\theta$ be the angle between the vectors $\vec{a}$ and $\vec{b}$.

The cosine of the angle between two vectors is given by the formula:

$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$

First, calculate the dot product $\vec{a} \cdot \vec{b}$:

$\vec{a} \cdot \vec{b} = (2\hat{i} - \hat{j} + \hat{k}) \cdot (3\hat{i} + 4\hat{j} - \hat{k})$

$\vec{a} \cdot \vec{b} = (2)(3) + (-1)(4) + (1)(-1)$

$\vec{a} \cdot \vec{b} = 6 - 4 - 1$

Next, calculate the magnitudes of the vectors $\vec{a}$ and $\vec{b}$.

$|\vec{a}| = \sqrt{(2)^2 + (-1)^2 + (1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$

$|\vec{b}| = \sqrt{(3)^2 + (4)^2 + (-1)^2} = \sqrt{9 + 16 + 1} = \sqrt{26}$

Now, substitute the values from (1), (2), and (3) into the formula for $\cos \theta$:

$\cos \theta = \frac{1}{(\sqrt{6})(\sqrt{26})}$

$\cos \theta = \frac{1}{\sqrt{6 \times 26}}$

$\cos \theta = \frac{1}{\sqrt{156}}$

We can simplify $\sqrt{156}$: $156 = 4 \times 39$.

$\sqrt{156} = \sqrt{4 \times 39} = \sqrt{4} \times \sqrt{39} = 2\sqrt{39}$

So, $\cos \theta = \frac{1}{2\sqrt{39}}$.

To find the angle $\theta$, we take the inverse cosine:

$\theta = \text{cos}^{-1}\left(\frac{1}{2\sqrt{39}}\right)$

The angle between the vectors is $\text{cos}^{-1}\left(\frac{1}{2\sqrt{39}}\right)$.

Given:

$\vec{a}+ \vec{b}+ \vec{c}=0$

To Show:

$\vec{a}× \vec{b}= \vec{b}× \vec{c}= \vec{c}× \vec{a}$

Proof:

We are given the vector equation:

Take the cross product of equation (1) with $\vec{a}$ from the left:

$\vec{a} \times (\vec{a} + \vec{b} + \vec{c}) = \vec{a} \times \vec{0}$

Using the distributive property of the cross product and the fact that the cross product of a vector with the zero vector is the zero vector ($\vec{a} \times \vec{0} = \vec{0}$):

$\vec{a} \times \vec{a} + \vec{a} \times \vec{b} + \vec{a} \times \vec{c} = \vec{0}$

Since the cross product of a vector with itself is the zero vector ($\vec{a} \times \vec{a} = \vec{0}$):

$\vec{0} + \vec{a} \times \vec{b} + \vec{a} \times \vec{c} = \vec{0}$

$\vec{a} \times \vec{b} + \vec{a} \times \vec{c} = \vec{0}$

Rearranging the terms:

$\vec{a} \times \vec{b} = -(\vec{a} \times \vec{c})$

Using the property that $\vec{x} \times \vec{y} = -(\vec{y} \times \vec{x})$, we have $-(\vec{a} \times \vec{c}) = \vec{c} \times \vec{a}$.

Now, take the cross product of equation (1) with $\vec{b}$ from the left:

$\vec{b} \times (\vec{a} + \vec{b} + \vec{c}) = \vec{b} \times \vec{0}$

Distribute and use the properties $\vec{b} \times \vec{b} = \vec{0}$ and $\vec{b} \times \vec{0} = \vec{0}$:

$\vec{b} \times \vec{a} + \vec{b} \times \vec{b} + \vec{b} \times \vec{c} = \vec{0}$

$\vec{b} \times \vec{a} + \vec{0} + \vec{b} \times \vec{c} = \vec{0}$

$\vec{b} \times \vec{a} + \vec{b} \times \vec{c} = \vec{0}$

Rearranging the terms:

$\vec{b} \times \vec{c} = -(\vec{b} \times \vec{a})$

Using the property that $\vec{x} \times \vec{y} = -(\vec{y} \times \vec{x})$, we have $-(\vec{b} \times \vec{a}) = \vec{a} \times \vec{b}$.

From equations (2) and (3), we have shown that $\vec{a} \times \vec{b} = \vec{c} \times \vec{a}$ and $\vec{a} \times \vec{b} = \vec{b} \times \vec{c}$.

Therefore, by transitivity, all three cross products are equal:

$\vec{a}× \vec{b}= \vec{b}× \vec{c}= \vec{c}× \vec{a}$

This completes the proof.

Geometric Interpretation:

The condition $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ means that the three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$, when placed head-to-tail, form a closed polygon. If no two vectors are collinear, they form a triangle.

Let's consider the case where $\vec{a}$, $\vec{b}$, and $\vec{c}$ form the sides of a triangle, taken in order. For instance, let $\vec{a} = \overrightarrow{AB}$, $\vec{b} = \overrightarrow{BC}$, and $\vec{c} = \overrightarrow{CA}$. Then $\vec{a} + \vec{b} + \vec{c} = \overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \overrightarrow{AA} = \vec{0}$.

The vector $\vec{a} \times \vec{b} = \overrightarrow{AB} \times \overrightarrow{BC}$. The magnitude of this vector is $|\vec{a} \times \vec{b}| = |\overrightarrow{AB}| |\overrightarrow{BC}| \sin(\text{angle between } \overrightarrow{AB} \text{ and } \overrightarrow{BC})$. The angle between $\overrightarrow{AB}$ and $\overrightarrow{BC}$ is $180^\circ - B$, where B is the internal angle of the triangle at vertex B. So, $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin(180^\circ - B) = |\vec{a}| |\vec{b}| \sin B$. The direction of $\vec{a} \times \vec{b}$ is perpendicular to the plane containing the triangle.

Similarly:

$\vec{b} \times \vec{c} = \overrightarrow{BC} \times \overrightarrow{CA}$. The angle between $\overrightarrow{BC}$ and $\overrightarrow{CA}$ is $180^\circ - C$, where C is the internal angle at vertex C. So, $|\vec{b} \times \vec{c}| = |\vec{b}| |\vec{c}| \sin(180^\circ - C) = |\vec{b}| |\vec{c}| \sin C$. The direction of $\vec{b} \times \vec{c}$ is perpendicular to the plane of the triangle.

$\vec{c} \times \vec{a} = \overrightarrow{CA} \times \overrightarrow{AB}$. The angle between $\overrightarrow{CA}$ and $\overrightarrow{AB}$ is $180^\circ - A$, where A is the internal angle at vertex A. So, $|\vec{c} \times \vec{a}| = |\vec{c}| |\vec{a}| \sin(180^\circ - A) = |\vec{c}| |\vec{a}| \sin A$. The direction of $\vec{c} \times \vec{a}$ is perpendicular to the plane of the triangle.

The result $\vec{a} \times \vec{b}= \vec{b}× \vec{c}= \vec{c}× \vec{a}$ implies that these three cross product vectors are identical. This means they have the same magnitude and the same direction.

The direction of each cross product is perpendicular to the plane of the triangle (assuming the vectors are not collinear). Since they form a closed triangle, they are coplanar, and their cross products are in the same direction (or opposite, depending on the order, but our result shows they are equal vectors, meaning the direction is the same for all three cross products taken cyclically).

The equality of magnitudes, $|\vec{a} \times \vec{b}| = |\vec{b} \times \vec{c}| = |\vec{c} \times \vec{a}|$, translates to:

$|\vec{a}| |\vec{b}| \sin B = |\vec{b}| |\vec{c}| \sin C = |\vec{c}| |\vec{a}| \sin A$

Let a, b, c be the lengths of the sides opposite to vertices A, B, C respectively. In our setup, $|\vec{a}| = c$ (side AB), $|\vec{b}| = a$ (side BC), $|\vec{c}| = b$ (side CA). The internal angles are A, B, C.

So the magnitude equality becomes:

$c \cdot a \cdot \sin B = a \cdot b \cdot \sin C = b \cdot c \cdot \sin A$

Dividing by $abc$ (assuming it's a non-degenerate triangle, so $a, b, c \neq 0$):

$\frac{ca \sin B}{abc} = \frac{ab \sin C}{abc} = \frac{bc \sin A}{abc}$

$\frac{\sin B}{b} = \frac{\sin C}{c} = \frac{\sin A}{a}$

This is the Law of Sines. Thus, the geometric interpretation is that if three vectors sum to the zero vector, they form a closed triangle (unless collinear), and the equality of their pairwise cyclic cross products is equivalent to the Law of Sines for the triangle formed by these vectors. Each equal cross product vector represents twice the area of the triangle, with a direction perpendicular to the plane of the triangle.

Given:

Vector $\vec{a} = 3\hat{i} + \hat{j} + 2\hat{k}$.

Vector $\vec{b} = 2\hat{i} - 2\hat{j} + 4\hat{k}$.

To Find:

The sine of the angle between vectors $\vec{a}$ and $\vec{b}$.

Solution:

Let $\theta$ be the angle between the vectors $\vec{a}$ and $\vec{b}$.

The sine of the angle between two vectors is given by the formula:

$\sin \theta = \frac{|\vec{a} \times \vec{b}|}{|\vec{a}| |\vec{b}|}$

First, calculate the cross product $\vec{a} \times \vec{b}$:

$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 2 \\ 2 & -2 & 4 \end{vmatrix}$

$\vec{a} \times \vec{b} = \hat{i}((1)(4) - (2)(-2)) - \hat{j}((3)(4) - (2)(2)) + \hat{k}((3)(-2) - (1)(2))$

$\vec{a} \times \vec{b} = \hat{i}(4 - (-4)) - \hat{j}(12 - 4) + \hat{k}(-6 - 2)$

$\vec{a} \times \vec{b} = \hat{i}(4 + 4) - \hat{j}(8) + \hat{k}(-8)$

Next, calculate the magnitude of the cross product vector from equation (1):

$|\vec{a} \times \vec{b}| = \sqrt{(8)^2 + (-8)^2 + (-8)^2}$

$|\vec{a} \times \vec{b}| = \sqrt{64 + 64 + 64}$

$|\vec{a} \times \vec{b}| = \sqrt{3 \times 64}$

Now, calculate the magnitudes of the individual vectors $\vec{a}$ and $\vec{b}$.

$|\vec{a}| = \sqrt{(3)^2 + (1)^2 + (2)^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$

$|\vec{b}| = \sqrt{(2)^2 + (-2)^2 + (4)^2} = \sqrt{4 + 4 + 16} = \sqrt{24}$

We can simplify $\sqrt{24}$: $\sqrt{24} = \sqrt{4 \times 6} = \sqrt{4} \times \sqrt{6} = 2\sqrt{6}$.

Finally, substitute the values from (2), (3), and (4) into the formula for $\sin \theta$:

$\sin \theta = \frac{|\vec{a} \times \vec{b}|}{|\vec{a}| |\vec{b}|} = \frac{8\sqrt{3}}{(\sqrt{14})(2\sqrt{6})}$

$\sin \theta = \frac{8\sqrt{3}}{2\sqrt{14 \times 6}}$

$\sin \theta = \frac{8\sqrt{3}}{2\sqrt{84}}$

Simplify the fraction and the square root in the denominator: $\sqrt{84} = \sqrt{4 \times 21} = 2\sqrt{21}$.

$\sin \theta = \frac{8\sqrt{3}}{2 \times 2\sqrt{21}} = \frac{8\sqrt{3}}{4\sqrt{21}}$

$\sin \theta = \frac{\cancel{8}^{2}\sqrt{3}}{\cancel{4}_{1}\sqrt{21}} = \frac{2\sqrt{3}}{\sqrt{21}}$

We can simplify $\sqrt{21}$: $\sqrt{21} = \sqrt{3 \times 7} = \sqrt{3}\sqrt{7}$.

$\sin \theta = \frac{2\sqrt{3}}{\sqrt{3}\sqrt{7}} = \frac{2\cancel{\sqrt{3}}}{\cancel{\sqrt{3}}\sqrt{7}}$

$\sin \theta = \frac{2}{\sqrt{7}}$

Rationalize the denominator:

$\sin \theta = \frac{2}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{2\sqrt{7}}{7}$

The sine of the angle between the vectors is $\frac{2\sqrt{7}}{7}$.

Given:

Position vector of point A: $\vec{a} = \hat{i} + \hat{j} - \hat{k}$.

Position vector of point B: $\vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}$.

Position vector of point C: $\vec{c} = 2\hat{i} - 3\hat{k}$. (Note: the $\hat{j}$ component is 0)

Position vector of point D: $\vec{d} = 3\hat{i} - 2\hat{j} + \hat{k}$.

To Find:

The projection of $\overrightarrow{AB}$ along $\overrightarrow{CD}$.

Solution:

First, we find the vector $\overrightarrow{AB}$. It is the vector from the initial point A to the terminal point B.

$\overrightarrow{AB} = \vec{b} - \vec{a}$

$\overrightarrow{AB} = (2\hat{i} - \hat{j} + 3\hat{k}) - (\hat{i} + \hat{j} - \hat{k})$

$\overrightarrow{AB} = (2 - 1)\hat{i} + (-1 - 1)\hat{j} + (3 - (-1))\hat{k}$

$\overrightarrow{AB} = 1\hat{i} - 2\hat{j} + (3 + 1)\hat{k}$

Next, we find the vector $\overrightarrow{CD}$. It is the vector from the initial point C to the terminal point D.

$\overrightarrow{CD} = \vec{d} - \vec{c}$

$\overrightarrow{CD} = (3\hat{i} - 2\hat{j} + \hat{k}) - (2\hat{i} + 0\hat{j} - 3\hat{k})$

$\overrightarrow{CD} = (3 - 2)\hat{i} + (-2 - 0)\hat{j} + (1 - (-3))\hat{k}$

$\overrightarrow{CD} = 1\hat{i} - 2\hat{j} + (1 + 3)\hat{k}$

The projection of vector $\vec{u}$ along vector $\vec{v}$ is given by the formula: $\frac{\vec{u} \cdot \vec{v}}{|\vec{v}|}$.

In this case, we need the projection of $\overrightarrow{AB}$ along $\overrightarrow{CD}$. So, $\vec{u} = \overrightarrow{AB}$ and $\vec{v} = \overrightarrow{CD}$.

First, calculate the dot product $\overrightarrow{AB} \cdot \overrightarrow{CD}$ using equations (1) and (2):

$\overrightarrow{AB} \cdot \overrightarrow{CD} = (\hat{i} - 2\hat{j} + 4\hat{k}) \cdot (\hat{i} - 2\hat{j} + 4\hat{k})$

$\overrightarrow{AB} \cdot \overrightarrow{CD} = (1)(1) + (-2)(-2) + (4)(4)$

$\overrightarrow{AB} \cdot \overrightarrow{CD} = 1 + 4 + 16$

Next, calculate the magnitude of the vector $\overrightarrow{CD}$ using equation (2):

$|\overrightarrow{CD}| = \sqrt{(1)^2 + (-2)^2 + (4)^2}$

$|\overrightarrow{CD}| = \sqrt{1 + 4 + 16}$

Now, substitute the values from (3) and (4) into the projection formula:

Projection of $\overrightarrow{AB}$ along $\overrightarrow{CD} = \frac{\overrightarrow{AB} \cdot \overrightarrow{CD}}{|\overrightarrow{CD}|}$

Projection $= \frac{21}{\sqrt{21}}$

We can simplify this expression:

Projection $= \frac{21}{\sqrt{21}} = \frac{\sqrt{21} \times \sqrt{21}}{\sqrt{21}} = \sqrt{21}$

The projection of $\overrightarrow{AB}$ along $\overrightarrow{CD}$ is $\sqrt{21}$.

Given:

Vertices of the triangle ABC are A(1, 2, 3), B(2, – 1, 4), and C(4, 5, – 1).

The position vectors of the vertices are:

$\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$

$\vec{b} = 2\hat{i} - \hat{j} + 4\hat{k}$

$\vec{c} = 4\hat{i} + 5\hat{j} - \hat{k}$

To Find:

The area of triangle ABC using vectors.

Solution:

The area of a triangle with vertices A, B, and C can be found using the magnitude of the cross product of two vectors representing two sides of the triangle originating from the same vertex. For example, the area is given by $\frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}|$.

First, find the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$.

$\overrightarrow{AB} = \vec{b} - \vec{a}$

$\overrightarrow{AB} = (2\hat{i} - \hat{j} + 4\hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k})$

$\overrightarrow{AB} = (2 - 1)\hat{i} + (-1 - 2)\hat{j} + (4 - 3)\hat{k}$

$\overrightarrow{AC} = \vec{c} - \vec{a}$

$\overrightarrow{AC} = (4\hat{i} + 5\hat{j} - \hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k})$

$\overrightarrow{AC} = (4 - 1)\hat{i} + (5 - 2)\hat{j} + (-1 - 3)\hat{k}$

Next, calculate the cross product $\overrightarrow{AB} \times \overrightarrow{AC}$ using equations (1) and (2):

$\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 1 \\ 3 & 3 & -4 \end{vmatrix}$

$\overrightarrow{AB} \times \overrightarrow{AC} = \hat{i}((-3)(-4) - (1)(3)) - \hat{j}((1)(-4) - (1)(3)) + \hat{k}((1)(3) - (-3)(3))$

$\overrightarrow{AB} \times \overrightarrow{AC} = \hat{i}(12 - 3) - \hat{j}(-4 - 3) + \hat{k}(3 - (-9))$

$\overrightarrow{AB} \times \overrightarrow{AC} = \hat{i}(9) - \hat{j}(-7) + \hat{k}(3 + 9)$

Now, calculate the magnitude of the cross product vector from equation (3):

$|\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{(9)^2 + (7)^2 + (12)^2}$

$|\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{81 + 49 + 144}$

The area of triangle ABC is half the magnitude of the cross product:

Substitute the magnitude from equation (4) into equation (5):

Area of $\triangle ABC = \frac{1}{2} \sqrt{274}$

The area of the triangle ABC is $\frac{\sqrt{274}}{2}$ square units.

Given:

Two parallelograms ABCD and ABEF on the same base AB and between the same parallel lines AB and CE (where C and E lie on the line parallel to AB).

To Prove:

Area of parallelogram ABCD = Area of parallelogram ABEF.

Proof:

Let the base vector $\overrightarrow{AB} = \vec{b}$.

Let the adjacent side vectors from vertex A be $\overrightarrow{AC} = \vec{a}_1$ for parallelogram ABCD, and $\overrightarrow{AE} = \vec{a}_2$ for parallelogram ABEF.

The area of a parallelogram formed by adjacent vectors $\vec{u}$ and $\vec{v}$ is given by the magnitude of their cross product, $|\vec{u} \times \vec{v}|$.

Area of parallelogram ABCD $= |\overrightarrow{AB} \times \overrightarrow{AC}| = |\vec{b} \times \vec{a}_1|$

Area of parallelogram ABEF $= |\overrightarrow{AB} \times \overrightarrow{AE}| = |\vec{b} \times \vec{a}_2|$

The parallelograms are between the same parallel lines AB and CE. This implies that the line passing through C and E is parallel to the line passing through A and B.

Therefore, the vector connecting C to E, $\overrightarrow{CE}$, must be parallel to the vector $\overrightarrow{AB} = \vec{b}$.

The vector $\overrightarrow{CE}$ can be expressed as the difference between the position vector of E relative to A and the position vector of C relative to A:

$\overrightarrow{CE} = \overrightarrow{AE} - \overrightarrow{AC} = \vec{a}_2 - \vec{a}_1$

Since $\overrightarrow{CE}$ is parallel to $\vec{b}$, we can write:

where k is some scalar.

From equation (3), we can express $\vec{a}_2$ as:

$\vec{a}_2 = \vec{a}_1 + k\vec{b}$

Now, substitute this expression for $\vec{a}_2$ into the formula for the area of parallelogram ABEF (equation 2):

Area (ABEF) $= |\vec{b} \times \vec{a}_2| = |\vec{b} \times (\vec{a}_1 + k\vec{b})|$

Using the distributive property of the cross product:

Area (ABEF) $= |\vec{b} \times \vec{a}_1 + \vec{b} \times (k\vec{b})|$

Using the property that a scalar multiple can be factored out of a cross product:

Area (ABEF) $= |\vec{b} \times \vec{a}_1 + k (\vec{b} \times \vec{b})|$

The cross product of a vector with itself is the zero vector, i.e., $\vec{b} \times \vec{b} = \vec{0}$.

Area (ABEF) $= |\vec{b} \times \vec{a}_1 + k \vec{0}|$

Area (ABEF) $= |\vec{b} \times \vec{a}_1 + \vec{0}|$

Area (ABEF) $= |\vec{b} \times \vec{a}_1|$

Comparing equation (1) and equation (4), we see that:

Area (ABCD) $= |\vec{b} \times \vec{a}_1|$

Area (ABEF) $= |\vec{b} \times \vec{a}_1|$

Therefore, the areas of the two parallelograms are equal.

Area of parallelogram ABCD = Area of parallelogram ABEF.

Hence Proved.

Given:

Triangle ABC with side lengths opposite to vertices A, B, C being a, b, and c respectively.

To Prove:

$\cos A = \frac{b^2 + c^2 − a^2}{2bc}$

Proof:

Consider the triangle ABC. Let the sides $\overrightarrow{AB}$ and $\overrightarrow{AC}$ be represented by vectors $\vec{c}$ and $\vec{b}$ respectively. The magnitude of $\overrightarrow{AB}$ is $|\vec{c}| = c$ and the magnitude of $\overrightarrow{AC}$ is $|\vec{b}| = b$.

The third side, $\overrightarrow{BC}$, is given by the vector difference of $\overrightarrow{AC}$ and $\overrightarrow{AB}$.

The magnitude of the side $\overrightarrow{BC}$ is $|\overrightarrow{BC}| = a$. Therefore, we have:

Square both sides of equation (2):

$a^2 = |\vec{b} - \vec{c}|^2$

Using the property $|\vec{v}|^2 = \vec{v} \cdot \vec{v}$, we can write:

Expand the dot product on the right side of equation (3) using the distributive property:

$a^2 = \vec{b} \cdot \vec{b} - \vec{b} \cdot \vec{c} - \vec{c} \cdot \vec{b} + \vec{c} \cdot \vec{c}$

Recall that $\vec{u} \cdot \vec{u} = |\vec{u}|^2$ and $\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}$. So, $\vec{b} \cdot \vec{b} = |\vec{b}|^2 = b^2$, $\vec{c} \cdot \vec{c} = |\vec{c}|^2 = c^2$, and $\vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{b}$.

$a^2 = |\vec{b}|^2 - 2(\vec{b} \cdot \vec{c}) + |\vec{c}|^2$

The dot product of two vectors is also given by $\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos\theta$, where $\theta$ is the angle between $\vec{u}$ and $\vec{v}$. The angle between vectors $\vec{b} = \overrightarrow{AC}$ and $\vec{c} = \overrightarrow{AB}$ is the angle A of the triangle.

So, $\vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos A = bc \cos A$.

Substitute this into equation (4):

$a^2 = b^2 - 2(bc \cos A) + c^2$

$a^2 = b^2 + c^2 - 2bc \cos A$

Now, rearrange the equation to solve for $\cos A$:

$2bc \cos A = b^2 + c^2 - a^2$

Divide by $2bc$ (since b and c are side lengths, they are non-zero):

This proves the Law of Cosines for angle A using vectors.

The laws for $\cos B$ and $\cos C$ can be derived similarly by considering other pairs of vectors representing the sides.

Hence Proved.

Given:

Three points A, B, C with position vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ respectively, determining the vertices of a triangle.

To Show:

The vector area of triangle ABC is given by $\frac{1}{2} \left[ \; \vec{b}× \vec{c}+ \vec{c}× \vec{a}+ \vec{a}× \vec{b} \; \right]$.

To Deduce:

The condition for the three points $\vec{a},\; \vec{b},\; \vec{c}$ to be collinear.

To Find:

The unit vector normal to the plane of the triangle.

Proof/Solution:

Let the vertices of the triangle be A, B, and C with position vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ respectively.

We can represent two sides of the triangle originating from vertex A by the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$.

The vector $\overrightarrow{AB}$ is the vector from point A to point B:

The vector $\overrightarrow{AC}$ is the vector from point A to point C:

The vector area of a triangle with adjacent sides $\vec{u}$ and $\vec{v}$ originating from a common vertex is given by $\frac{1}{2}(\vec{u} \times \vec{v})$. Using $\overrightarrow{AB}$ and $\overrightarrow{AC}$ as the adjacent sides, the vector area of triangle ABC is:

Substitute the expressions for $\overrightarrow{AB}$ and $\overrightarrow{AC}$ from equations (1) and (2) into equation (3):

Vector Area $= \frac{1}{2} [(\vec{b} - \vec{a}) \times (\vec{c} - \vec{a})]$

Expand the cross product using the distributive property:

Vector Area $= \frac{1}{2} [\vec{b} \times \vec{c} - \vec{b} \times \vec{a} - \vec{a} \times \vec{c} + \vec{a} \times \vec{a}]$

Using the properties of the cross product: $\vec{u} \times \vec{v} = -(\vec{v} \times \vec{u})$ and $\vec{u} \times \vec{u} = \vec{0}$.

So, $-\vec{b} \times \vec{a} = \vec{a} \times \vec{b}$, $-\vec{a} \times \vec{c} = \vec{c} \times \vec{a}$, and $\vec{a} \times \vec{a} = \vec{0}$.

Substitute these into the expression for the vector area:

Vector Area $= \frac{1}{2} [\vec{b} \times \vec{c} + \vec{a} \times \vec{b} + \vec{c} \times \vec{a} + \vec{0}]$

This shows that the vector area of the triangle ABC is given by $\frac{1}{2} \left[ \; \vec{b}× \vec{c}+ \vec{c}× \vec{a}+ \vec{a}× \vec{b} \; \right]$.

Deduction for Collinearity:

If the three points A, B, and C are collinear, they do not form a triangle. In this case, the area of the triangle is zero.

The magnitude of the vector area is equal to the scalar area. If the scalar area is zero, the vector area must be the zero vector.

So, for collinear points, the vector area is $\vec{0}$:

Multiplying by 2, the condition for collinearity of points with position vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ is:

$\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} = \vec{0}$

This condition implies that the vector $\overrightarrow{AB}$ is parallel to $\overrightarrow{AC}$, which means $\overrightarrow{AC} = \lambda \overrightarrow{AB}$ for some scalar $\lambda$. When this condition holds, the points A, B, and C lie on the same line.

Unit Vector Normal to the Plane:

The vector area of the triangle is a vector perpendicular to the plane containing the triangle.

Let $\vec{V} = \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}$. This vector is normal to the plane of the triangle.

The unit vector normal to the plane is obtained by dividing the vector $\vec{V}$ by its magnitude $|\vec{V}|$. There are two possible unit normal vectors (in opposite directions).

The unit normal vector $\hat{n}$ is given by:

The magnitude of the vector $\vec{V}$ is $|\vec{V}| = |\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}|$. From equation (4), we know that $|\vec{V}| = 2 \times (\text{Area of } \triangle ABC)$.

So the unit normal vector can also be written as:

$\hat{n} = \pm \frac{\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}}{2 \times (\text{Area of } \triangle ABC)}$

Let the diagonals of the parallelogram be $\vec{a}$ and $\vec{b}$.

Let the adjacent sides of the parallelogram be $\vec{p}$ and $\vec{q}$.

Part 1: Showing the area formula

Given:

Diagonals of a parallelogram are $\vec{a}$ and $\vec{b}$.

To Prove:

Area of the parallelogram is $\frac{1}{2} \left| \; \vec{a} \times \vec{b} \; \right|$.

Proof:

Let the adjacent sides of the parallelogram be $\vec{p}$ and $\vec{q}$.

The diagonals $\vec{a}$ and $\vec{b}$ can be expressed in terms of the adjacent sides:

$\vec{a} = \vec{p} + \vec{q}$

$\vec{b} = \vec{q} - \vec{p}$

Adding the two equations, we get:

$\vec{a} + \vec{b} = (\vec{p} + \vec{q}) + (\vec{q} - \vec{p})$

$\vec{a} + \vec{b} = 2\vec{q}$

$\vec{q} = \frac{1}{2}(\vec{a} + \vec{b})$

Subtracting the second equation from the first, we get:

$\vec{a} - \vec{b} = (\vec{p} + \vec{q}) - (\vec{q} - \vec{p})$

$\vec{a} - \vec{b} = 2\vec{p}$

$\vec{p} = \frac{1}{2}(\vec{a} - \vec{b})$

The area of the parallelogram with adjacent sides $\vec{p}$ and $\vec{q}$ is given by $\left| \; \vec{p} \times \vec{q} \; \right|$.

Let's calculate the cross product $\vec{p} \times \vec{q}$:

$\vec{p} \times \vec{q} = \left( \frac{1}{2}(\vec{a} - \vec{b}) \right) \times \left( \frac{1}{2}(\vec{a} + \vec{b}) \right)$

$\vec{p} \times \vec{q} = \frac{1}{4} [(\vec{a} - \vec{b}) \times (\vec{a} + \vec{b})]$

Using the distributive property of the cross product:

$(\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = \vec{a} \times \vec{a} + \vec{a} \times \vec{b} - \vec{b} \times \vec{a} - \vec{b} \times \vec{b}$

We know that the cross product of a vector with itself is the zero vector ($\vec{v} \times \vec{v} = \vec{0}$), and $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$.

So,

$\vec{a} \times \vec{a} = \vec{0}$

$\vec{b} \times \vec{b} = \vec{0}$

$-\vec{b} \times \vec{a} = -(-(\vec{a} \times \vec{b})) = \vec{a} \times \vec{b}$

Substituting these into the expansion:

$(\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = \vec{0} + \vec{a} \times \vec{b} + \vec{a} \times \vec{b} - \vec{0}$

$(\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = 2 (\vec{a} \times \vec{b})$

Therefore,

$\vec{p} \times \vec{q} = \frac{1}{4} [2 (\vec{a} \times \vec{b})]$

$\vec{p} \times \vec{q} = \frac{1}{2} (\vec{a} \times \vec{b})$

The area of the parallelogram is the magnitude of $\vec{p} \times \vec{q}$:

Area $= \left| \; \vec{p} \times \vec{q} \; \right| = \left| \; \frac{1}{2} (\vec{a} \times \vec{b}) \; \right|$

Since $\frac{1}{2}$ is a scalar constant, we can write:

Area $= \frac{1}{2} \left| \; \vec{a} \times \vec{b} \; \right|$

This proves the formula for the area of a parallelogram in terms of its diagonals.

Part 2: Finding the area for given diagonals

Given:

Diagonals are $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 3\hat{j} - \hat{k}$.

To Find:

Area of the parallelogram.

Solution:

We use the formula for the area of a parallelogram given its diagonals $\vec{a}$ and $\vec{b}$, which is Area $= \frac{1}{2} \left| \; \vec{a} \times \vec{b} \; \right|$.

First, we calculate the cross product $\vec{a} \times \vec{b}$.

$\vec{a} \times \vec{b} = (2\hat{i} - \hat{j} + \hat{k}) \times (\hat{i} + 3\hat{j} - \hat{k})$

We can compute this as a determinant:

$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 1 & 3 & -1 \end{vmatrix}$

$= \hat{i}((-1)(-1) - (1)(3)) - \hat{j}((2)(-1) - (1)(1)) + \hat{k}((2)(3) - (-1)(1))$

$= \hat{i}(1 - 3) - \hat{j}(-2 - 1) + \hat{k}(6 + 1)$

$= \hat{i}(-2) - \hat{j}(-3) + \hat{k}(7)$

$\vec{a} \times \vec{b} = -2\hat{i} + 3\hat{j} + 7\hat{k}$

Next, we find the magnitude of $\vec{a} \times \vec{b}$.

$\left| \; \vec{a} \times \vec{b} \; \right| = \left| \; -2\hat{i} + 3\hat{j} + 7\hat{k} \; \right|$

$\left| \; \vec{a} \times \vec{b} \; \right| = \sqrt{(-2)^2 + (3)^2 + (7)^2}$

$\left| \; \vec{a} \times \vec{b} \; \right| = \sqrt{4 + 9 + 49}$

$\left| \; \vec{a} \times \vec{b} \; \right| = \sqrt{62}$

Finally, the area of the parallelogram is half of this magnitude:

Area $= \frac{1}{2} \left| \; \vec{a} \times \vec{b} \; \right| = \frac{1}{2} \sqrt{62}$ square units.

The area of the parallelogram with the given diagonals is $\frac{\sqrt{62}}{2}$ square units.

Given:

Vectors $\vec{a}= \hat{i}+ \hat{j}+ \hat{k}$ and $\vec{b}= \hat{j}− \hat{k}$.

Conditions $\vec{a} \times \vec{c} = \vec{b}$ and $\vec{a} \cdot \vec{c} = 3$.

To Find:

A vector $\vec{c}$.

Solution:

Let the vector $\vec{c}$ be represented as $\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$, where $c_1, c_2, c_3$ are scalar components.

Using the condition $\vec{a} \cdot \vec{c} = 3$:

$(\hat{i} + \hat{j} + \hat{k}) \cdot (c_1\hat{i} + c_2\hat{j} + c_3\hat{k}) = 3$

$ (1)(c_1) + (1)(c_2) + (1)(c_3) = 3 $

Using the condition $\vec{a} \times \vec{c} = \vec{b}$:

First, calculate the cross product $\vec{a} \times \vec{c}$:

$\vec{a} \times \vec{c} = (\hat{i} + \hat{j} + \hat{k}) \times (c_1\hat{i} + c_2\hat{j} + c_3\hat{k})$

$\vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ c_1 & c_2 & c_3 \end{vmatrix}$

$= \hat{i}((1)(c_3) - (1)(c_2)) - \hat{j}((1)(c_3) - (1)(c_1)) + \hat{k}((1)(c_2) - (1)(c_1))$

$= (c_3 - c_2)\hat{i} - (c_3 - c_1)\hat{j} + (c_2 - c_1)\hat{k}$

$= (c_3 - c_2)\hat{i} + (c_1 - c_3)\hat{j} + (c_2 - c_1)\hat{k}$

Now, equate this to the given vector $\vec{b} = \hat{j} - \hat{k}$ (which can be written as $0\hat{i} + 1\hat{j} - 1\hat{k}$):

$(c_3 - c_2)\hat{i} + (c_1 - c_3)\hat{j} + (c_2 - c_1)\hat{k} = 0\hat{i} + 1\hat{j} - 1\hat{k}$

Comparing the coefficients of $\hat{i}, \hat{j}, \hat{k}$ on both sides, we get a system of linear equations:

From Equation (2), we have $c_3 = c_2$.

From Equation (3), we have $c_1 = c_3 + 1$.

Substituting $c_3 = c_2$ into the expression for $c_1$, we get $c_1 = c_2 + 1$.

Now, substitute $c_3 = c_2$ into Equation (1):

$c_1 + c_2 + c_2 = 3$

We have two equations with two variables $c_1$ and $c_2$:

Equation (5): $c_1 + 2c_2 = 3$

Equation from substituting (2) into (3): $c_1 = c_2 + 1$ (Let's call this Equation (6) as used in thought process, although it's derived)

Substitute Equation (6) into Equation (5):

$(c_2 + 1) + 2c_2 = 3$

$3c_2 + 1 = 3$

$3c_2 = 3 - 1$

$3c_2 = 2$

$c_2 = \frac{2}{3}$

Now find $c_3$ using $c_3 = c_2$ (from Equation 2):

$c_3 = \frac{2}{3}$

Now find $c_1$ using $c_1 = c_2 + 1$ (from Equation 6):

$c_1 = \frac{2}{3} + 1 = \frac{2}{3} + \frac{3}{3} = \frac{5}{3}$

Thus, the components of vector $\vec{c}$ are $c_1 = \frac{5}{3}$, $c_2 = \frac{2}{3}$, and $c_3 = \frac{2}{3}$.

The vector $\vec{c}$ is:

$\vec{c} = \frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}$

Given:

Vector $\vec{v} = \hat{i} - 2\hat{j} + 2\hat{k}$.

Desired magnitude = 9.

To Find:

A vector in the direction of $\vec{v}$ with magnitude 9.

Solution:

To find a vector in the direction of $\vec{v}$ with a specific magnitude, we first find the unit vector in the direction of $\vec{v}$. The unit vector $\hat{v}$ is given by $\frac{\vec{v}}{|\vec{v}|}$.

First, calculate the magnitude of the given vector $\vec{v}$:

$|\vec{v}| = \left| \; \hat{i} - 2\hat{j} + 2\hat{k} \; \right|$

$|\vec{v}| = \sqrt{(1)^2 + (-2)^2 + (2)^2}$

$|\vec{v}| = \sqrt{1 + 4 + 4}$

$|\vec{v}| = \sqrt{9}$

$|\vec{v}| = 3$

Now, find the unit vector in the direction of $\vec{v}$:

$\hat{v} = \frac{\vec{v}}{|\vec{v}|} = \frac{\hat{i} - 2\hat{j} + 2\hat{k}}{3}$

The vector in the direction of $\vec{v}$ with magnitude 9 is obtained by multiplying the unit vector by the desired magnitude:

Required Vector = $9 \times \hat{v}$

Required Vector = $9 \times \left( \frac{\hat{i} - 2\hat{j} + 2\hat{k}}{3} \right)$

Required Vector = $\frac{9}{3} (\hat{i} - 2\hat{j} + 2\hat{k})$

Required Vector = $3 (\hat{i} - 2\hat{j} + 2\hat{k})$

Comparing this result with the given options:

(A) $\hat{i}− 2\hat{j}+ 2\hat{k}$ (Magnitude is 3)

(B) $\frac{\hat{i} − 2\hat{j} + 2\hat{k}}{3}$ (Magnitude is 1)

(C) $3 \left( \hat{i}− 2\hat{j}+ 2\hat{k} \right)$ (Magnitude is $|3 (\hat{i} - 2\hat{j} + 2\hat{k})| = 3 | \hat{i} - 2\hat{j} + 2\hat{k}| = 3 \times 3 = 9$)

(D) $9 \left( \hat{i}− 2\hat{j}+ 2\hat{k} \right)$ (Magnitude is $|9 (\hat{i} - 2\hat{j} + 2\hat{k})| = 9 | \hat{i} - 2\hat{j} + 2\hat{k}| = 9 \times 3 = 27$)

The vector $3(\hat{i} - 2\hat{j} + 2\hat{k})$ has a magnitude of 9 and is in the same direction as $\hat{i} - 2\hat{j} + 2\hat{k}$.

Therefore, the correct option is (C).

Answer: (C)

Given:

Position vector of the first point $\vec{p_1} = 2\vec{a} - 3\vec{b}$.

Position vector of the second point $\vec{p_2} = \vec{a} + \vec{b}$.

The ratio of division is $m:n = 3:1$.

To Find:

The position vector of the point dividing the line segment joining the two points in the given ratio.

Solution:

We use the section formula for internal division. If a point divides the line segment joining two points with position vectors $\vec{p_1}$ and $\vec{p_2}$ in the ratio $m:n$ internally, its position vector $\vec{r}$ is given by:

$\vec{r} = \frac{n\vec{p_1} + m\vec{p_2}}{m+n}$

In this problem, $\vec{p_1} = 2\vec{a} - 3\vec{b}$, $\vec{p_2} = \vec{a} + \vec{b}$, $m = 3$, and $n = 1$.

Substitute these values into the section formula:

$\vec{r} = \frac{1 \cdot (2\vec{a} - 3\vec{b}) + 3 \cdot (\vec{a} + \vec{b})}{3+1}$

$\vec{r} = \frac{2\vec{a} - 3\vec{b} + 3(\vec{a} + \vec{b})}{4}$

$\vec{r} = \frac{2\vec{a} - 3\vec{b} + 3\vec{a} + 3\vec{b}}{4}$

Combine the terms with $\vec{a}$ and $\vec{b}$:

$\vec{r} = \frac{(2\vec{a} + 3\vec{a}) + (-3\vec{b} + 3\vec{b})}{4}$

$\vec{r} = \frac{5\vec{a} + 0\vec{b}}{4}$

$\vec{r} = \frac{5\vec{a}}{4}$

Comparing this result with the given options, we find that it matches option (D).

Answer: (D)

Given:

Initial point P has coordinates $(2, 5, 0)$.

Terminal point Q has coordinates $(-3, 7, 4)$.

To Find:

The vector with initial point P and terminal point Q.

Solution:

The position vector of the initial point P is $\vec{OP} = 2\hat{i} + 5\hat{j} + 0\hat{k}$.

The position vector of the terminal point Q is $\vec{OQ} = -3\hat{i} + 7\hat{j} + 4\hat{k}$.

The vector from the initial point P to the terminal point Q is given by $\vec{PQ} = \vec{OQ} - \vec{OP}$.

$\vec{PQ} = (-3\hat{i} + 7\hat{j} + 4\hat{k}) - (2\hat{i} + 5\hat{j} + 0\hat{k})$

$\vec{PQ} = (-3 - 2)\hat{i} + (7 - 5)\hat{j} + (4 - 0)\hat{k}$

$\vec{PQ} = -5\hat{i} + 2\hat{j} + 4\hat{k}$

Comparing this result with the given options:

(A) $-\hat{i}+ 12\hat{j}+ 4\hat{k}$

(B) $5\hat{i}+ 2\hat{j}− 4\hat{k}$

(C) $-5\hat{i}+ 2\hat{j}+ 4\hat{k}$

(D) $\hat{i}+ \hat{j}+ \hat{k}$

The calculated vector matches option (C).

Answer: (C)

Given:

Magnitude of vector $\vec{a}$ is $|\vec{a}| = \sqrt{3}$.

Magnitude of vector $\vec{b}$ is $|\vec{b}| = 4$.

Dot product $\vec{a} \cdot \vec{b} = 2\sqrt{3}$.

To Find:

The angle $\theta$ between the vectors $\vec{a}$ and $\vec{b}$.

Solution:

The dot product of two vectors $\vec{a}$ and $\vec{b}$ is defined as:

$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$

where $\theta$ is the angle between the vectors $\vec{a}$ and $\vec{b}$.

We are given the values for $|\vec{a}|$, $|\vec{b}|$, and $\vec{a} \cdot \vec{b}$. Substitute these values into the formula:

$2\sqrt{3} = (\sqrt{3})(4) \cos \theta$

$2\sqrt{3} = 4\sqrt{3} \cos \theta$

Now, solve for $\cos \theta$:

$\cos \theta = \frac{2\sqrt{3}}{4\sqrt{3}}$

Cancel out the $\sqrt{3}$ term from the numerator and denominator:

$\cos \theta = \frac{2}{4}$

Simplify the fraction:

$\cos \theta = \frac{1}{2}$

To find the angle $\theta$, we need to find the inverse cosine of $\frac{1}{2}$.

$\theta = \cos^{-1}\left(\frac{1}{2}\right)$

The angle in the standard range $[0, \pi]$ whose cosine is $\frac{1}{2}$ is $\frac{\pi}{3}$.

$\theta = \frac{\pi}{3}$ radians.

Comparing this result with the given options, we see that it matches option (B).

Answer: (B)

Given:

Vector $\vec{a}= 2\hat{i}+ λ\hat{j}+ \hat{k}$.

Vector $\vec{b}= \hat{i}+ 2\hat{j}+ 3\hat{k}$.

The vectors are orthogonal.

To Find:

The value of $\lambda$.

Solution:

Two vectors $\vec{a}$ and $\vec{b}$ are orthogonal if and only if their dot product is zero, i.e., $\vec{a} \cdot \vec{b} = 0$.

Calculate the dot product of $\vec{a}$ and $\vec{b}$:

$\vec{a} \cdot \vec{b} = (2\hat{i} + \lambda\hat{j} + \hat{k}) \cdot (\hat{i} + 2\hat{j} + 3\hat{k})$

$\vec{a} \cdot \vec{b} = (2)(1) + (\lambda)(2) + (1)(3)$

$\vec{a} \cdot \vec{b} = 2 + 2\lambda + 3$

$\vec{a} \cdot \vec{b} = 5 + 2\lambda$

Since the vectors are orthogonal, set the dot product equal to zero:

$5 + 2\lambda = 0$

Solve for $\lambda$:

$2\lambda = -5$

$\lambda = -\frac{5}{2}$

Comparing this result with the given options, we find that it matches option (D).

Answer: (D)

Given:

Vector $\vec{a}= 3\hat{i}− 6\hat{j}+ \hat{k}$.

Vector $\vec{b}= 2\hat{i}− 4\hat{j}+ λ\hat{k}$.

Vectors $\vec{a}$ and $\vec{b}$ are parallel.

To Find:

The value of $\lambda$.

Solution:

Two non-zero vectors $\vec{a}$ and $\vec{b}$ are parallel if and only if their corresponding components are proportional. That is, if $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$ are parallel, then $\frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3}$.

For the given vectors $\vec{a}= 3\hat{i}− 6\hat{j}+ \hat{k}$ and $\vec{b}= 2\hat{i}− 4\hat{j}+ λ\hat{k}$, we have the components:

$a_1=3, a_2=-6, a_3=1$

$b_1=2, b_2=-4, b_3=\lambda$

Since the vectors are parallel, their components must be proportional:

$\frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3}$

$\frac{3}{2} = \frac{-6}{-4} = \frac{1}{\lambda}$

Let's verify the proportionality of the first two ratios:

$\frac{-6}{-4} = \frac{6}{4} = \frac{3}{2}$. The first two ratios are indeed equal.

Now, we can use any of the equal ratios and set it equal to the third ratio to solve for $\lambda$. Using the first ratio:

To solve for $\lambda$, we can cross-multiply:

$3 \times \lambda = 2 \times 1$

$3\lambda = 2$

$\lambda = \frac{2}{3}$

Alternatively, using the second ratio:

Simplify the left side: $\frac{3}{2} = \frac{1}{\lambda}$.

Cross-multiply: $3\lambda = 2$, which gives $\lambda = \frac{2}{3}$.

Both comparisons give the same value for $\lambda$.

Thus, the value of $\lambda$ for which the vectors $\vec{a}$ and $\vec{b}$ are parallel is $\frac{2}{3}$.

Comparing this result with the given options, we find that it matches option (A).

Answer: (A)

Given:

Position vector of point A: $\vec{OA} = \vec{a} = 2\hat{i} - 3\hat{j} + 2\hat{k}$.

Position vector of point B: $\vec{OB} = \vec{b} = 2\hat{i} + 3\hat{j} + \hat{k}$.

Origin O.

To Find:

Area of triangle OAB.

Solution:

The area of a triangle formed by two vectors $\vec{u}$ and $\vec{v}$ as adjacent sides originating from a common vertex is given by the formula:

Area $= \frac{1}{2} \left| \; \vec{u} \times \vec{v} \; \right|$

In triangle OAB, the sides originating from the origin O are $\vec{OA}$ and $\vec{OB}$.

We are given $\vec{OA} = \vec{a} = 2\hat{i} - 3\hat{j} + 2\hat{k}$ and $\vec{OB} = \vec{b} = 2\hat{i} + 3\hat{j} + \hat{k}$.

So, the area of triangle OAB is Area $= \frac{1}{2} \left| \; \vec{a} \times \vec{b} \; \right|$.

First, we calculate the cross product $\vec{a} \times \vec{b}$:

$\vec{a} \times \vec{b} = (2\hat{i} - 3\hat{j} + 2\hat{k}) \times (2\hat{i} + 3\hat{j} + \hat{k})$

We can compute this using the determinant form:

$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 2 \\ 2 & 3 & 1 \end{vmatrix}$

$= \hat{i}((-3)(1) - (2)(3)) - \hat{j}((2)(1) - (2)(2)) + \hat{k}((2)(3) - (-3)(2))$

$= \hat{i}(-3 - 6) - \hat{j}(2 - 4) + \hat{k}(6 - (-6))$

$= \hat{i}(-9) - \hat{j}(-2) + \hat{k}(6 + 6)$

$= -9\hat{i} + 2\hat{j} + 12\hat{k}$

Next, we find the magnitude of the cross product $\left| \; \vec{a} \times \vec{b} \; \right|$:

$\left| \; \vec{a} \times \vec{b} \; \right| = \left| \; -9\hat{i} + 2\hat{j} + 12\hat{k} \; \right|$

$\left| \; \vec{a} \times \vec{b} \; \right| = \sqrt{(-9)^2 + (2)^2 + (12)^2}$

$\left| \; \vec{a} \times \vec{b} \; \right| = \sqrt{81 + 4 + 144}$

$\left| \; \vec{a} \times \vec{b} \; \right| = \sqrt{229}$

Finally, the area of triangle OAB is half of this magnitude:

Area $= \frac{1}{2} \left| \; \vec{a} \times \vec{b} \; \right| = \frac{1}{2} \sqrt{229}$ square units.

Comparing this result with the given options, we find that it matches option (D).

Answer: (D)

Given:

The expression $\left( \vec{a}× \hat{i} \right)^2 + \left( \vec{a}× \hat{j} \right)^2 + \left( \vec{a}× \hat{k} \right)^2$ for any vector $\vec{a}$.

To Find:

The value of the given expression.

Solution:

Let the vector $\vec{a}$ be represented in component form as $\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$.

The square notation $\vec{v}^2$ for a vector $\vec{v}$ represents the dot product $\vec{v} \cdot \vec{v}$, which is equal to the square of its magnitude, $|\vec{v}|^2$. Therefore, $(\vec{a} \times \hat{i})^2 = |\vec{a} \times \hat{i}|^2$, and similarly for the other terms.

First, calculate the cross product $\vec{a} \times \hat{i}$:

$\vec{a} \times \hat{i} = (x\hat{i} + y\hat{j} + z\hat{k}) \times \hat{i}$

Using the properties of the cross product ($\hat{i} \times \hat{i} = \vec{0}$, $\hat{j} \times \hat{i} = -\hat{k}$, $\hat{k} \times \hat{i} = \hat{j}$):

$\vec{a} \times \hat{i} = x(\hat{i} \times \hat{i}) + y(\hat{j} \times \hat{i}) + z(\hat{k} \times \hat{i})$

$\vec{a} \times \hat{i} = x(\vec{0}) + y(-\hat{k}) + z(\hat{j})$

$\vec{a} \times \hat{i} = z\hat{j} - y\hat{k}$

The magnitude squared is:

$|\vec{a} \times \hat{i}|^2 = |z\hat{j} - y\hat{k}|^2 = z^2 + (-y)^2 = y^2 + z^2$

Next, calculate the cross product $\vec{a} \times \hat{j}$:

$\vec{a} \times \hat{j} = (x\hat{i} + y\hat{j} + z\hat{k}) \times \hat{j}$

Using the properties of the cross product ($\hat{i} \times \hat{j} = \hat{k}$, $\hat{j} \times \hat{j} = \vec{0}$, $\hat{k} \times \hat{j} = -\hat{i}$):

$\vec{a} \times \hat{j} = x(\hat{i} \times \hat{j}) + y(\hat{j} \times \hat{j}) + z(\hat{k} \times \hat{j})$

$\vec{a} \times \hat{j} = x(\hat{k}) + y(\vec{0}) + z(-\hat{i})$

$\vec{a} \times \hat{j} = -z\hat{i} + x\hat{k}$

The magnitude squared is:

$|\vec{a} \times \hat{j}|^2 = |-z\hat{i} + x\hat{k}|^2 = (-z)^2 + x^2 = x^2 + z^2$

Finally, calculate the cross product $\vec{a} \times \hat{k}$:

$\vec{a} \times \hat{k} = (x\hat{i} + y\hat{j} + z\hat{k}) \times \hat{k}$

Using the properties of the cross product ($\hat{i} \times \hat{k} = -\hat{j}$, $\hat{j} \times \hat{k} = \hat{i}$, $\hat{k} \times \hat{k} = \vec{0}$):

$\vec{a} \times \hat{k} = x(\hat{i} \times \hat{k}) + y(\hat{j} \times \hat{k}) + z(\hat{k} \times \hat{k})$

$\vec{a} \times \hat{k} = x(-\hat{j}) + y(\hat{i}) + z(\vec{0})$

$\vec{a} \times \hat{k} = y\hat{i} - x\hat{j}$

The magnitude squared is:

$|\vec{a} \times \hat{k}|^2 = |y\hat{i} - x\hat{j}|^2 = y^2 + (-x)^2 = x^2 + y^2$

Now, sum the three squared magnitudes:

$\left( \vec{a}× \hat{i} \right)^2 + \left( \vec{a}× \hat{j} \right)^2 + \left( \vec{a}× \hat{k} \right)^2 = (y^2 + z^2) + (x^2 + z^2) + (x^2 + y^2)$

$= y^2 + z^2 + x^2 + z^2 + x^2 + y^2$

$= 2x^2 + 2y^2 + 2z^2$

$= 2(x^2 + y^2 + z^2)$

The magnitude of vector $\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$ is $|\vec{a}| = \sqrt{x^2 + y^2 + z^2}$.

So, $|\vec{a}|^2 = x^2 + y^2 + z^2$.

Also, $\vec{a}^2 = \vec{a} \cdot \vec{a} = |\vec{a}|^2$.

Substituting $|\vec{a}|^2$ back into the sum:

$2(x^2 + y^2 + z^2) = 2|\vec{a}|^2 = 2\vec{a}^2$

Thus, the value of the expression $\left( \vec{a}× \hat{i} \right)^2 + \left( \vec{a}× \hat{j} \right)^2 + \left( \vec{a}× \hat{k} \right)^2$ is $2\vec{a}^2$.

Comparing this result with the given options, we find that it matches option (D).

Answer: (D)

Given:

$|\vec{a}| = 10$

$|\vec{b}| = 2$

$\vec{a} \cdot \vec{b} = 12$

To Find:

The value of $|\vec{a} \times \vec{b}|$.

Solution:

We know the relationship between the dot product, the magnitudes, and the angle between two vectors:

$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$

where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.

Substitute the given values into the formula:

$12 = (10)(2) \cos \theta$

$12 = 20 \cos \theta$

Solving for $\cos \theta$:

$\cos \theta = \frac{\cancel{12}^{3}}{\cancel{20}_{5}}$

$\cos \theta = \frac{3}{5}$

Now, we use the identity $\sin^2 \theta + \cos^2 \theta = 1$ to find $\sin \theta$.

$\sin^2 \theta = 1 - \cos^2 \theta$

$\sin^2 \theta = 1 - \left(\frac{3}{5}\right)^2$

$\sin^2 \theta = 1 - \frac{9}{25}$

$\sin^2 \theta = \frac{25 - 9}{25} = \frac{16}{25}$

$\sin \theta = \pm\sqrt{\frac{16}{25}} = \pm\frac{4}{5}$

Since the angle $\theta$ between two vectors is typically taken in the range $0 \leq \theta \leq \pi$, $\sin \theta$ is non-negative in this range. Thus, we take the positive value:

$\sin \theta = \frac{4}{5}$

The magnitude of the cross product of two vectors is given by:

$|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$

Substitute the given magnitudes and the value of $\sin \theta$ we found:

$|\vec{a} \times \vec{b}| = (10)(2) \left(\frac{4}{5}\right)$

$|\vec{a} \times \vec{b}| = 20 \times \frac{4}{5}$

$|\vec{a} \times \vec{b}| = \cancel{20}^{4} \times \frac{4}{\cancel{5}_{1}}$

$|\vec{a} \times \vec{b}| = 4 \times 4$

$|\vec{a} \times \vec{b}| = 16$

The value of $|\vec{a} \times \vec{b}|$ is 16.

Comparing this result with the given options, we find that it matches option (D).

Answer: (D)

Given:

Three vectors: $\vec{u} = λ\hat{i}+ \hat{j}+ 2\hat{k}$, $\vec{v} = \hat{i}+ λ\hat{j}− \hat{k}$, and $\vec{w} = 2\hat{i}− \hat{j}+ λ\hat{k}$.

The vectors are coplanar.

To Find:

The value of $\lambda$ for which the vectors are coplanar.

Solution:

Three vectors $\vec{u}$, $\vec{v}$, and $\vec{w}$ are coplanar if and only if their scalar triple product is zero. The scalar triple product $[\vec{u} \vec{v} \vec{w}]$ is given by the determinant formed by the components of the vectors:

$[\vec{u} \vec{v} \vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w}) = \begin{vmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{vmatrix}$

For the given vectors:

$\vec{u} = \lambda\hat{i} + 1\hat{j} + 2\hat{k}$

$\vec{v} = 1\hat{i} + \lambda\hat{j} - 1\hat{k}$

$\vec{w} = 2\hat{i} - 1\hat{j} + \lambda\hat{k}$

The condition for coplanarity is:

$\begin{vmatrix} \lambda & 1 & 2 \\ 1 & \lambda & -1 \\ 2 & -1 & \lambda \end{vmatrix} = 0$

Expanding the determinant along the first row:

$\lambda \begin{vmatrix} \lambda & -1 \\ -1 & \lambda \end{vmatrix} - 1 \begin{vmatrix} 1 & -1 \\ 2 & \lambda \end{vmatrix} + 2 \begin{vmatrix} 1 & \lambda \\ 2 & -1 \end{vmatrix} = 0$

Calculate the $2 \times 2$ determinants:

$\begin{vmatrix} \lambda & -1 \\ -1 & \lambda \end{vmatrix} = (\lambda)(\lambda) - (-1)(-1) = \lambda^2 - 1$

$\begin{vmatrix} 1 & -1 \\ 2 & \lambda \end{vmatrix} = (1)(\lambda) - (-1)(2) = \lambda + 2$

$\begin{vmatrix} 1 & \lambda \\ 2 & -1 \end{vmatrix} = (1)(-1) - (\lambda)(2) = -1 - 2\lambda$

Substitute these values back into the equation:

$\lambda(\lambda^2 - 1) - 1(\lambda + 2) + 2(-1 - 2\lambda) = 0$

$\lambda^3 - \lambda - \lambda - 2 - 2 - 4\lambda = 0$

$\lambda^3 - 6\lambda - 4 = 0$

We need to find the value of $\lambda$ from the given options that satisfies this equation.

Let's test option (A) $\lambda = -2$:

$(-2)^3 - 6(-2) - 4 = -8 + 12 - 4 = 4 - 4 = 0$

Since the equation is satisfied for $\lambda = -2$, this is the required value.

Let's verify other options (optional, but good practice):

For $\lambda = 0$: $(0)^3 - 6(0) - 4 = -4 \neq 0$

For $\lambda = 1$: $(1)^3 - 6(1) - 4 = 1 - 6 - 4 = -9 \neq 0$

For $\lambda = -1$: $(-1)^3 - 6(-1) - 4 = -1 + 6 - 4 = 1 \neq 0$

Thus, the only value among the options that makes the vectors coplanar is $\lambda = -2$.

Answer: (A)

Given:

$\vec{a}$, $\vec{b}$, $\vec{c}$ are unit vectors.

This implies $|\vec{a}| = 1$, $|\vec{b}| = 1$, and $|\vec{c}| = 1$.

Also given the vector sum: $\vec{a} + \vec{b} + \vec{c} = \vec{0}$.

To Find:

The value of $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$.

Solution:

We start with the given condition $\vec{a} + \vec{b} + \vec{c} = \vec{0}$.

Take the dot product of this equation with itself:

$(\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = \vec{0} \cdot \vec{0}$

$(\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = 0$

Expand the left side using the distributive property of the dot product:

$\vec{a} \cdot (\vec{a} + \vec{b} + \vec{c}) + \vec{b} \cdot (\vec{a} + \vec{b} + \vec{c}) + \vec{c} \cdot (\vec{a} + \vec{b} + \vec{c}) = 0$

$\vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b} + \vec{c} \cdot \vec{c} = 0$

Using the properties of the dot product $\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}$ and $\vec{v} \cdot \vec{v} = |\vec{v}|^2$:

$|\vec{a}|^2 + \vec{a} \cdot \vec{b} + \vec{c} \cdot \vec{a} + \vec{a} \cdot \vec{b} + |\vec{b}|^2 + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} + \vec{b} \cdot \vec{c} + |\vec{c}|^2 = 0$

Group the terms:

$|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + (\vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{b}) + (\vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{b}) + (\vec{c} \cdot \vec{a} + \vec{a} \cdot \vec{c}) = 0$

$|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b}) + 2(\vec{b} \cdot \vec{c}) + 2(\vec{c} \cdot \vec{a}) = 0$

$|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$

Since $\vec{a}$, $\vec{b}$, and $\vec{c}$ are unit vectors, $|\vec{a}| = 1$, $|\vec{b}| = 1$, and $|\vec{c}| = 1$. Substitute these magnitudes:

$1^2 + 1^2 + 1^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$

$1 + 1 + 1 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$

$3 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$

Now, solve for the required expression $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$:

$2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = -3$

$\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{3}{2}$

Comparing this result with the given options, we find that it matches option (C).

Answer: (C)

Given:

Two vectors $\vec{a}$ and $\vec{b}$.

To Find:

The projection vector of $\vec{a}$ on $\vec{b}$.

Solution:

The projection vector of $\vec{a}$ on $\vec{b}$, denoted as $\text{proj}_{\vec{b}} \vec{a}$, is a vector component of $\vec{a}$ along the direction of $\vec{b}$.

It can be found by multiplying the scalar projection of $\vec{a}$ on $\vec{b}$ by the unit vector in the direction of $\vec{b}$.

The scalar projection of $\vec{a}$ on $\vec{b}$ is given by:

$\text{comp}_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$

The unit vector in the direction of $\vec{b}$ is given by:

$\hat{b} = \frac{\vec{b}}{|\vec{b}|}$

The projection vector is the product of the scalar projection and the unit vector:

$\text{proj}_{\vec{b}} \vec{a} = \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \right) \hat{b}$

Substitute the expression for $\hat{b}$:

$\text{proj}_{\vec{b}} \vec{a} = \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \right) \left( \frac{\vec{b}}{|\vec{b}|} \right)$

$\text{proj}_{\vec{b}} \vec{a} = \frac{(\vec{a} \cdot \vec{b}) \vec{b}}{|\vec{b}| \cdot |\vec{b}|}$

$\text{proj}_{\vec{b}} \vec{a} = \frac{(\vec{a} \cdot \vec{b}) \vec{b}}{|\vec{b}|^2}$

This can also be written as:

$\text{proj}_{\vec{b}} \vec{a} = \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \right) \vec{b}$

Comparing this formula with the given options:

(A) $\left( \frac{\vec{a} \;.\; \vec{b}}{\left| \vec{b} \right|^2} \right) \vec{b}$ - Matches the derived formula.

(B) $\frac{\vec{a} \;.\; \vec{b}}{\left| \vec{b} \right|}$ - This is the scalar projection, not the vector projection.

(C) $\frac{\vec{a} \;.\; \vec{b}}{\left| \vec{a} \right|}$ - This is the scalar projection of $\vec{b}$ on $\vec{a}$.

(D) $\left( \frac{\vec{a} \;.\; \vec{b}}{\left| \vec{a} \right|^2} \right) \hat{b}$ - This is incorrect.

The correct formula for the projection vector of $\vec{a}$ on $\vec{b}$ is $\left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \right) \vec{b}$.

Answer: (A)

Given:

Three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$.

Magnitude of $\vec{a}$ is $|\vec{a}| = 2$.

Magnitude of $\vec{b}$ is $|\vec{b}| = 3$.

Magnitude of $\vec{c}$ is $|\vec{c}| = 5$.

Vector sum $\vec{a} + \vec{b} + \vec{c} = \vec{0}$.

To Find:

The value of $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$.

Solution:

We are given the condition $\vec{a} + \vec{b} + \vec{c} = \vec{0}$.

Take the dot product of this equation with itself:

$(\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = \vec{0} \cdot \vec{0}$

$(\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = 0$

Expand the left side:

$\vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b} + \vec{c} \cdot \vec{c} = 0$

Using the properties of the dot product ($\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}$ and $\vec{v} \cdot \vec{v} = |\vec{v}|^2$):

$|\vec{a}|^2 + \vec{a} \cdot \vec{b} + \vec{c} \cdot \vec{a} + \vec{a} \cdot \vec{b} + |\vec{b}|^2 + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} + \vec{b} \cdot \vec{c} + |\vec{c}|^2 = 0$

Combine like terms:

$|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b}) + 2(\vec{b} \cdot \vec{c}) + 2(\vec{c} \cdot \vec{a}) = 0$

$|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$

Substitute the given magnitudes $|\vec{a}| = 2$, $|\vec{b}| = 3$, and $|\vec{c}| = 5$:

$2^2 + 3^2 + 5^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$

$4 + 9 + 25 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$

$38 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$

Solve for the required expression $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$:

$2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = -38$

$\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = \frac{-38}{2}$

$\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -19$

The value of $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$ is $-19$.

Comparing this result with the given options, we find that it matches option (C).

Answer: (C)

Given:

Magnitude of vector $\vec{a}$ is $|\vec{a}| = 4$.

The range of the scalar $\lambda$ is $-3 \leq \lambda \leq 3$.

To Find:

The range of the magnitude of the vector $λ\vec{a}$, which is $|λ\vec{a}|$.

Solution:

We use the property of vector magnitude that states for a scalar $k$ and a vector $\vec{v}$, the magnitude of the scaled vector is $|k\vec{v}| = |k| |\vec{v}|$.

Applying this property to the vector $λ\vec{a}$, we have:

$|λ\vec{a}| = |λ| |\vec{a}|$

We are given that $|\vec{a}| = 4$. Substitute this value into the equation:

$|λ\vec{a}| = |λ| \times 4$

$|λ\vec{a}| = 4|λ|$

Now, we need to determine the range of $|λ|$ given the range of $\lambda$. The range of $\lambda$ is $-3 \leq \lambda \leq 3$.

The absolute value $|λ|$ represents the distance of $\lambda$ from zero on the number line.

The minimum value of $|λ|$ in the interval $[-3, 3]$ occurs when $\lambda = 0$, so $|0| = 0$.

The maximum value of $|λ|$ in the interval $[-3, 3]$ occurs at the endpoints $\lambda = -3$ or $\lambda = 3$. In both cases, $|-3| = 3$ and $|3| = 3$.

So, the range of $|λ|$ is $0 \leq |λ| \leq 3$.

Now, we can find the range of $4|λ|$ by multiplying the inequality by 4:

$4 \times 0 \leq 4|λ| \leq 4 \times 3$

$0 \leq 4|λ| \leq 12$

Since $|λ\vec{a}| = 4|λ|$, the range of $|λ\vec{a}|$ is $[0, 12]$.

Comparing this result with the given options:

(A) [0, 8]

(B) [– 12, 8] (Magnitudes are always non-negative)

(C) [0, 12]

(D) [8, 12]

The calculated range matches option (C).

Answer: (C)

Given:

Vector $\vec{a}= 2\hat{i}+ \hat{j}+ 2\hat{k}$.

Vector $\vec{b}= \hat{j}+ \hat{k}$.

To Find:

The number of vectors of unit length perpendicular to both $\vec{a}$ and $\vec{b}$.

Solution:

A vector that is perpendicular to two given vectors $\vec{a}$ and $\vec{b}$ is proportional to their cross product $\vec{a} \times \vec{b}$.

First, calculate the cross product $\vec{a} \times \vec{b}$:

$\vec{a} \times \vec{b} = (2\hat{i} + \hat{j} + 2\hat{k}) \times (\hat{j} + \hat{k})$

We can write $\vec{b}$ as $0\hat{i} + \hat{j} + \hat{k}$.

$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 2 \\ 0 & 1 & 1 \end{vmatrix}$

Expanding the determinant:

$= \hat{i}((1)(1) - (2)(1)) - \hat{j}((2)(1) - (2)(0)) + \hat{k}((2)(1) - (1)(0))$

$= \hat{i}(1 - 2) - \hat{j}(2 - 0) + \hat{k}(2 - 0)$

$= -1\hat{i} - 2\hat{j} + 2\hat{k}$

So, $\vec{a} \times \vec{b} = -\hat{i} - 2\hat{j} + 2\hat{k}$.

Any vector perpendicular to both $\vec{a}$ and $\vec{b}$ must be parallel to $\vec{a} \times \vec{b}$. The directions perpendicular to both vectors are given by $\vec{a} \times \vec{b}$ and $-(\vec{a} \times \vec{b})$.

Let $\vec{v} = \vec{a} \times \vec{b} = -\hat{i} - 2\hat{j} + 2\hat{k}$.

To find the unit vectors in these directions, we need to find the magnitude of $\vec{v}$.

$|\vec{v}| = |-\hat{i} - 2\hat{j} + 2\hat{k}|$

$|\vec{v}| = \sqrt{(-1)^2 + (-2)^2 + (2)^2}$

$|\vec{v}| = \sqrt{1 + 4 + 4}$

$|\vec{v}| = \sqrt{9}$

$|\vec{v}| = 3$

The unit vectors perpendicular to both $\vec{a}$ and $\vec{b}$ are the unit vectors in the direction of $\vec{v}$ and $-\vec{v}$.

Unit vector 1: $\hat{u}_1 = \frac{\vec{v}}{|\vec{v}|} = \frac{-\hat{i} - 2\hat{j} + 2\hat{k}}{3} = -\frac{1}{3}\hat{i} - \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}$

Unit vector 2: $\hat{u}_2 = \frac{-\vec{v}}{|\vec{v}|} = \frac{- (-\hat{i} - 2\hat{j} + 2\hat{k})}{3} = \frac{\hat{i} + 2\hat{j} - 2\hat{k}}{3} = \frac{1}{3}\hat{i} + \frac{2}{3}\hat{j} - \frac{2}{3}\hat{k}$

These are the only two distinct unit vectors that are perpendicular to both $\vec{a}$ and $\vec{b}$. They point in exactly opposite directions.

Thus, the number of vectors of unit length perpendicular to the given vectors is two.

Comparing this result with the given options, we find that it matches option (B).

Answer: (B)

Let $\vec{a}$ and $\vec{b}$ be two non-collinear vectors originating from the same point O.

Explanation:

The sum of two vectors, $\vec{a} + \vec{b}$, represents the diagonal of the parallelogram formed by $\vec{a}$ and $\vec{b}$ as adjacent sides, starting from the same origin point O.

Let this parallelogram be OACB, where $\vec{OA} = \vec{a}$ and $\vec{OB} = \vec{b}$. The diagonal is $\vec{OC} = \vec{OA} + \vec{OB} = \vec{a} + \vec{b}$.

In a parallelogram, the diagonal bisects the angles at the vertices *only if* the parallelogram is a rhombus.

A parallelogram is a rhombus if and only if its adjacent sides have equal length.

In our case, the adjacent sides originating from O are represented by the vectors $\vec{a}$ and $\vec{b}$.

For the parallelogram OACB to be a rhombus, the magnitudes of the adjacent sides must be equal:

Which means:

$|\vec{a}| = |\vec{b}|$

Therefore, the vector $\vec{a} + \vec{b}$ bisects the angle between the non-collinear vectors $\vec{a}$ and $\vec{b}$ if and only if their magnitudes are equal, i.e., $|\vec{a}| = |\vec{b}|$.

Alternatively, using unit vectors:

The direction of the angle bisector between $\vec{a}$ and $\vec{b}$ is given by the unit vector $\frac{\vec{a}}{|\vec{a}|} + \frac{\vec{b}}{|\vec{b}|}$.

The vector $\vec{a} + \vec{b}$ bisects the angle between $\vec{a}$ and $\vec{b}$ if its direction is the same as the direction of the angle bisector. This means $\vec{a} + \vec{b}$ must be parallel to $\frac{\vec{a}}{|\vec{a}|} + \frac{\vec{b}}{|\vec{b}|}$.

For two vectors to be parallel, one must be a scalar multiple of the other. So, there exists a scalar $k$ such that:

$\vec{a} + \vec{b} = k \left( \frac{\vec{a}}{|\vec{a}|} + \frac{\vec{b}}{|\vec{b}|} \right)$

$\vec{a} + \vec{b} = \frac{k}{|\vec{a}|} \vec{a} + \frac{k}{|\vec{b}|} \vec{b}$

Since $\vec{a}$ and $\vec{b}$ are non-collinear, they are linearly independent. For the equality to hold, the coefficients of $\vec{a}$ and $\vec{b}$ on both sides must be equal:

From the first equation, $k = |\vec{a}|$. From the second equation, $k = |\vec{b}|$.

Therefore, $|\vec{a}| = |\vec{b}|$.

So, the blank should be filled with the condition $|\vec{a}| = |\vec{b}|$.

The vector $\vec{a}+ \vec{b}$ bisects the angle between the non-collinear vectors $\vec{a}$ and $\vec{b}$ if $\left| \vec{a} \right| = \left| \vec{b} \right|$.

The value of $\vec{a} \;.\; \left( \vec{b}× \vec{c} \right)$ is 0.

Explanation:

We are given that $\vec{r}$ is a non-zero vector such that $\vec{r} \cdot \vec{a} = 0$, $\vec{r} \cdot \vec{b} = 0$, and $\vec{r} \cdot \vec{c} = 0$.

The dot product of two non-zero vectors is zero if and only if the vectors are orthogonal (perpendicular).

So, the condition $\vec{r} \cdot \vec{a} = 0$ means $\vec{r}$ is perpendicular to $\vec{a}$.

The condition $\vec{r} \cdot \vec{b} = 0$ means $\vec{r}$ is perpendicular to $\vec{b}$.

The condition $\vec{r} \cdot \vec{c} = 0$ means $\vec{r}$ is perpendicular to $\vec{c}$.

Since $\vec{r}$ is perpendicular to both $\vec{b}$ and $\vec{c}$, $\vec{r}$ must be parallel to the cross product of $\vec{b}$ and $\vec{c}$, which is $\vec{b} \times \vec{c}$.

Therefore, $\vec{r} = k (\vec{b} \times \vec{c})$ for some scalar $k$.

Since $\vec{r}$ is a non-zero vector, $k$ must be non-zero ($k \neq 0$). Also, $\vec{b} \times \vec{c}$ must be non-zero, implying $\vec{b}$ and $\vec{c}$ are not parallel.

We are also given that $\vec{r}$ is perpendicular to $\vec{a}$. So, the dot product of $\vec{r}$ and $\vec{a}$ is zero:

Substitute $\vec{r} = k (\vec{b} \times \vec{c})$ into this equation:

$(k (\vec{b} \times \vec{c})) \cdot \vec{a} = 0$

$k (\vec{b} \times \vec{c} \cdot \vec{a}) = 0$

The scalar triple product has the property that the dot and cross can be interchanged, so $\vec{b} \times \vec{c} \cdot \vec{a} = \vec{a} \cdot (\vec{b} \times \vec{c})$.

Thus, the equation becomes:

$k (\vec{a} \cdot (\vec{b} \times \vec{c})) = 0$

Since $\vec{r}$ is a non-zero vector, $k$ must be non-zero ($k \neq 0$).

For the product $k (\vec{a} \cdot (\vec{b} \times \vec{c}))$ to be zero when $k \neq 0$, the other factor must be zero.

Therefore, $\vec{a} \cdot (\vec{b} \times \vec{c}) = 0$.

This result implies that the vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ are coplanar, which is consistent with them all being perpendicular to the same non-zero vector $\vec{r}$.

The value of $\vec{a} \;.\; \left( \vec{b}× \vec{c} \right)$ is 0.

Let the adjacent sides of the parallelogram be $\vec{a}$ and $\vec{b}$.

Given $\vec{a}= 3\hat{i}− 2\hat{j}+ 2\hat{k}$ and $\vec{b}= −\hat{i}− 2\hat{k}$.

The diagonals of the parallelogram are given by the vectors $\vec{d_1} = \vec{a} + \vec{b}$ and $\vec{d_2} = \vec{a} - \vec{b}$.

Calculate $\vec{d_1}$:

$\vec{d_1} = (3\hat{i} - 2\hat{j} + 2\hat{k}) + (-\hat{i} + 0\hat{j} - 2\hat{k})$

$\vec{d_1} = (3 - 1)\hat{i} + (-2 + 0)\hat{j} + (2 - 2)\hat{k}$

$\vec{d_1} = 2\hat{i} - 2\hat{j} + 0\hat{k}$

Calculate $\vec{d_2}$:

$\vec{d_2} = (3\hat{i} - 2\hat{j} + 2\hat{k}) - (-\hat{i} + 0\hat{j} - 2\hat{k})$

$\vec{d_2} = (3 - (-1))\hat{i} + (-2 - 0)\hat{j} + (2 - (-2))\hat{k}$

$\vec{d_2} = 4\hat{i} - 2\hat{j} + 4\hat{k}$

To find the angle $\theta$ between the diagonals $\vec{d_1}$ and $\vec{d_2}$, we use the dot product formula:

$\cos \theta = \frac{\vec{d_1} \cdot \vec{d_2}}{|\vec{d_1}| |\vec{d_2}|}$

Calculate the dot product $\vec{d_1} \cdot \vec{d_2}$:

$\vec{d_1} \cdot \vec{d_2} = (2\hat{i} - 2\hat{j} + 0\hat{k}) \cdot (4\hat{i} - 2\hat{j} + 4\hat{k})$

$\vec{d_1} \cdot \vec{d_2} = (2)(4) + (-2)(-2) + (0)(4)$

$\vec{d_1} \cdot \vec{d_2} = 8 + 4 + 0 = 12$

Calculate the magnitudes of $\vec{d_1}$ and $\vec{d_2}$:

$|\vec{d_1}| = \sqrt{2^2 + (-2)^2 + 0^2} = \sqrt{4 + 4 + 0} = \sqrt{8} = 2\sqrt{2}$

$|\vec{d_2}| = \sqrt{4^2 + (-2)^2 + 4^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$

Substitute these values into the cosine formula:

$\cos \theta = \frac{12}{(2\sqrt{2})(6)} = \frac{12}{12\sqrt{2}} = \frac{1}{\sqrt{2}}$

Since $\cos \theta = \frac{1}{\sqrt{2}}$ is positive, the angle $\theta$ is acute.

The angle whose cosine is $\frac{1}{\sqrt{2}}$ is $\frac{\pi}{4}$ radians or $45^\circ$.

$\theta = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}$

The acute angle between the diagonals is $\frac{\pi}{4}$.

The blank should be filled with $\frac{\pi}{4}$.

The acute angle between its diagonals is $\frac{\pi}{4}$.

Let the given conditions be:

1. $\left| k\vec{a} \right| < \left| \vec{a} \right|$

2. $k\vec{a}+ \frac{1}{2} \vec{a}$ is parallel to $\vec{a}$

Consider the first condition:

Using the property of vector magnitude, $|c\vec{v}| = |c| |\vec{v}|$, we have:

$|k| |\vec{a}| < |\vec{a}|$

We need to consider two cases for vector $\vec{a}$.

Case 1: $\vec{a} = \vec{0}$.

If $\vec{a} = \vec{0}$, then $|\vec{a}| = 0$. The inequality becomes $|k| \cdot 0 < 0$, which simplifies to $0 < 0$. This is a false statement. So, no value of $k$ satisfies the first condition if $\vec{a}$ is the zero vector.

Case 2: $\vec{a} \neq \vec{0}$.

If $\vec{a}$ is a non-zero vector, then $|\vec{a}| > 0$. We can divide both sides of the inequality by $|\vec{a}|$.

$\frac{|k| |\vec{a}|}{|\vec{a}|} < \frac{|\vec{a}|}{|\vec{a}|}$

$|k| < 1$

This inequality holds for $-1 < k < 1$.

Consider the second condition:

First, simplify the vector $k\vec{a}+ \frac{1}{2} \vec{a}$:

$k\vec{a}+ \frac{1}{2} \vec{a} = \left( k + \frac{1}{2} \right) \vec{a}$

A vector $c\vec{a}$ is parallel to $\vec{a}$ for any scalar $c$, provided $\vec{a}$ is non-zero. If $\vec{a} = \vec{0}$, then $c\vec{0} = \vec{0}$, which is considered parallel to any vector, including $\vec{0}$.

So, the vector $\left( k + \frac{1}{2} \right) \vec{a}$ is always parallel to $\vec{a}$ for any scalar $k$ and any vector $\vec{a}$.

This second condition holds true for all real values of $k$.

We need to find the values of $k$ for which both conditions hold true.

If $\vec{a} = \vec{0}$, the first condition is never true, so no value of $k$ satisfies both conditions.

If $\vec{a} \neq \vec{0}$, the first condition is true for $-1 < k < 1$, and the second condition is true for all real $k$. The values of $k$ that satisfy both are the intersection of these two sets, which is $-1 < k < 1$.

The phrasing "The values of k for which ... holds true are ______" implies finding the set of k for which the statement is true for an arbitrary vector $\vec{a}$. In typical contexts for such problems, this refers to the case where $\vec{a}$ is a non-zero vector, and the result for $k$ is independent of the specific non-zero vector chosen. As shown, the range $-1 < k < 1$ is independent of the specific non-zero $\vec{a}$.

Thus, the values of k for which the given conditions hold true (for a non-zero vector $\vec{a}$) are those in the interval $(-1, 1)$.

The blank should be filled with the range of $k$.

The values of k for which $\left| k\vec{a} \right| < \left| \vec{a} \right|$ and $k\vec{a}+ \frac{1}{2} \vec{a}$ is parallel to $\vec{a}$ holds true are $-1 < k < 1$.

The value of the expression $\left| \vec{a}× \vec{b} \right|^2 + \left( \vec{a} \;.\; \vec{b} \right)^2$ is $\left| \vec{a} \right|^2 \left| \vec{b} \right|^2$.

Explanation:

Let $\theta$ be the angle between the vectors $\vec{a}$ and $\vec{b}$.

The magnitude of the cross product $\vec{a} \times \vec{b}$ is given by:

$|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$

Squaring the magnitude of the cross product:

The dot product $\vec{a} \cdot \vec{b}$ is given by:

$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$

Squaring the dot product:

Now, consider the given expression:

$\left| \vec{a}× \vec{b} \right|^2 + \left( \vec{a} \;.\; \vec{b} \right)^2$

Substitute the squared terms we found:

$= |\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta + |\vec{a}|^2 |\vec{b}|^2 \cos^2 \theta$

Factor out the common term $|\vec{a}|^2 |\vec{b}|^2$:

$= |\vec{a}|^2 |\vec{b}|^2 (\sin^2 \theta + \cos^2 \theta)$

Using the fundamental trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$:

$= |\vec{a}|^2 |\vec{b}|^2 (1)$

$= |\vec{a}|^2 |\vec{b}|^2$

The value of the expression is $|\vec{a}|^2 |\vec{b}|^2$. Note that $|\vec{a}|^2$ is often written as $\vec{a}^2$ and $|\vec{b}|^2$ as $\vec{b}^2$, but using magnitude notation is standard.

Given:

$\left| \vec{a}× \vec{b} \right|^2 + \left| \vec{a} \;.\; \vec{b} \right|^2 = 144$

$\left| \vec{a} \right| = 4$

To Find:

The value of $\left| \vec{b} \right|$.

Solution:

We know the identity relating the magnitudes of the cross product and dot product with the magnitudes of the vectors:

$\left| \vec{a}× \vec{b} \right|^2 + \left( \vec{a} \;.\; \vec{b} \right)^2 = \left| \vec{a} \right|^2 \left| \vec{b} \right|^2$

This identity was discussed in the previous question (Question 38).

We are given that $\left| \vec{a}× \vec{b} \right|^2 + \left| \vec{a} \;.\; \vec{b} \right|^2 = 144$.

Using the identity, we can write:

We are also given that $\left| \vec{a} \right| = 4$.

Squaring the magnitude of $\vec{a}$:

$\left| \vec{a} \right|^2 = 4^2 = 16$

Substitute the value of $\left| \vec{a} \right|^2$ into Equation (1):

$16 \left| \vec{b} \right|^2 = 144$

Now, solve for $\left| \vec{b} \right|^2$:

$\left| \vec{b} \right|^2 = \frac{144}{16}$

$\left| \vec{b} \right|^2 = 9$

To find $\left| \vec{b} \right|$, take the square root of both sides:

$\left| \vec{b} \right| = \sqrt{9}$

Since the magnitude of a vector is always non-negative, we take the positive square root:

$\left| \vec{b} \right| = 3$

The value of $\left| \vec{b} \right|$ is 3.

The blank should be filled with 3.

If $\left| \vec{a}× \vec{b} \right|^2 + \left| \vec{a} \;.\; \vec{b} \right|^2 = 144$ and $\left| \vec{a} \right| = 4$, then $\left| \vec{b} \right|$ is equal to 3.

Given:

$\vec{a}$ is any non-zero vector.

The expression $(\vec{a} \cdot \hat{i}) \hat{i} + (\vec{a} \cdot \hat{j}) \hat{j} + (\vec{a} \cdot \hat{k}) \hat{k}$.

To Evaluate:

The value of the given expression.

Solution:

Let the vector $\vec{a}$ be expressed in terms of its components along the standard basis vectors $\hat{i}$, $\hat{j}$, and $\hat{k}$.

Let $\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$, where $x, y, z$ are the scalar components of $\vec{a}$ along the x, y, and z axes, respectively.

We can find the scalar components $x, y, z$ by taking the dot product of $\vec{a}$ with the unit vectors $\hat{i}$, $\hat{j}$, and $\hat{k}$.

The component along $\hat{i}$ is:

$\vec{a} \cdot \hat{i} = (x\hat{i} + y\hat{j} + z\hat{k}) \cdot \hat{i}$

Using the orthogonality property of the basis vectors ($\hat{i} \cdot \hat{i} = 1$, $\hat{j} \cdot \hat{i} = 0$, $\hat{k} \cdot \hat{i} = 0$):

$\vec{a} \cdot \hat{i} = x(\hat{i} \cdot \hat{i}) + y(\hat{j} \cdot \hat{i}) + z(\hat{k} \cdot \hat{i})$

$\vec{a} \cdot \hat{i} = x(1) + y(0) + z(0) = x$

The component along $\hat{j}$ is:

$\vec{a} \cdot \hat{j} = (x\hat{i} + y\hat{j} + z\hat{k}) \cdot \hat{j}$

Using the orthogonality property ($\hat{i} \cdot \hat{j} = 0$, $\hat{j} \cdot \hat{j} = 1$, $\hat{k} \cdot \hat{j} = 0$):

$\vec{a} \cdot \hat{j} = x(\hat{i} \cdot \hat{j}) + y(\hat{j} \cdot \hat{j}) + z(\hat{k} \cdot \hat{j})$

$\vec{a} \cdot \hat{j} = x(0) + y(1) + z(0) = y$

The component along $\hat{k}$ is:

$\vec{a} \cdot \hat{k} = (x\hat{i} + y\hat{j} + z\hat{k}) \cdot \hat{k}$

Using the orthogonality property ($\hat{i} \cdot \hat{k} = 0$, $\hat{j} \cdot \hat{k} = 0$, $\hat{k} \cdot \hat{k} = 1$):

$\vec{a} \cdot \hat{k} = x(\hat{i} \cdot \hat{k}) + y(\hat{j} \cdot \hat{k}) + z(\hat{k} \cdot \hat{k})$

$\vec{a} \cdot \hat{k} = x(0) + y(0) + z(1) = z$

Now substitute these values back into the given expression:

$(\vec{a} \cdot \hat{i}) \hat{i} + (\vec{a} \cdot \hat{j}) \hat{j} + (\vec{a} \cdot \hat{k}) \hat{k} = (x) \hat{i} + (y) \hat{j} + (z) \hat{k}$

$= x\hat{i} + y\hat{j} + z\hat{k}$

This resulting vector is exactly the original vector $\vec{a}$.

Thus, for any non-zero vector $\vec{a}$, the expression $(\vec{a} \cdot \hat{i}) \hat{i} + (\vec{a} \cdot \hat{j}) \hat{j} + (\vec{a} \cdot \hat{k}) \hat{k}$ is equal to $\vec{a}$.

The blank should be filled with $\vec{a}$.

If $\vec{a}$ is any non - zero vector, then $\left( \vec{a} \;.\; \hat{i} \right) \hat{i} + \left( \vec{a}. \hat{j} \right) \hat{j} + \left( \vec{a}. \hat{k} \right) \hat{k}$ equals $\vec{a}$.

False

Explanation:

The statement "If $\left| \vec{a} \right| = \left| \vec{b} \right|$, then necessarily it implies $\vec{a} = ± \vec{b}$" is incorrect.

While it is true that if $\vec{a} = \vec{b}$, then $|\vec{a}| = |\vec{b}|$, and if $\vec{a} = -\vec{b}$, then $|\vec{a}| = |-\vec{b}| = |-1| |\vec{b}| = |\vec{b}|$, the converse is not necessarily true.

Having the same magnitude only means the vectors are of the same length. They can still have different directions.

For example, consider two vectors in a 2D plane:

Let $\vec{a} = 3\hat{i}$

The magnitude of $\vec{a}$ is $|\vec{a}| = \sqrt{3^2 + 0^2} = 3$.

Let $\vec{b} = 3\hat{j}$

The magnitude of $\vec{b}$ is $|\vec{b}| = \sqrt{0^2 + 3^2} = 3$.

Here, we have $|\vec{a}| = |\vec{b}| = 3$.

However, $\vec{a} = 3\hat{i}$ and $\vec{b} = 3\hat{j}$ are perpendicular vectors, so $\vec{a} \neq \vec{b}$.

Also, $-\vec{b} = -3\hat{j}$, and $\vec{a} = 3\hat{i} \neq -3\hat{j} = -\vec{b}$.

Since we found an example where $|\vec{a}| = |\vec{b}|$ but $\vec{a} \neq \pm \vec{b}$, the statement is false.

True

Explanation:

By definition, the position vector of a point P with respect to an origin O is the vector $\vec{OP}$. The initial point of this vector is the origin O, and the terminal point is the point P. Thus, a position vector always originates from the origin.

True

Explanation:

We are given the condition $\left| \vec{a}+ \vec{b} \right| = \left| \vec{a} − \vec{b} \right|$.

Let's square both sides of the equation:

Using the property that $|\vec{v}|^2 = \vec{v} \cdot \vec{v}$, we can rewrite the equation as:

Expand the dot products using the distributive property and the commutative property ($\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$):

Left side:

$(\vec{a}+ \vec{b}) \cdot (\vec{a}+ \vec{b}) = \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b}$

$= |\vec{a}|^2 + \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{b} + |\vec{b}|^2$

$= |\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b})$

Right side:

$(\vec{a} − \vec{b}) \cdot (\vec{a} − \vec{b}) = \vec{a} \cdot \vec{a} - \vec{a} \cdot \vec{b} - \vec{b} \cdot \vec{a} + (-\vec{b}) \cdot (-\vec{b})$

$= |\vec{a}|^2 - \vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{b} + |\vec{b}|^2$

$= |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b})$

Equating the expanded sides:

$|\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b})$

Subtract $|\vec{a}|^2 + |\vec{b}|^2$ from both sides:

$2(\vec{a} \cdot \vec{b}) = -2(\vec{a} \cdot \vec{b})$

Add $2(\vec{a} \cdot \vec{b})$ to both sides:

$2(\vec{a} \cdot \vec{b}) + 2(\vec{a} \cdot \vec{b}) = 0$

$4(\vec{a} \cdot \vec{b}) = 0$

Divide by 4:

The dot product of two vectors is zero if and only if the vectors are orthogonal (perpendicular). This is true even if one or both vectors are the zero vector (in which case they are considered orthogonal by convention). The condition $|\vec{a}+ \vec{b}| = |\vec{a} − \vec{b}|$ implies that the diagonals of the parallelogram formed by $\vec{a}$ and $\vec{b}$ have equal length. A parallelogram with equal diagonals is a rectangle, and the adjacent sides of a rectangle are orthogonal.

Therefore, if $\left| \vec{a}+ \vec{b} \right| = \left| \vec{a} − \vec{b} \right|$, then the vectors $\vec{a}$ and $\vec{b}$ are orthogonal.

False

Explanation:

The notation $\vec{v}^2$ for a vector $\vec{v}$ is commonly used to represent the dot product of the vector with itself, which is equal to the square of its magnitude: $\vec{v}^2 = \vec{v} \cdot \vec{v} = |\vec{v}|^2$.

Let's expand the left side of the given formula using the definition of the square of a vector and the properties of the dot product:

$\left( \vec{a} + \vec{b} \right)^2 = (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b})$

Using the distributive property of the dot product:

$= \vec{a} \cdot (\vec{a} + \vec{b}) + \vec{b} \cdot (\vec{a} + \vec{b})$

$= \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b}$

Using the property that the dot product is commutative ($\vec{b} \cdot \vec{a} = \vec{a} \cdot \vec{b}$) and $\vec{v} \cdot \vec{v} = |\vec{v}|^2 = \vec{v}^2$:

$= |\vec{a}|^2 + \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{b} + |\vec{b}|^2$

Or, using the $\vec{v}^2$ notation:

Now, compare this correct expansion to the given formula:

Given formula: $\left( \vec{a} + \vec{b} \right)^2 = \vec{a}^2 + \vec{b}^2 + 2\vec{a}× \vec{b}$

Correct expansion: $\left( \vec{a} + \vec{b} \right)^2 = \vec{a}^2 + \vec{b}^2 + 2(\vec{a} \cdot \vec{b})$

For the given formula to be valid, it must imply that:

$\vec{a}^2 + \vec{b}^2 + 2(\vec{a} \cdot \vec{b}) = \vec{a}^2 + \vec{b}^2 + 2\vec{a}× \vec{b}$

Subtracting $\vec{a}^2 + \vec{b}^2$ from both sides:

$2(\vec{a} \cdot \vec{b}) = 2(\vec{a} \times \vec{b})$

Dividing by 2:

$\vec{a} \cdot \vec{b} = \vec{a} \times \vec{b}$

The dot product $\vec{a} \cdot \vec{b}$ is a scalar quantity, while the cross product $\vec{a} \times \vec{b}$ is a vector quantity. A scalar quantity can only be equal to a vector quantity if both are the zero quantity (the scalar 0 and the zero vector $\vec{0}$).

So, this equality would require $\vec{a} \cdot \vec{b} = 0$ AND $\vec{a} \times \vec{b} = \vec{0}$.

For non-zero vectors $\vec{a}$ and $\vec{b}$:

$\vec{a} \cdot \vec{b} = 0$ implies $\vec{a}$ and $\vec{b}$ are orthogonal (perpendicular).

$\vec{a} \times \vec{b} = \vec{0}$ implies $\vec{a}$ and $\vec{b}$ are parallel (collinear).

Non-zero vectors cannot be simultaneously orthogonal and parallel. This only happens if one or both vectors are zero (which is excluded by the question stating "non-zero vectors").

Therefore, the equality $\vec{a} \cdot \vec{b} = \vec{a} \times \vec{b}$ is generally not true for non-zero vectors $\vec{a}$ and $\vec{b}$.

The correct formula for the square of the sum of two vectors is $(\vec{a} + \vec{b})^2 = \vec{a}^2 + \vec{b}^2 + 2(\vec{a} \cdot \vec{b})$.

The given formula incorrectly replaces the dot product term with a cross product term.

False

Explanation:

Let $\vec{a}$ and $\vec{b}$ be two adjacent sides of a rhombus, originating from the same vertex.

By definition, a rhombus is a quadrilateral with all four sides of equal length. Therefore, the magnitudes of the adjacent sides are equal:

The dot product of $\vec{a}$ and $\vec{b}$ is given by the formula:

$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$

where $\theta$ is the angle between the vectors $\vec{a}$ and $\vec{b}$ (the angle between the adjacent sides of the rhombus).

The statement claims that $\vec{a} \cdot \vec{b} = 0$. According to the dot product formula, $\vec{a} \cdot \vec{b} = 0$ if and only if $|\vec{a}| |\vec{b}| \cos \theta = 0$.

Since $\vec{a}$ and $\vec{b}$ are sides of a rhombus, they are typically non-zero vectors (assuming a non-degenerate rhombus), so $|\vec{a}| \neq 0$ and $|\vec{b}| \neq 0$.

Therefore, $\vec{a} \cdot \vec{b} = 0$ implies $\cos \theta = 0$.

For $\cos \theta = 0$, the angle $\theta$ must be $\frac{\pi}{2}$ radians or $90^\circ$ (assuming $0 \le \theta \le \pi$).

So, the condition $\vec{a} \cdot \vec{b} = 0$ implies that the angle between the adjacent sides of the rhombus is $90^\circ$. A rhombus with a $90^\circ$ angle between adjacent sides is a square.

Thus, the statement $\vec{a} \cdot \vec{b} = 0$ is true for a rhombus only if that rhombus is a square.

However, a rhombus does not necessarily have to be a square. For example, a rhombus can have angles of $60^\circ$ and $120^\circ$. In such a case, the angle between adjacent sides is $60^\circ$ or $120^\circ$, and $\cos 60^\circ = \frac{1}{2} \neq 0$, and $\cos 120^\circ = -\frac{1}{2} \neq 0$. Therefore, $\vec{a} \cdot \vec{b} \neq 0$ for such rhombuses.

Since the statement is not true for all rhombuses, it is false.