Chapter 11 Three Dimensional Geometry

Given:

A line makes angles with the positive directions of x, y, and z-axes as follows:

Angle with x-axis, $\alpha = 90^\circ$

Angle with y-axis, $\beta = 60^\circ$

Angle with z-axis, $\gamma = 30^\circ$

To Find:

The direction cosines of the line.

Solution:

The direction cosines of a line that makes angles $\alpha$, $\beta$, and $\gamma$ with the positive directions of the x, y, and z-axes respectively are given by $l = \cos\alpha$, $m = \cos\beta$, and $n = \cos\gamma$.

Using the given angles:

$l = \cos(90^\circ)$

$m = \cos(60^\circ)$

$n = \cos(30^\circ)$

The direction cosines of the line are $(l, m, n) = \left(0, \frac{1}{2}, \frac{\sqrt{3}}{2}\right)$.

We can verify this by checking if the sum of the squares of the direction cosines is equal to 1:

$l^2 + m^2 + n^2 = 0^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2$

$l^2 + m^2 + n^2 = 0 + \frac{1}{4} + \frac{3}{4}$

The sum of the squares is 1, which confirms that these are valid direction cosines.

The direction cosines of the line are $\left(0, \frac{1}{2}, \frac{\sqrt{3}}{2}\right)$.

Given:

The direction ratios of the line are 2, –1, –2.

To Find:

The direction cosines of the line.

Solution:

Let the direction ratios of the line be $a=2$, $b=-1$, and $c=-2$.

The direction cosines $l$, $m$, and $n$ are related to the direction ratios $a$, $b$, and $c$ by the formulas:

First, we calculate the magnitude $\sqrt{a^2+b^2+c^2}$.

$\sqrt{a^2+b^2+c^2} = \sqrt{(2)^2 + (-1)^2 + (-2)^2}$

$\sqrt{a^2+b^2+c^2} = \sqrt{4 + 1 + 4}$

$\sqrt{a^2+b^2+c^2} = \sqrt{9}$

Now, we find the direction cosines by dividing each direction ratio by this magnitude.

$l = \frac{2}{3}$

$m = \frac{-1}{3} = -\frac{1}{3}$

$n = \frac{-2}{3} = -\frac{2}{3}$

The direction cosines of the line are $\left(\frac{2}{3}, -\frac{1}{3}, -\frac{2}{3}\right)$.

We can verify the property that the sum of the squares of the direction cosines is 1:

$l^2+m^2+n^2 = \left(\frac{2}{3}\right)^2 + \left(-\frac{1}{3}\right)^2 + \left(-\frac{2}{3}\right)^2$

$l^2+m^2+n^2 = \frac{4}{9} + \frac{1}{9} + \frac{4}{9}$

The property holds true.

The direction cosines of the line are $\left(\frac{2}{3}, -\frac{1}{3}, -\frac{2}{3}\right)$.

Given:

Two points on the line are P(–2, 4, –5) and Q(1, 2, 3).

To Find:

The direction cosines of the line passing through these two points.

Solution:

Let the two points be $P(x_1, y_1, z_1) = (-2, 4, -5)$ and $Q(x_2, y_2, z_2) = (1, 2, 3)$.

The direction ratios of the line passing through points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are given by $a = x_2 - x_1$, $b = y_2 - y_1$, and $c = z_2 - z_1$.

$a = 1 - (-2) = 1 + 2 = 3$

$b = 2 - 4 = -2$

$c = 3 - (-5) = 3 + 5 = 8$

The direction ratios of the line are 3, -2, 8.

To find the direction cosines $l, m, n$ from the direction ratios $a, b, c$, we use the formulas:

First, calculate the magnitude $\sqrt{a^2+b^2+c^2}$.

$\sqrt{a^2+b^2+c^2} = \sqrt{(3)^2 + (-2)^2 + (8)^2}$

$\sqrt{a^2+b^2+c^2} = \sqrt{9 + 4 + 64}$

$\sqrt{a^2+b^2+c^2} = \sqrt{77}$

Now, find the direction cosines:

$l = \frac{3}{\sqrt{77}}$

$m = \frac{-2}{\sqrt{77}} = -\frac{2}{\sqrt{77}}$

$n = \frac{8}{\sqrt{77}}$

The direction cosines of the line passing through the given points are $\left(\frac{3}{\sqrt{77}}, -\frac{2}{\sqrt{77}}, \frac{8}{\sqrt{77}}\right)$.

Note that the direction ratios and direction cosines can also be calculated in the reverse order (from Q to P), which would give the opposite signs for the direction ratios and direction cosines, representing the line in the opposite direction. The set of direction ratios would be (-3, 2, -8) and the direction cosines would be $\left(-\frac{3}{\sqrt{77}}, \frac{2}{\sqrt{77}}, -\frac{8}{\sqrt{77}}\right)$. Both sets are valid for representing the direction of the line.

The direction cosines of the line are $\left(\frac{3}{\sqrt{77}}, -\frac{2}{\sqrt{77}}, \frac{8}{\sqrt{77}}\right)$ or $\left(-\frac{3}{\sqrt{77}}, \frac{2}{\sqrt{77}}, -\frac{8}{\sqrt{77}}\right)$.

To Find:

The direction cosines of the x-axis, y-axis, and z-axis.

Solution:

The direction cosines of a line are the cosines of the angles that the line makes with the positive directions of the x, y, and z-axes.

Direction Cosines of the x-axis:

The x-axis makes an angle of $0^\circ$ with the positive x-axis, $90^\circ$ with the positive y-axis, and $90^\circ$ with the positive z-axis.

Angle with x-axis, $\alpha = 0^\circ$

Angle with y-axis, $\beta = 90^\circ$

Angle with z-axis, $\gamma = 90^\circ$

The direction cosines are:

$l = \cos\alpha = \cos(0^\circ) = 1$

$m = \cos\beta = \cos(90^\circ) = 0$

$n = \cos\gamma = \cos(90^\circ) = 0$

The direction cosines of the x-axis are $(1, 0, 0)$.

Direction Cosines of the y-axis:

The y-axis makes an angle of $90^\circ$ with the positive x-axis, $0^\circ$ with the positive y-axis, and $90^\circ$ with the positive z-axis.

Angle with x-axis, $\alpha = 90^\circ$

Angle with y-axis, $\beta = 0^\circ$

Angle with z-axis, $\gamma = 90^\circ$

The direction cosines are:

$l = \cos\alpha = \cos(90^\circ) = 0$

$m = \cos\beta = \cos(0^\circ) = 1$

$n = \cos\gamma = \cos(90^\circ) = 0$

The direction cosines of the y-axis are $(0, 1, 0)$.

Direction Cosines of the z-axis:

The z-axis makes an angle of $90^\circ$ with the positive x-axis, $90^\circ$ with the positive y-axis, and $0^\circ$ with the positive z-axis.

Angle with x-axis, $\alpha = 90^\circ$

Angle with y-axis, $\beta = 90^\circ$

Angle with z-axis, $\gamma = 0^\circ$

The direction cosines are:

$l = \cos\alpha = \cos(90^\circ) = 0$

$m = \cos\beta = \cos(90^\circ) = 0$

$n = \cos\gamma = \cos(0^\circ) = 1$

The direction cosines of the z-axis are $(0, 0, 1)$.

Summary of direction cosines:

• Direction cosines of the x-axis: (1, 0, 0)
• Direction cosines of the y-axis: (0, 1, 0)
• Direction cosines of the z-axis: (0, 0, 1)

Given:

The points are A(2, 3, – 4), B(1, – 2, 3), and C(3, 8, – 11).

To Show:

The points A, B, and C are collinear.

Solution (Method 1: Using Direction Ratios):

To show that the three points A, B, and C are collinear, we can show that the direction ratios of the line segment AB are proportional to the direction ratios of the line segment BC.

The direction ratios of a line segment joining points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are given by $(x_2-x_1, y_2-y_1, z_2-z_1)$.

For the line segment AB, with A(2, 3, -4) and B(1, -2, 3):

Direction ratios of AB = $(1 - 2, -2 - 3, 3 - (-4))$

Direction ratios of AB = $(-1, -5, 7)$

For the line segment BC, with B(1, -2, 3) and C(3, 8, -11):

Direction ratios of BC = $(3 - 1, 8 - (-2), -11 - 3)$

Direction ratios of BC = $(2, 10, -14)$

Now, we check if the direction ratios of AB and BC are proportional. This means we check if there is a constant $k$ such that the direction ratios of BC are $k$ times the direction ratios of AB.

Since $\frac{2}{-1} = \frac{10}{-5} = \frac{-14}{7} = -2$, the direction ratios of AB and BC are proportional.

This indicates that the vector $\vec{AB}$ is parallel to the vector $\vec{BC}$. Since point B is common to both vectors (or line segments), the points A, B, and C must lie on the same line.

Therefore, the points A, B, and C are collinear.

Alternate Solution (Method 2: Using Distance Formula):

To show that three points are collinear using the distance formula, we need to show that the sum of the distances between two pairs of points is equal to the distance between the third pair.

The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.

Calculate the distance AB:

$AB = \sqrt{(1-2)^2 + (-2-3)^2 + (3-(-4))^2}$

$AB = \sqrt{(-1)^2 + (-5)^2 + (7)^2}$

$AB = \sqrt{1 + 25 + 49} = \sqrt{75} = \sqrt{25 \times 3}$

Calculate the distance BC:

$BC = \sqrt{(3-1)^2 + (8-(-2))^2 + (-11-3)^2}$

$BC = \sqrt{(2)^2 + (10)^2 + (-14)^2}$

$BC = \sqrt{4 + 100 + 196} = \sqrt{300} = \sqrt{100 \times 3}$

Calculate the distance AC:

$AC = \sqrt{(3-2)^2 + (8-3)^2 + (-11-(-4))^2}$

$AC = \sqrt{(1)^2 + (5)^2 + (-7)^2}$

$AC = \sqrt{1 + 25 + 49} = \sqrt{75} = \sqrt{25 \times 3}$

Now, check if the sum of two distances equals the third distance:

$AB + AC = 5\sqrt{3} + 5\sqrt{3} = 10\sqrt{3}$

We see that $AB + AC = BC$.

Since the sum of the lengths of two segments equals the length of the third segment, the three points A, B, and C must lie on the same line, with point B lying between A and C.

Therefore, the points A, B, and C are collinear.

Given:

A line makes angles with the x, y, and z-axes as follows:

Angle with x-axis, $\alpha = 90^\circ$

Angle with y-axis, $\beta = 135^\circ$

Angle with z-axis, $\gamma = 45^\circ$

To Find:

The direction cosines of the line.

Solution:

The direction cosines of a line are given by the cosines of the angles it makes with the positive directions of the x, y, and z-axes. Let the direction cosines be $l$, $m$, and $n$.

$l = \cos\alpha$

$m = \cos\beta$

$n = \cos\gamma$

Substitute the given angle values:

$l = \cos(90^\circ)$

$m = \cos(135^\circ) = \cos(180^\circ - 45^\circ) = -\cos(45^\circ)$

$n = \cos(45^\circ)$

The direction cosines of the line are $(l, m, n) = \left(0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$.

We can verify this result by checking if the sum of the squares of the direction cosines is 1:

$l^2 + m^2 + n^2 = (0)^2 + \left(-\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2$

$l^2 + m^2 + n^2 = 0 + \frac{1}{2} + \frac{1}{2}$

The condition $l^2 + m^2 + n^2 = 1$ is satisfied, confirming our direction cosines are correct.

The direction cosines of the line are $\left(0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$.

Given:

A line makes equal angles with the coordinate axes (x, y, and z).

To Find:

The direction cosines of the line.

Solution:

Let the angles that the line makes with the positive directions of the x-axis, y-axis, and z-axis be $\alpha$, $\beta$, and $\gamma$ respectively.

Since the line makes equal angles with the coordinate axes, we have:

Let the common angle be $\theta$. So, $\alpha = \beta = \gamma = \theta$.

The direction cosines of the line are $l = \cos\alpha$, $m = \cos\beta$, and $n = \cos\gamma$.

Substituting the equal angles, we get:

Thus, the direction cosines are equal, i.e., $l = m = n$.

We know that the sum of the squares of the direction cosines of any line is equal to 1. That is:

Substitute $l=m=n$ into equation (i):

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

$3(\cos\theta)^2 = 1$

$3\cos^2\theta = 1$

Divide by 3:

Taking the square root of both sides:

Since $l = m = n = \cos\theta$, the direction cosines are:

$l = \pm\frac{1}{\sqrt{3}}$

$m = \pm\frac{1}{\sqrt{3}}$

$n = \pm\frac{1}{\sqrt{3}}$

As the angles $\alpha, \beta, \gamma$ are equal, their cosines must have the same sign. Therefore, the direction cosines are either all positive or all negative.

The possible sets of direction cosines are $\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$ and $\left(-\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right)$.

This can be represented concisely as $\pm \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$.

The direction cosines of a line which makes equal angles with the coordinate axes are $\pm \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$.

Given:

The direction ratios of a line are –18, 12, – 4.

To Find:

The direction cosines of the line.

Solution:

Let the direction ratios of the line be $a = -18$, $b = 12$, and $c = -4$.

To find the direction cosines $l$, $m$, and $n$, we need to normalize the direction ratios by dividing them by the magnitude of the vector formed by these ratios. The magnitude is given by $\sqrt{a^2+b^2+c^2}$.

Calculate the magnitude:

$\sqrt{a^2+b^2+c^2} = \sqrt{(-18)^2 + (12)^2 + (-4)^2}$

$\sqrt{a^2+b^2+c^2} = \sqrt{324 + 144 + 16}$

$\sqrt{a^2+b^2+c^2} = \sqrt{484}$

So, $\sqrt{a^2+b^2+c^2} = 22$.

Now, we find the direction cosines by dividing each direction ratio by the magnitude (22):

$l = \frac{a}{\sqrt{a^2+b^2+c^2}} = \frac{-18}{22}$

$l = -\frac{\cancel{18}^{9}}{\cancel{22}_{11}} = -\frac{9}{11}$

$m = \frac{b}{\sqrt{a^2+b^2+c^2}} = \frac{12}{22}$

$m = \frac{\cancel{12}^{6}}{\cancel{22}_{11}} = \frac{6}{11}$

$n = \frac{c}{\sqrt{a^2+b^2+c^2}} = \frac{-4}{22}$

$n = -\frac{\cancel{4}^{2}}{\cancel{22}_{11}} = -\frac{2}{11}$

The direction cosines of the line are $\left(-\frac{9}{11}, \frac{6}{11}, -\frac{2}{11}\right)$.

We can verify the sum of the squares of the direction cosines:

$l^2+m^2+n^2 = \left(-\frac{9}{11}\right)^2 + \left(\frac{6}{11}\right)^2 + \left(-\frac{2}{11}\right)^2$

$l^2+m^2+n^2 = \frac{81}{121} + \frac{36}{121} + \frac{4}{121}$

The property $l^2+m^2+n^2=1$ is satisfied.

The direction cosines of the line are $\left(-\frac{9}{11}, \frac{6}{11}, -\frac{2}{11}\right)$.

Given:

The points are A(2, 3, 4), B(– 1, – 2, 1), and C(5, 8, 7).

To Show:

The points A, B, and C are collinear.

Solution (Method 1: Using Direction Ratios):

To show that the three points A, B, and C are collinear, we can demonstrate that the direction ratios of the line segment AB are proportional to the direction ratios of the line segment BC.

The direction ratios of the line segment joining points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are given by $(x_2-x_1, y_2-y_1, z_2-z_1)$.

For the line segment AB, with A(2, 3, 4) and B(-1, -2, 1):

Direction ratios of AB = $(-1 - 2, -2 - 3, 1 - 4)$

Direction ratios of AB = $(-3, -5, -3)$

For the line segment BC, with B(-1, -2, 1) and C(5, 8, 7):

Direction ratios of BC = $(5 - (-1), 8 - (-2), 7 - 1)$

Direction ratios of BC = $(5 + 1, 8 + 2, 6)$

Direction ratios of BC = $(6, 10, 6)$

Now, we check for proportionality between the direction ratios of AB and BC. We look for a constant $k$ such that $(6, 10, 6) = k(-3, -5, -3)$.

Since the ratios are equal ($\frac{6}{-3} = \frac{10}{-5} = \frac{6}{-3} = -2$), the direction ratios of AB and BC are proportional.

This implies that the line segment AB is parallel to the line segment BC. As point B is common to both segments, the points A, B, and C must lie on the same straight line.

Therefore, the points A, B, and C are collinear.

Alternate Solution (Method 2: Using Distance Formula):

To show that three points are collinear using the distance formula, we can check if the sum of the distances between two pairs of points is equal to the distance between the third pair.

The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.

Calculate the distance AB:

$AB = \sqrt{(-1-2)^2 + (-2-3)^2 + (1-4)^2}$

$AB = \sqrt{(-3)^2 + (-5)^2 + (-3)^2}$

$AB = \sqrt{9 + 25 + 9} = \sqrt{43}$

Calculate the distance BC:

$BC = \sqrt{(5-(-1))^2 + (8-(-2))^2 + (7-1)^2}$

$BC = \sqrt{(6)^2 + (10)^2 + (6)^2}$

$BC = \sqrt{36 + 100 + 36} = \sqrt{172} = \sqrt{4 \times 43}$

Calculate the distance AC:

$AC = \sqrt{(5-2)^2 + (8-3)^2 + (7-4)^2}$

$AC = \sqrt{(3)^2 + (5)^2 + (3)^2}$

$AC = \sqrt{9 + 25 + 9} = \sqrt{43}$

Now, check the sums of distances:

$AB + AC = \sqrt{43} + \sqrt{43} = 2\sqrt{43}$

We observe that $AB + AC = BC$.

Since the sum of the lengths of two segments (AB and AC) equals the length of the third segment (BC), the points A, B, and C lie on the same line, with B lying between A and C is incorrect based on calculation; rather, A and C are on either side of B, and $AB + BC = AC$ or $AC + CB = AB$ or $BA+AC = BC$. Our calculation shows $AB+AC=BC$ is incorrect. Let's recheck.

Distances are $AB = \sqrt{43}$, $BC = 2\sqrt{43}$, $AC = \sqrt{43}$.

Let's see if $AB + BC = AC$ or $AC + CB = AB$ or $BA + AC = BC$.

$AB + BC = \sqrt{43} + 2\sqrt{43} = 3\sqrt{43}$. This is not equal to $AC = \sqrt{43}$.

$AC + BC = \sqrt{43} + 2\sqrt{43} = 3\sqrt{43}$. This is not equal to $AB = \sqrt{43}$.

Let's check $AB + AC = BC$. This is $\sqrt{43} + \sqrt{43} = 2\sqrt{43}$, which is indeed equal to $BC$.

The order of points is A, C, B or B, C, A or A, B, C etc. The condition for collinearity using distances is that the sum of the lengths of the two shorter segments equals the length of the longest segment.

Here, the lengths are $\sqrt{43}$, $2\sqrt{43}$, $\sqrt{43}$. The two shorter lengths are $AB = \sqrt{43}$ and $AC = \sqrt{43}$. The longest length is $BC = 2\sqrt{43}$.

Sum of the two shorter lengths $= AB + AC = \sqrt{43} + \sqrt{43} = 2\sqrt{43}$.

This sum is equal to the longest length $BC = 2\sqrt{43}$.

Since the sum of the distances AB and AC is equal to the distance BC, the points A, C, and B are collinear, with point C lying between A and B.

Therefore, the points A, B, and C are collinear.

Given:

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

To Find:

The direction cosines of the sides of the triangle (AB, BC, and CA).

Solution:

To find the direction cosines of each side, we first find the direction ratios of the line segment forming the side and then normalize them.

The direction ratios of a line segment joining points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are $(x_2-x_1, y_2-y_1, z_2-z_1)$. The direction cosines are $\left(\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}}\right)$, where $(a, b, c)$ are the direction ratios.

Side AB:

Points are A(3, 5, -4) and B(-1, 1, 2).

Direction ratios of AB $(a_1, b_1, c_1) = (-1 - 3, 1 - 5, 2 - (-4))$

$(a_1, b_1, c_1) = (-4, -4, 6)$

Magnitude of the direction ratios: $\sqrt{(-4)^2 + (-4)^2 + (6)^2} = \sqrt{16 + 16 + 36} = \sqrt{68}$

$\sqrt{68} = \sqrt{4 \times 17} = 2\sqrt{17}$

Direction cosines of AB:

$l_{AB} = \frac{-4}{2\sqrt{17}} = -\frac{2}{\sqrt{17}}$

$m_{AB} = \frac{-4}{2\sqrt{17}} = -\frac{2}{\sqrt{17}}$

$n_{AB} = \frac{6}{2\sqrt{17}} = \frac{3}{\sqrt{17}}$

Direction cosines of side AB are $\left(-\frac{2}{\sqrt{17}}, -\frac{2}{\sqrt{17}}, \frac{3}{\sqrt{17}}\right)$.

Side BC:

Points are B(-1, 1, 2) and C(-5, -5, -2).

Direction ratios of BC $(a_2, b_2, c_2) = (-5 - (-1), -5 - 1, -2 - 2)$

$(a_2, b_2, c_2) = (-5 + 1, -6, -4)$

$(a_2, b_2, c_2) = (-4, -6, -4)$

Magnitude of the direction ratios: $\sqrt{(-4)^2 + (-6)^2 + (-4)^2} = \sqrt{16 + 36 + 16} = \sqrt{68}$

$\sqrt{68} = 2\sqrt{17}$

Direction cosines of BC:

$l_{BC} = \frac{-4}{2\sqrt{17}} = -\frac{2}{\sqrt{17}}$

$m_{BC} = \frac{-6}{2\sqrt{17}} = -\frac{3}{\sqrt{17}}$

$n_{BC} = \frac{-4}{2\sqrt{17}} = -\frac{2}{\sqrt{17}}$

Direction cosines of side BC are $\left(-\frac{2}{\sqrt{17}}, -\frac{3}{\sqrt{17}}, -\frac{2}{\sqrt{17}}\right)$.

Side CA:

Points are C(-5, -5, -2) and A(3, 5, -4).

Direction ratios of CA $(a_3, b_3, c_3) = (3 - (-5), 5 - (-5), -4 - (-2))$

$(a_3, b_3, c_3) = (3 + 5, 5 + 5, -4 + 2)$

$(a_3, b_3, c_3) = (8, 10, -2)$

Magnitude of the direction ratios: $\sqrt{(8)^2 + (10)^2 + (-2)^2} = \sqrt{64 + 100 + 4} = \sqrt{168}$

$\sqrt{168} = \sqrt{4 \times 42} = 2\sqrt{42}$

Direction cosines of CA:

$l_{CA} = \frac{8}{2\sqrt{42}} = \frac{4}{\sqrt{42}}$

$m_{CA} = \frac{10}{2\sqrt{42}} = \frac{5}{\sqrt{42}}$

$n_{CA} = \frac{-2}{2\sqrt{42}} = -\frac{1}{\sqrt{42}}$

Direction cosines of side CA are $\left(\frac{4}{\sqrt{42}}, \frac{5}{\sqrt{42}}, -\frac{1}{\sqrt{42}}\right)$.

The direction cosines of the sides of the triangle are:

• For side AB: $\left(-\frac{2}{\sqrt{17}}, -\frac{2}{\sqrt{17}}, \frac{3}{\sqrt{17}}\right)$
• For side BC: $\left(-\frac{2}{\sqrt{17}}, -\frac{3}{\sqrt{17}}, -\frac{2}{\sqrt{17}}\right)$
• For side CA: $\left(\frac{4}{\sqrt{42}}, \frac{5}{\sqrt{42}}, -\frac{1}{\sqrt{42}}\right)$

Note: The direction cosines of the side BA, CB, or AC would have opposite signs but represent the same lines.

Given:

The line passes through the point (5, 2, – 4).

The line is parallel to the vector $3\hat{i} + 2\hat{j} − 8\hat{k}$.

To Find:

The vector equation and the Cartesian equations of the line.

Solution:

Let the given point be $A(5, 2, -4)$. The position vector of this point is $\vec{a} = 5\hat{i} + 2\hat{j} - 4\hat{k}$.

The line is parallel to the given vector. Let this parallel vector be $\vec{b} = 3\hat{i} + 2\hat{j} - 8\hat{k}$.

The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{b}$ is given by:

Substitute the given values of $\vec{a}$ and $\vec{b}$:

This is the vector equation of the line.

To find the Cartesian equations, let $\vec{r}$ be the position vector of any point $(x, y, z)$ on the line. So, $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.

Substitute this into the vector equation:

$x\hat{i} + y\hat{j} + z\hat{k} = (5\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda (3\hat{i} + 2\hat{j} - 8\hat{k})$

$x\hat{i} + y\hat{j} + z\hat{k} = (5 + 3\lambda)\hat{i} + (2 + 2\lambda)\hat{j} + (-4 - 8\lambda)\hat{k}$

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

From each of these equations, we can express $\lambda$:

From (1): $x - 5 = 3\lambda \implies \lambda = \frac{x - 5}{3}$

From (2): $y - 2 = 2\lambda \implies \lambda = \frac{y - 2}{2}$

From (3): $z + 4 = -8\lambda \implies \lambda = \frac{z + 4}{-8}$

Since all these expressions are equal to $\lambda$, we can equate them:

These are the Cartesian equations of the line.

Note that the denominators (3, 2, -8) are the direction ratios of the line, which are the components of the parallel vector $\vec{b}$. The numerators $(x-5, y-2, z-(-4))$ come from subtracting the coordinates of the point (5, 2, -4) from $(x, y, z)$.

The vector equation of the line is $\vec{r} = (5\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda (3\hat{i} + 2\hat{j} - 8\hat{k})$.

The Cartesian equations of the line are $\frac{x - 5}{3} = \frac{y - 2}{2} = \frac{z + 4}{-8}$.

Given:

The equations of two lines are:

[where $\lambda$ and $\mu$ are scalar parameters]

To Find:

The angle between the pair of lines.

Solution:

The angle between two lines is the angle between their direction vectors.

The vector equation of a line is of the form $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\vec{b}$ is a vector parallel to the line (direction vector).

For Line 1, the direction vector is $\vec{b_1} = \hat{i} + 2\hat{j} + 2\hat{k}$.

For Line 2, the direction vector is $\vec{b_2} = 3\hat{i} + 2\hat{j} + 6\hat{k}$.

Let $\theta$ be the angle between the lines. The angle between the lines is the angle between their direction vectors $\vec{b_1}$ and $\vec{b_2}$.

The cosine of the angle between two vectors $\vec{u}$ and $\vec{v}$ is given by:

In our case, $\vec{u} = \vec{b_1}$ and $\vec{v} = \vec{b_2}$.

Calculate the dot product $\vec{b_1} \cdot \vec{b_2}$:

$\vec{b_1} \cdot \vec{b_2} = (\hat{i} + 2\hat{j} + 2\hat{k}) \cdot (3\hat{i} + 2\hat{j} + 6\hat{k})$

$\vec{b_1} \cdot \vec{b_2} = (1)(3) + (2)(2) + (2)(6)$

$\vec{b_1} \cdot \vec{b_2} = 3 + 4 + 12 = 19$

Calculate the magnitudes of $\vec{b_1}$ and $\vec{b_2}$:

$|\vec{b_1}| = \sqrt{(1)^2 + (2)^2 + (2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9}$

$|\vec{b_2}| = \sqrt{(3)^2 + (2)^2 + (6)^2} = \sqrt{9 + 4 + 36} = \sqrt{49}$

Now, substitute these values into the formula for $\cos\theta$:

To find the angle $\theta$, we take the inverse cosine of $\frac{19}{21}$.

The angle between the pair of lines is the acute angle, which is given by $\cos^{-1}\left(\frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| |\vec{b_2}|}\right)$. Since $\vec{b_1} \cdot \vec{b_2} = 19 > 0$, the angle is acute and the absolute value is not strictly necessary here, but it's good practice to include it in the general formula for the angle between lines.

The angle between the pair of lines is $\cos^{-1}\left(\frac{19}{21}\right)$.

Given:

The Cartesian equations of two lines are:

To Find:

The angle between the pair of lines.

Solution:

The Cartesian equation of a line is of the form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$, where $(a, b, c)$ are the direction ratios of the line.

For Line 1, the direction ratios are $(a_1, b_1, c_1) = (3, 5, 4)$.

For Line 2, the direction ratios are $(a_2, b_2, c_2) = (1, 1, 2)$.

The direction vector for Line 1 can be taken as $\vec{b_1} = 3\hat{i} + 5\hat{j} + 4\hat{k}$.

The direction vector for Line 2 can be taken as $\vec{b_2} = \hat{i} + \hat{j} + 2\hat{k}$.

Let $\theta$ be the angle between the lines. The angle between the lines is the angle between their direction vectors $\vec{b_1}$ and $\vec{b_2}$.

The cosine of the angle $\theta$ between two vectors $\vec{u}$ and $\vec{v}$ is given by:

In our case, $\vec{u} = \vec{b_1}$ and $\vec{v} = \vec{b_2}$.

Calculate the dot product $\vec{b_1} \cdot \vec{b_2}$:

$\vec{b_1} \cdot \vec{b_2} = (3\hat{i} + 5\hat{j} + 4\hat{k}) \cdot (\hat{i} + \hat{j} + 2\hat{k})$

$\vec{b_1} \cdot \vec{b_2} = (3)(1) + (5)(1) + (4)(2)$

$\vec{b_1} \cdot \vec{b_2} = 3 + 5 + 8 = 16$

Calculate the magnitudes of $\vec{b_1}$ and $\vec{b_2}$:

$|\vec{b_1}| = \sqrt{(3)^2 + (5)^2 + (4)^2} = \sqrt{9 + 25 + 16} = \sqrt{50}$

$\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$

$|\vec{b_2}| = \sqrt{(1)^2 + (1)^2 + (2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$

Now, substitute these values into the formula for $\cos\theta$:

$\cos\theta = \frac{16}{5\sqrt{12}}$

Simplify the denominator: $\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$.

$\cos\theta = \frac{16}{5(2\sqrt{3})}$

$\cos\theta = \frac{16}{10\sqrt{3}}$

Simplify the fraction:

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{3}$:

$\cos\theta = \frac{8\sqrt{3}}{5\sqrt{3}\sqrt{3}} = \frac{8\sqrt{3}}{5(3)}$

The angle $\theta$ is given by $\cos^{-1}\left(\frac{8\sqrt{3}}{15}\right)$.

The angle between the pair of lines is $\cos^{-1}\left(\frac{8\sqrt{3}}{15}\right)$.

Given:

The vector equations of two lines $l_1$ and $l_2$ are:

and

[where $\lambda$ and $\mu$ are scalar parameters]

To Find:

The shortest distance between the lines $l_1$ and $l_2$.

Solution:

The vector equation of a line is of the form $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\vec{a}$ is the position vector of a point on the line and $\vec{b}$ is a vector parallel to the line.

For line $l_1$:

For line $l_2$:

Since $\vec{b_1}$ is not a scalar multiple of $\vec{b_2}$ (check ratios: $2/3 \neq -1/-5 \neq 1/2$), the lines are not parallel. Thus, they are skew lines.

The shortest distance between two skew lines $\vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu \vec{b_2}$ is given by the formula:

First, calculate $\vec{a_2} - \vec{a_1}$:

$\vec{a_2} - \vec{a_1} = (2\hat{i} + \hat{j} - \hat{k}) - (\hat{i} + \hat{j})$

$\vec{a_2} - \vec{a_1} = (2-1)\hat{i} + (1-1)\hat{j} + (-1-0)\hat{k}$

Next, calculate the cross product $\vec{b_1} \times \vec{b_2}$:

$\vec{b_1} \times \vec{b_2} = \hat{i}[(-1)(2) - (1)(-5)] - \hat{j}[(2)(2) - (1)(3)] + \hat{k}[(2)(-5) - (-1)(3)]$

$\vec{b_1} \times \vec{b_2} = \hat{i}[-2 + 5] - \hat{j}[4 - 3] + \hat{k}[-10 + 3]$

Now, calculate the magnitude of the cross product $|\vec{b_1} \times \vec{b_2}|$:

$|\vec{b_1} \times \vec{b_2}| = \sqrt{(3)^2 + (-1)^2 + (-7)^2}$

$|\vec{b_1} \times \vec{b_2}| = \sqrt{9 + 1 + 49}$

Next, calculate the dot product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$:

$(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (\hat{i} - \hat{k}) \cdot (3\hat{i} - \hat{j} - 7\hat{k})$

$(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (1)(3) + (0)(-1) + (-1)(-7)$

$(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = 3 + 0 + 7$

Finally, substitute the values into the shortest distance formula (i):

$d = \frac{|10|}{|\sqrt{59}|}$

To rationalize the denominator, we can multiply the numerator and denominator by $\sqrt{59}$:

$d = \frac{10 \times \sqrt{59}}{\sqrt{59} \times \sqrt{59}}$

The shortest distance between the two lines is $\frac{10}{\sqrt{59}}$ units or $\frac{10\sqrt{59}}{59}$ units.

Given:

The vector equations of two lines $l_1$ and $l_2$ are:

and

[where $\lambda$ and $\mu$ are scalar parameters]

To Find:

The distance between the lines $l_1$ and $l_2$.

Solution:

The vector equation of a line is of the form $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\vec{a}$ is the position vector of a point on the line and $\vec{b}$ is a vector parallel to the line.

For line $l_1$:

For line $l_2$:

We observe that $\vec{b_1} = \vec{b_2} = 2\hat{i} + 3\hat{j} + 6\hat{k}$. This means the two lines are parallel.

The distance between two parallel lines $\vec{r} = \vec{a_1} + \lambda \vec{b}$ and $\vec{r} = \vec{a_2} + \mu \vec{b}$ is given by the formula:

Here, $\vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}$.

First, calculate $\vec{a_2} - \vec{a_1}$:

$\vec{a_2} - \vec{a_1} = (3\hat{i} + 3\hat{j} - 5\hat{k}) - (\hat{i} + 2\hat{j} - 4\hat{k})$

$\vec{a_2} - \vec{a_1} = (3-1)\hat{i} + (3-2)\hat{j} + (-5-(-4))\hat{k}$

$\vec{a_2} - \vec{a_1} = 2\hat{i} + \hat{j} + (-5+4)\hat{k}$

Next, calculate the cross product $(\vec{a_2} - \vec{a_1}) \times \vec{b}$:

$(\vec{a_2} - \vec{a_1}) \times \vec{b} = \hat{i}[(1)(6) - (-1)(3)] - \hat{j}[(2)(6) - (-1)(2)] + \hat{k}[(2)(3) - (1)(2)]$

$(\vec{a_2} - \vec{a_1}) \times \vec{b} = \hat{i}[6 + 3] - \hat{j}[12 + 2] + \hat{k}[6 - 2]$

Now, calculate the magnitude of this cross product:

$|(\vec{a_2} - \vec{a_1}) \times \vec{b}| = \sqrt{(9)^2 + (-14)^2 + (4)^2}$

$|(\vec{a_2} - \vec{a_1}) \times \vec{b}| = \sqrt{81 + 196 + 16}$

Calculate the magnitude of the direction vector $\vec{b}$:

$|\vec{b}| = \sqrt{(2)^2 + (3)^2 + (6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49}$

Finally, substitute the values into the distance formula (i):

$d = \frac{|\sqrt{293}|}{|7|}$

The distance between the two parallel lines is $\frac{\sqrt{293}}{7}$ units.

Given:

The direction cosines of three lines are:

Line 1: $(l_1, m_1, n_1) = \left(\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}\right)$

Line 2: $(l_2, m_2, n_2) = \left(\frac{4}{13}, \frac{12}{13}, \frac{3}{13}\right)$

Line 3: $(l_3, m_3, n_3) = \left(\frac{3}{13}, \frac{-4}{13}, \frac{12}{13}\right)$

To Show:

The three lines are mutually perpendicular.

Solution:

Two lines with direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ are perpendicular if and only if $l_1l_2 + m_1m_2 + n_1n_2 = 0$.

We need to check this condition for each pair of lines.

Check Perpendicularity between Line 1 and Line 2:

We need to check if $l_1l_2 + m_1m_2 + n_1n_2 = 0$ for $(l_1, m_1, n_1) = \left(\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}\right)$ and $(l_2, m_2, n_2) = \left(\frac{4}{13}, \frac{12}{13}, \frac{3}{13}\right)$.

$l_1l_2 + m_1m_2 + n_1n_2 = \left(\frac{12}{13}\right)\left(\frac{4}{13}\right) + \left(\frac{-3}{13}\right)\left(\frac{12}{13}\right) + \left(\frac{-4}{13}\right)\left(\frac{3}{13}\right)$

$= \frac{48}{169} + \frac{-36}{169} + \frac{-12}{169}$

$= \frac{48 - 36 - 12}{169} = \frac{48 - 48}{169} = \frac{0}{169}$

Since the dot product of their direction cosines is 0, Line 1 is perpendicular to Line 2.

Check Perpendicularity between Line 2 and Line 3:

We need to check if $l_2l_3 + m_2m_3 + n_2n_3 = 0$ for $(l_2, m_2, n_2) = \left(\frac{4}{13}, \frac{12}{13}, \frac{3}{13}\right)$ and $(l_3, m_3, n_3) = \left(\frac{3}{13}, \frac{-4}{13}, \frac{12}{13}\right)$.

$l_2l_3 + m_2m_3 + n_2n_3 = \left(\frac{4}{13}\right)\left(\frac{3}{13}\right) + \left(\frac{12}{13}\right)\left(\frac{-4}{13}\right) + \left(\frac{3}{13}\right)\left(\frac{12}{13}\right)$

$= \frac{12}{169} + \frac{-48}{169} + \frac{36}{169}$

$= \frac{12 - 48 + 36}{169} = \frac{48 - 48}{169} = \frac{0}{169}$

Since the dot product of their direction cosines is 0, Line 2 is perpendicular to Line 3.

Check Perpendicularity between Line 3 and Line 1:

We need to check if $l_3l_1 + m_3m_1 + n_3n_1 = 0$ for $(l_3, m_3, n_3) = \left(\frac{3}{13}, \frac{-4}{13}, \frac{12}{13}\right)$ and $(l_1, m_1, n_1) = \left(\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}\right)$.

$l_3l_1 + m_3m_1 + n_3n_1 = \left(\frac{3}{13}\right)\left(\frac{12}{13}\right) + \left(\frac{-4}{13}\right)\left(\frac{-3}{13}\right) + \left(\frac{12}{13}\right)\left(\frac{-4}{13}\right)$

$= \frac{36}{169} + \frac{12}{169} + \frac{-48}{169}$

$= \frac{36 + 12 - 48}{169} = \frac{48 - 48}{169} = \frac{0}{169}$

Since the dot product of their direction cosines is 0, Line 3 is perpendicular to Line 1.

Since every pair of the three lines is perpendicular, the three lines are mutually perpendicular.

Given:

Line $L_1$ passes through points A(1, – 1, 2) and B(3, 4, – 2).

Line $L_2$ passes through points C(0, 3, 2) and D(3, 5, 6).

To Show:

Line $L_1$ is perpendicular to Line $L_2$.

Solution:

Two lines are perpendicular if the dot product of their direction ratios is zero.

First, find the direction ratios of Line $L_1$ passing through A(1, -1, 2) and B(3, 4, -2).

Direction ratios of $L_1$, $(a_1, b_1, c_1) = (x_2 - x_1, y_2 - y_1, z_2 - z_1)$

$(a_1, b_1, c_1) = (3 - 1, 4 - (-1), -2 - 2)$

$(a_1, b_1, c_1) = (2, 4 + 1, -4)$

Next, find the direction ratios of Line $L_2$ passing through C(0, 3, 2) and D(3, 5, 6).

Direction ratios of $L_2$, $(a_2, b_2, c_2) = (x_2 - x_1, y_2 - y_1, z_2 - z_1)$

$(a_2, b_2, c_2) = (3 - 0, 5 - 3, 6 - 2)$

Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are perpendicular if $a_1a_2 + b_1b_2 + c_1c_2 = 0$.

Check the dot product of the direction ratios:

$a_1a_2 + b_1b_2 + c_1c_2 = (2)(3) + (5)(2) + (-4)(4)$

$= 6 + 10 - 16$

$= 16 - 16$

Since the sum of the products of the corresponding direction ratios is 0, the two lines are perpendicular.

Hence, the line through the points (1, – 1, 2), (3, 4, – 2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).

Given:

Line $L_1$ passes through points A(4, 7, 8) and B(2, 3, 4).

Line $L_2$ passes through points C(– 1, – 2, 1) and D(1, 2, 5).

To Show:

Line $L_1$ is parallel to Line $L_2$.

Solution:

Two lines are parallel if their direction ratios are proportional, or if their direction cosines are equal (or negatives of each other).

First, find the direction ratios of Line $L_1$ passing through A(4, 7, 8) and B(2, 3, 4).

Direction ratios of $L_1$, $(a_1, b_1, c_1) = (x_2 - x_1, y_2 - y_1, z_2 - z_1)$

$(a_1, b_1, c_1) = (2 - 4, 3 - 7, 4 - 8)$

Next, find the direction ratios of Line $L_2$ passing through C(– 1, – 2, 1) and D(1, 2, 5).

Direction ratios of $L_2$, $(a_2, b_2, c_2) = (x_2 - x_1, y_2 - y_1, z_2 - z_1)$

$(a_2, b_2, c_2) = (1 - (-1), 2 - (-2), 5 - 1)$

$(a_2, b_2, c_2) = (1 + 1, 2 + 2, 4)$

Now, check if the direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are proportional. We look for a constant $k$ such that $a_2 = ka_1$, $b_2 = kb_1$, and $c_2 = kc_1$.

Check the ratios of corresponding components:

Since $\frac{a_2}{a_1} = \frac{b_2}{b_1} = \frac{c_2}{c_1} = -1$, the direction ratios of the two lines are proportional. The constant of proportionality is $k = -1$.

This indicates that the direction vector of Line $L_2$ is $-1$ times the direction vector of Line $L_1$, meaning they are parallel vectors.

Since their direction ratios are proportional, the lines are parallel.

Hence, the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (– 1, – 2, 1), (1, 2, 5).

Given:

The line passes through the point (1, 2, 3).

The line is parallel to the vector $3\hat{i} + 2\hat{j} −2\hat{k}$.

To Find:

The equation of the line (both vector and Cartesian forms).

Solution:

Let the given point be $A(1, 2, 3)$. The position vector of this point is $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$.

The line is parallel to the given vector. Let this parallel vector be $\vec{b} = 3\hat{i} + 2\hat{j} - 2\hat{k}$.

Vector Equation of the line:

The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{b}$ is given by:

Substitute the given values of $\vec{a}$ and $\vec{b}$:

This is the vector equation of the line.

Cartesian Equations of the line:

To find the Cartesian equations, let $\vec{r}$ be the position vector of any point $(x, y, z)$ on the line. So, $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.

Substitute this into the vector equation:

$x\hat{i} + y\hat{j} + z\hat{k} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda (3\hat{i} + 2\hat{j} - 2\hat{k})$

$x\hat{i} + y\hat{j} + z\hat{k} = (1 + 3\lambda)\hat{i} + (2 + 2\lambda)\hat{j} + (3 - 2\lambda)\hat{k}$

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

From each of these equations, we can express $\lambda$:

$\lambda = \frac{x - 1}{3}$

$\lambda = \frac{y - 2}{2}$

$\lambda = \frac{z - 3}{-2}$

Since all these expressions are equal to $\lambda$, we can equate them:

These are the Cartesian equations of the line.

The denominators (3, 2, -2) are the direction ratios of the line, which are the components of the parallel vector $\vec{b}$. The numerators $(x-1, y-2, z-3)$ come from subtracting the coordinates of the point (1, 2, 3) from $(x, y, z)$.

The vector equation of the line is $\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda (3\hat{i} + 2\hat{j} - 2\hat{k})$.

The Cartesian equations of the line are $\frac{x - 1}{3} = \frac{y - 2}{2} = \frac{z - 3}{-2}$.

Given:

The line passes through the point with position vector $\vec{a} = 2\hat{i} − \hat{j} + 4\hat{k}$.

The line is in the direction of the vector $\vec{b} = \hat{i} + 2\hat{j} − \hat{k}$.

To Find:

The equation of the line in vector form and in Cartesian form.

Solution:

The position vector of the point through which the line passes is $\vec{a} = 2\hat{i} - \hat{j} + 4\hat{k}$.

The line is in the direction of (parallel to) the vector $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$.

Vector Equation of the line:

The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{b}$ is given by:

Substitute the given values of $\vec{a}$ and $\vec{b}$:

This is the vector equation of the line.

Cartesian Equations of the line:

To find the Cartesian equations, let $\vec{r}$ be the position vector of any point $(x, y, z)$ on the line. So, $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.

Substitute this into the vector equation:

$x\hat{i} + y\hat{j} + z\hat{k} = (2\hat{i} - \hat{j} + 4\hat{k}) + \lambda (\hat{i} + 2\hat{j} - \hat{k})$

$x\hat{i} + y\hat{j} + z\hat{k} = (2 + \lambda)\hat{i} + (-1 + 2\lambda)\hat{j} + (4 - \lambda)\hat{k}$

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

From each of these equations, we can express $\lambda$:

$\lambda = x - 2$

$\lambda = \frac{y + 1}{2}$

$\lambda = \frac{z - 4}{-1}$

Since all these expressions are equal to $\lambda$, we can equate them:

These are the Cartesian equations of the line.

The denominators (1, 2, -1) are the direction ratios of the line, which are the components of the parallel vector $\vec{b}$. The numerators $(x-2, y-(-1), z-4)$ come from subtracting the coordinates of the point (2, -1, 4) from $(x, y, z)$.

The vector equation of the line is $\vec{r} = (2\hat{i} - \hat{j} + 4\hat{k}) + \lambda (\hat{i} + 2\hat{j} - \hat{k})$.

The Cartesian equations of the line are $\frac{x - 2}{1} = \frac{y + 1}{2} = \frac{z - 4}{-1}$.

Given:

The line passes through the point (– 2, 4, – 5).

The line is parallel to the line given by $\frac{x + 3}{3} = \frac{y − 4}{5} = \frac{z + 8}{6}$.

To Find:

The Cartesian equation of the line.

Solution:

The line passes through the point $(x_1, y_1, z_1) = (-2, 4, -5)$.

The given line is $\frac{x + 3}{3} = \frac{y − 4}{5} = \frac{z + 8}{6}$.

The Cartesian equation of a line is of the form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$, where $(x_1, y_1, z_1)$ is a point on the line and $(a, b, c)$ are the direction ratios of the line.

The denominators in the given equation are the direction ratios of that line. So, the direction ratios of the line $\frac{x + 3}{3} = \frac{y − 4}{5} = \frac{z + 8}{6}$ are (3, 5, 6).

Since the line we need to find is parallel to this given line, it will have the same (or proportional) direction ratios.

We can take the direction ratios of our line as $(a, b, c) = (3, 5, 6)$.

Now we have the point $(x_1, y_1, z_1) = (-2, 4, -5)$ and the direction ratios $(a, b, c) = (3, 5, 6)$.

Substitute these values into the Cartesian equation of a line:

This is the Cartesian equation of the required line.

The Cartesian equation of the line is $\frac{x + 2}{3} = \frac{y - 4}{5} = \frac{z + 5}{6}$.

Given:

The Cartesian equation of a line is $\frac{x − 5}{3} = \frac{ y + 4}{7} = \frac{z − 6}{2}$.

To Find:

The vector form of the equation of the line.

Solution:

The Cartesian equation of a line is of the form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$.

From the given equation $\frac{x − 5}{3} = \frac{ y + 4}{7} = \frac{z − 6}{2}$, we can identify the coordinates of a point on the line and the direction ratios of the line.

The point $(x_1, y_1, z_1)$ through which the line passes is given by $(x_1, y_1, z_1) = (5, -4, 6)$ because the numerators are $(x-5)$, $(y-(-4))$, and $(z-6)$.

The direction ratios of the line $(a, b, c)$ are given by the denominators $(3, 7, 2)$.

The position vector of the point $(5, -4, 6)$ is $\vec{a} = 5\hat{i} - 4\hat{j} + 6\hat{k}$.

The vector parallel to the line (direction vector) is $\vec{b} = 3\hat{i} + 7\hat{j} + 2\hat{k}$.

The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{b}$ is given by:

Substitute the identified values of $\vec{a}$ and $\vec{b}$:

This is the vector form of the equation of the line.

The vector form of the equation of the line is $\vec{r} = (5\hat{i} - 4\hat{j} + 6\hat{k}) + \lambda (3\hat{i} + 7\hat{j} + 2\hat{k})$.

Given:

Two pairs of lines are given in vector form.

To Find:

The angle between each pair of lines.

Solution:

The angle between two lines with vector equations $\vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu \vec{b_2}$ is the angle between their direction vectors $\vec{b_1}$ and $\vec{b_2}$. The cosine of the angle $\theta$ between $\vec{b_1}$ and $\vec{b_2}$ is given by:

Part (i):

Line 1: $\vec{r} = 2\hat{i} − 5\hat{j} + \hat{k} + λ (3\hat{i} + 2\hat{j} + 6\hat{k})$

Line 2: $\vec{r} = 7\hat{i} − 6\hat{k} + µ (\hat{i} + 2\hat{j} + 2\hat{k})$

The direction vectors are:

Calculate the dot product $\vec{b_1} \cdot \vec{b_2}$:

$\vec{b_1} \cdot \vec{b_2} = (3)(1) + (2)(2) + (6)(2)$

$\vec{b_1} \cdot \vec{b_2} = 3 + 4 + 12 = 19$

Calculate the magnitudes of $\vec{b_1}$ and $\vec{b_2}$:

$|\vec{b_1}| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$

$|\vec{b_2}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$

Substitute these values into the formula for $\cos\theta$:

The angle $\theta$ is given by $\cos^{-1}\left(\frac{19}{21}\right)$.

The angle between the lines in part (i) is $\cos^{-1}\left(\frac{19}{21}\right)$.

Part (ii):

Line 1: $\vec{r} = 3\hat{i} + \hat{j} − 2\hat{k} + λ (\hat{i} − \hat{j} − 2\hat{k})$

Line 2: $\vec{r} = 2\hat{i} − \hat{j} − 56\hat{k} + µ (3\hat{i} − 5\hat{j} − 4\hat{k})$

The direction vectors are:

Calculate the dot product $\vec{b_1} \cdot \vec{b_2}$:

$\vec{b_1} \cdot \vec{b_2} = (1)(3) + (-1)(-5) + (-2)(-4)$

$\vec{b_1} \cdot \vec{b_2} = 3 + 5 + 8 = 16$

Calculate the magnitudes of $\vec{b_1}$ and $\vec{b_2}$:

$|\vec{b_1}| = \sqrt{1^2 + (-1)^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$

$|\vec{b_2}| = \sqrt{3^2 + (-5)^2 + (-4)^2} = \sqrt{9 + 25 + 16} = \sqrt{50}$

$|\vec{b_2}| = \sqrt{25 \times 2} = 5\sqrt{2}$

Substitute these values into the formula for $\cos\theta$:

$\cos\theta = \frac{16}{5\sqrt{12}}$

Simplify the denominator: $\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$.

$\cos\theta = \frac{16}{5(2\sqrt{3})} = \frac{16}{10\sqrt{3}}$

Simplify the fraction and rationalize the denominator:

The angle $\theta$ is given by $\cos^{-1}\left(\frac{8\sqrt{3}}{15}\right)$.

The angle between the lines in part (ii) is $\cos^{-1}\left(\frac{8\sqrt{3}}{15}\right)$.

Given:

Two pairs of lines are given in Cartesian form.

To Find:

The angle between each pair of lines.

Solution:

The Cartesian equation of a line is of the form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$, where $(a, b, c)$ are the direction ratios of the line.

The angle between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is given by the angle between their direction vectors $\vec{b_1} = a_1\hat{i} + b_1\hat{j} + c_1\hat{k}$ and $\vec{b_2} = a_2\hat{i} + b_2\hat{j} + c_2\hat{k}$. The cosine of the angle $\theta$ is:

Part (i):

Line 1: $\frac{x − 2}{2} = \frac{y − 1}{5} = \frac{z + 3}{−3}$

Line 2: $\frac{x + 2}{−1} = \frac{y − 4}{8} = \frac{z − 5}{4}$

The direction ratios are:

Calculate $a_1a_2 + b_1b_2 + c_1c_2$:

$a_1a_2 + b_1b_2 + c_1c_2 = (2)(-1) + (5)(8) + (-3)(4)$

$= -2 + 40 - 12 = 26$

Calculate the magnitudes of the direction ratio vectors:

$\sqrt{a_1^2+b_1^2+c_1^2} = \sqrt{2^2 + 5^2 + (-3)^2} = \sqrt{4 + 25 + 9} = \sqrt{38}$

$\sqrt{a_2^2+b_2^2+c_2^2} = \sqrt{(-1)^2 + 8^2 + 4^2} = \sqrt{1 + 64 + 16} = \sqrt{81} = 9$

Substitute these values into the formula for $\cos\theta$:

To rationalize the denominator, multiply numerator and denominator by $\sqrt{38}$:

$\cos\theta = \frac{26\sqrt{38}}{9\sqrt{38}\sqrt{38}} = \frac{26\sqrt{38}}{9 \times 38}$

Simplify the fraction:

The angle $\theta$ is given by $\cos^{-1}\left(\frac{13\sqrt{38}}{171}\right)$.

The angle between the lines in part (i) is $\cos^{-1}\left(\frac{13\sqrt{38}}{171}\right)$.

Part (ii):

Line 1: $\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$

Line 2: $\frac{x − 5}{4} = \frac{y − 2}{1} = \frac{z − 3}{8}$

The direction ratios are:

Calculate $a_1a_2 + b_1b_2 + c_1c_2$:

$a_1a_2 + b_1b_2 + c_1c_2 = (2)(4) + (2)(1) + (1)(8)$

$= 8 + 2 + 8 = 18$

Calculate the magnitudes of the direction ratio vectors:

$\sqrt{a_1^2+b_1^2+c_1^2} = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$

$\sqrt{a_2^2+b_2^2+c_2^2} = \sqrt{4^2 + 1^2 + 8^2} = \sqrt{16 + 1 + 64} = \sqrt{81} = 9$

Substitute these values into the formula for $\cos\theta$:

Simplify the fraction:

The angle $\theta$ is given by $\cos^{-1}\left(\frac{2}{3}\right)$.

The angle between the lines in part (ii) is $\cos^{-1}\left(\frac{2}{3}\right)$.

Given:

The equations of two lines are:

The lines are at right angles (perpendicular).

To Find:

The value of p.

Solution:

For two lines to be perpendicular, the dot product of their direction ratios must be zero.

First, we need to rewrite the given equations in the standard Cartesian form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$ to identify the direction ratios.

For Line 1: $\frac{1 − x}{3} = \frac{7y − 14}{2p} = \frac{z − 3}{2}$

Rewrite the first term: $\frac{1 − x}{3} = \frac{-(x - 1)}{3} = \frac{x - 1}{-3}$.

Rewrite the second term: $\frac{7y − 14}{2p} = \frac{7(y - 2)}{2p} = \frac{y - 2}{2p/7}$.

The third term is already in standard form: $\frac{z - 3}{2}$.

So, the standard form of Line 1 is $\frac{x - 1}{-3} = \frac{y - 2}{2p/7} = \frac{z - 3}{2}$.

The direction ratios of Line 1 are $(a_1, b_1, c_1) = \left(-3, \frac{2p}{7}, 2\right)$.

For Line 2: $\frac{7 − 7x}{3p} = \frac{y − 5}{1} = \frac{6 − z}{5}$

Rewrite the first term: $\frac{7 − 7x}{3p} = \frac{7(1 - x)}{3p} = \frac{7(-(x - 1))}{3p} = \frac{-(x - 1) \times 7}{3p} = \frac{x - 1}{-3p/7}$.

Rewrite the third term: $\frac{6 − z}{5} = \frac{-(z - 6)}{5} = \frac{z - 6}{-5}$.

The second term is already in standard form: $\frac{y - 5}{1}$.

So, the standard form of Line 2 is $\frac{x - 1}{-3p/7} = \frac{y - 5}{1} = \frac{z - 6}{-5}$.

The direction ratios of Line 2 are $(a_2, b_2, c_2) = \left(-\frac{3p}{7}, 1, -5\right)$.

The lines are at right angles, which means they are perpendicular. The condition for perpendicularity using direction ratios is $a_1a_2 + b_1b_2 + c_1c_2 = 0$.

Substitute the direction ratios of Line 1 and Line 2:

$(-3)\left(-\frac{3p}{7}\right) + \left(\frac{2p}{7}\right)(1) + (2)(-5) = 0$

$\frac{9p}{7} + \frac{2p}{7} - 10 = 0$

Combine the terms with p:

$\frac{9p + 2p}{7} - 10 = 0$

$\frac{11p}{7} - 10 = 0$

Add 10 to both sides:

$\frac{11p}{7} = 10$

Multiply both sides by 7:

$11p = 70$

Divide by 11:

The value of p is $\frac{70}{11}$.

Given:

The equations of two lines are:

To Show:

The two lines are perpendicular to each other.

Solution:

Two lines with Cartesian equations $\frac{x - x_1}{a_1} = \frac{y - y_1}{b_1} = \frac{z - z_1}{c_1}$ and $\frac{x - x_2}{a_2} = \frac{y - y_2}{b_2} = \frac{z - z_2}{c_2}$ are perpendicular if the sum of the products of their corresponding direction ratios is zero, i.e., $a_1a_2 + b_1b_2 + c_1c_2 = 0$.

From the equation of Line 1, $\frac{x − 5}{7} = \frac{y + 2}{−5} = \frac{z}{1}$, the direction ratios are $(a_1, b_1, c_1) = (7, -5, 1)$.

From the equation of Line 2, $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$, the direction ratios are $(a_2, b_2, c_2) = (1, 2, 3)$.

Now, calculate the sum of the products of the corresponding direction ratios:

$a_1a_2 + b_1b_2 + c_1c_2 = (7)(1) + (-5)(2) + (1)(3)$

$= 7 - 10 + 3$

$= 10 - 10$

Since the sum of the products of the corresponding direction ratios is 0, the two lines are perpendicular to each other.

Hence, the lines $\frac{x − 5}{7} = \frac{y + 2}{−5} = \frac{z}{1}$ and $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ are perpendicular to each other.

Given:

The vector equations of two lines are:

[where $\lambda$ and $\mu$ are scalar parameters]

To Find:

The shortest distance between the two lines.

Solution:

The vector equation of a line is of the form $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\vec{a}$ is the position vector of a point on the line and $\vec{b}$ is a vector parallel to the line.

For Line 1:

For Line 2:

Since $\vec{b_1}$ is not a scalar multiple of $\vec{b_2}$ (check ratios: $1/2 \neq -1/1 \neq 1/2$), the lines are not parallel. Thus, they are skew lines.

The shortest distance between two skew lines $\vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu \vec{b_2}$ is given by the formula:

First, calculate $\vec{a_2} - \vec{a_1}$:

$\vec{a_2} - \vec{a_1} = (2\hat{i} − \hat{j} - \hat{k}) - (\hat{i} + 2\hat{j} + \hat{k})$

$\vec{a_2} - \vec{a_1} = (2-1)\hat{i} + (-1-2)\hat{j} + (-1-1)\hat{k}$

Next, calculate the cross product $\vec{b_1} \times \vec{b_2}$:

$\vec{b_1} \times \vec{b_2} = \hat{i}[(-1)(2) - (1)(1)] - \hat{j}[(1)(2) - (1)(2)] + \hat{k}[(1)(1) - (-1)(2)]$

$\vec{b_1} \times \vec{b_2} = \hat{i}[-2 - 1] - \hat{j}[2 - 2] + \hat{k}[1 + 2]$

Now, calculate the magnitude of the cross product $|\vec{b_1} \times \vec{b_2}|$:

$|\vec{b_1} \times \vec{b_2}| = \sqrt{(-3)^2 + (0)^2 + (3)^2}$

$|\vec{b_1} \times \vec{b_2}| = \sqrt{9 + 0 + 9}$

$|\vec{b_1} \times \vec{b_2}| = \sqrt{18} = \sqrt{9 \times 2}$

Next, calculate the dot product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$:

$(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (\hat{i} - 3\hat{j} - 2\hat{k}) \cdot (-3\hat{i} + 0\hat{j} + 3\hat{k})$

$(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (1)(-3) + (-3)(0) + (-2)(3)$

$(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = -3 + 0 - 6$

Finally, substitute the values into the shortest distance formula (i):

$d = \frac{|-9|}{|3\sqrt{2}|}$

$d = \frac{9}{3\sqrt{2}}$

Simplify the fraction:

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{2}$:

$d = \frac{3 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \frac{3\sqrt{2}}{2}$

The shortest distance between the two lines is $\frac{3}{\sqrt{2}}$ units or $\frac{3\sqrt{2}}{2}$ units.

Given:

The Cartesian equations of two lines are:

To Find:

The shortest distance between the two lines.

Solution:

From the given Cartesian equations $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$, we can identify a point on each line and their direction ratios.

For Line 1: $\frac{x - (-1)}{7} = \frac{y - (-1)}{-6} = \frac{z - (-1)}{1}$.

A point on Line 1 is $(x_1, y_1, z_1) = (-1, -1, -1)$. The position vector of this point is $\vec{a_1} = -\hat{i} - \hat{j} - \hat{k}$.

The direction ratios of Line 1 are $(a_1, b_1, c_1) = (7, -6, 1)$. The direction vector is $\vec{b_1} = 7\hat{i} - 6\hat{j} + \hat{k}$.

For Line 2: $\frac{x - 3}{1} = \frac{y - 5}{-2} = \frac{z - 7}{1}$.

A point on Line 2 is $(x_2, y_2, z_2) = (3, 5, 7)$. The position vector of this point is $\vec{a_2} = 3\hat{i} + 5\hat{j} + 7\hat{k}$.

The direction ratios of Line 2 are $(a_2, b_2, c_2) = (1, -2, 1)$. The direction vector is $\vec{b_2} = \hat{i} - 2\hat{j} + \hat{k}$.

Since the direction ratios are not proportional ($7/1 \neq -6/-2 \neq 1/1$), the lines are not parallel. Thus, they are skew lines.

The shortest distance between two skew lines $\vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu \vec{b_2}$ is given by the formula:

First, calculate $\vec{a_2} - \vec{a_1}$:

$\vec{a_2} - \vec{a_1} = (3\hat{i} + 5\hat{j} + 7\hat{k}) - (-\hat{i} - \hat{j} - \hat{k})$

$\vec{a_2} - \vec{a_1} = (3-(-1))\hat{i} + (5-(-1))\hat{j} + (7-(-1))\hat{k}$

$\vec{a_2} - \vec{a_1} = (3+1)\hat{i} + (5+1)\hat{j} + (7+1)\hat{k}$

Next, calculate the cross product $\vec{b_1} \times \vec{b_2}$:

$\vec{b_1} \times \vec{b_2} = \hat{i}[(-6)(1) - (1)(-2)] - \hat{j}[(7)(1) - (1)(1)] + \hat{k}[(7)(-2) - (-6)(1)]$

$\vec{b_1} \times \vec{b_2} = \hat{i}[-6 + 2] - \hat{j}[7 - 1] + \hat{k}[-14 + 6]$

Now, calculate the magnitude of the cross product $|\vec{b_1} \times \vec{b_2}|$:

$|\vec{b_1} \times \vec{b_2}| = \sqrt{(-4)^2 + (-6)^2 + (-8)^2}$

$|\vec{b_1} \times \vec{b_2}| = \sqrt{16 + 36 + 64}$

$|\vec{b_1} \times \vec{b_2}| = \sqrt{116} = \sqrt{4 \times 29}$

Next, calculate the dot product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$:

$(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (4\hat{i} + 6\hat{j} + 8\hat{k}) \cdot (-4\hat{i} - 6\hat{j} - 8\hat{k})$

$(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (4)(-4) + (6)(-6) + (8)(-8)$

$(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = -16 - 36 - 64$

Finally, substitute the values into the shortest distance formula (i):

$d = \frac{|-116|}{|2\sqrt{29}|}$

$d = \frac{116}{2\sqrt{29}}$

Simplify the fraction:

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{29}$:

$d = \frac{58 \times \sqrt{29}}{\sqrt{29} \times \sqrt{29}} = \frac{58\sqrt{29}}{29}$

Simplify the fraction:

The shortest distance between the two lines is $2\sqrt{29}$ units.

Given:

The vector equations of two lines are:

[where $\lambda$ and $\mu$ are scalar parameters]

To Find:

The shortest distance between the two lines.

Solution:

The vector equation of a line is of the form $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\vec{a}$ is the position vector of a point on the line and $\vec{b}$ is a vector parallel to the line.

For Line 1:

For Line 2:

Since $\vec{b_1}$ is not a scalar multiple of $\vec{b_2}$ (check ratios: $1/2 \neq -3/3 \neq 2/1$), the lines are not parallel. Thus, they are skew lines.

The shortest distance between two skew lines $\vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu \vec{b_2}$ is given by the formula:

First, calculate $\vec{a_2} - \vec{a_1}$:

$\vec{a_2} - \vec{a_1} = (4\hat{i} + 5\hat{j} + 6\hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k})$

$\vec{a_2} - \vec{a_1} = (4-1)\hat{i} + (5-2)\hat{j} + (6-3)\hat{k}$

Next, calculate the cross product $\vec{b_1} \times \vec{b_2}$:

$\vec{b_1} \times \vec{b_2} = \hat{i}[(-3)(1) - (2)(3)] - \hat{j}[(1)(1) - (2)(2)] + \hat{k}[(1)(3) - (-3)(2)]$

$\vec{b_1} \times \vec{b_2} = \hat{i}[-3 - 6] - \hat{j}[1 - 4] + \hat{k}[3 + 6]$

Now, calculate the magnitude of the cross product $|\vec{b_1} \times \vec{b_2}|$:

$|\vec{b_1} \times \vec{b_2}| = \sqrt{(-9)^2 + (3)^2 + (9)^2}$

$|\vec{b_1} \times \vec{b_2}| = \sqrt{81 + 9 + 81}$

$|\vec{b_1} \times \vec{b_2}| = \sqrt{171} = \sqrt{9 \times 19}$

Next, calculate the dot product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$:

$(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (3\hat{i} + 3\hat{j} + 3\hat{k}) \cdot (-9\hat{i} + 3\hat{j} + 9\hat{k})$

$(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (3)(-9) + (3)(3) + (3)(9)$

$(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = -27 + 9 + 27$

Finally, substitute the values into the shortest distance formula (i):

$d = \frac{|9|}{|3\sqrt{19}|}$

$d = \frac{9}{3\sqrt{19}}$

Simplify the fraction:

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{19}$:

$d = \frac{3 \times \sqrt{19}}{\sqrt{19} \times \sqrt{19}} = \frac{3\sqrt{19}}{19}$

The shortest distance between the two lines is $\frac{3}{\sqrt{19}}$ units or $\frac{3\sqrt{19}}{19}$ units.

Given:

The vector equations of two lines are:

[where $t$ and $s$ are scalar parameters]

To Find:

The shortest distance between the two lines.

Solution:

First, we rewrite the given vector equations in the standard form $\vec{r} = \vec{a} + \lambda \vec{b}$.

For Line 1: $\vec{r} = (1−t)\hat{i} + (t−2)\hat{j} + (3−2t)\hat{k}$

Group terms without $t$ and terms with $t$:

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

Comparing this with $\vec{r} = \vec{a_1} + t \vec{b_1}$, we have:

For Line 2: $\vec{r} = (s+1)\hat{i} + (2s−1)\hat{j} − (2s+1)\hat{k}$

Group terms without $s$ and terms with $s$:

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

Comparing this with $\vec{r} = \vec{a_2} + s \vec{b_2}$, we have:

We check if the lines are parallel by comparing their direction vectors $\vec{b_1}$ and $\vec{b_2}$. The direction ratios are $(-1, 1, -2)$ and $(1, 2, -2)$. Since the ratios of corresponding components $\frac{1}{-1} = -1$, $\frac{2}{1} = 2$, $\frac{-2}{-2} = 1$ are not equal, the vectors $\vec{b_1}$ and $\vec{b_2}$ are not parallel, and thus the lines are skew.

The shortest distance between two skew lines $\vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu \vec{b_2}$ is given by the formula:

First, calculate $\vec{a_2} - \vec{a_1}$:

$\vec{a_2} - \vec{a_1} = (\hat{i} - \hat{j} - \hat{k}) - (\hat{i} - 2\hat{j} + 3\hat{k})$

$\vec{a_2} - \vec{a_1} = (1-1)\hat{i} + (-1 - (-2))\hat{j} + (-1-3)\hat{k}$

Next, calculate the cross product $\vec{b_1} \times \vec{b_2}$:

$\vec{b_1} \times \vec{b_2} = \hat{i}[(1)(-2) - (-2)(2)] - \hat{j}[(-1)(-2) - (-2)(1)] + \hat{k}[(-1)(2) - (1)(1)]$

$\vec{b_1} \times \vec{b_2} = \hat{i}[-2 + 4] - \hat{j}[2 + 2] + \hat{k}[-2 - 1]$

Now, calculate the magnitude of the cross product $|\vec{b_1} \times \vec{b_2}|$:

$|\vec{b_1} \times \vec{b_2}| = \sqrt{(2)^2 + (-4)^2 + (-3)^2}$

$|\vec{b_1} \times \vec{b_2}| = \sqrt{4 + 16 + 9}$

Next, calculate the dot product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$:

$(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (0\hat{i} + \hat{j} - 4\hat{k}) \cdot (2\hat{i} - 4\hat{j} - 3\hat{k})$

$(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (0)(2) + (1)(-4) + (-4)(-3)$

$(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = 0 - 4 + 12$

Finally, substitute the values into the shortest distance formula (i):

$d = \frac{|8|}{|\sqrt{29}|}$

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{29}$:

$d = \frac{8 \times \sqrt{29}}{\sqrt{29} \times \sqrt{29}} = \frac{8\sqrt{29}}{29}$

The shortest distance between the two lines is $\frac{8}{\sqrt{29}}$ units or $\frac{8\sqrt{29}}{29}$ units.

Given:

The direction ratios of the first line are $(a_1, b_1, c_1) = (a, b, c)$.

The direction ratios of the second line are $(a_2, b_2, c_2) = (b – c, c – a, a – b)$.

To Find:

The angle between the two lines.

Solution:

Let $\theta$ be the angle between the two lines. The cosine of the angle between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is given by the formula:

Substitute the given direction ratios into the numerator:

$a_1a_2 + b_1b_2 + c_1c_2 = (a)(b – c) + (b)(c – a) + (c)(a – b)$

$= ab - ac + bc - ba + ca - cb$

$= (ab - ba) + (-ac + ca) + (bc - cb)$

So, the numerator of the $\cos\theta$ formula is 0.

Assuming the lines are well-defined (i.e., the direction ratios are not all zero, so the denominators are non-zero), the equation $\cos\theta = 0$ implies that the angle between the lines is $90^\circ$ or $\frac{\pi}{2}$ radians.

The condition for two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ to be perpendicular is $a_1a_2 + b_1b_2 + c_1c_2 = 0$. Since we found that $a_1a_2 + b_1b_2 + c_1c_2 = 0$, the lines are perpendicular.

The angle between the two lines is $90^\circ$ or $\frac{\pi}{2}$ radians.

Given:

The line passes through the origin (0, 0, 0).

The line is parallel to the x-axis.

To Find:

The equation of the line (vector and Cartesian forms).

Solution:

The line passes through the origin, so a point on the line is $(0, 0, 0)$. The position vector of this point is $\vec{a} = 0\hat{i} + 0\hat{j} + 0\hat{k} = \vec{0}$.

The line is parallel to the x-axis. The direction ratios of the x-axis are (1, 0, 0). A vector parallel to the x-axis can be taken as $\vec{b} = 1\hat{i} + 0\hat{j} + 0\hat{k} = \hat{i}$.

Vector Equation of the line:

The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{b}$ is given by:

Substitute the identified values of $\vec{a}$ and $\vec{b}$:

This is the vector equation of the line.

Cartesian Equations of the line:

To find the Cartesian equations, let $\vec{r}$ be the position vector of any point $(x, y, z)$ on the line. So, $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.

Substitute this into the vector equation:

$x\hat{i} + y\hat{j} + z\hat{k} = \lambda \hat{i} + 0\hat{j} + 0\hat{k}$

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

The Cartesian equations are obtained by setting the coefficients of the zero vectors to zero.

The x-coordinate is $x = \lambda$, which can be any real number. This indicates that the line extends infinitely along the x-axis.

The Cartesian equations of the line are $y = 0, z = 0$.

The vector equation of the line is $\vec{r} = \lambda \hat{i}$.

The Cartesian equations of the line are $y = 0, z = 0$.

Given:

The equations of two lines are:

The lines are perpendicular.

To Find:

The value of k.

Solution:

The Cartesian equation of a line is of the form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$, where $(a, b, c)$ are the direction ratios of the line.

From the equation of Line 1, $\frac{x − 1}{−3} = \frac{y − 2}{2k} = \frac{z − 3}{2}$, the direction ratios are $(a_1, b_1, c_1) = (-3, 2k, 2)$.

From the equation of Line 2, $\frac{x − 1}{3k} = \frac{y − 1}{1} = \frac{z − 6}{−5}$, the direction ratios are $(a_2, b_2, c_2) = (3k, 1, -5)$.

Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are perpendicular if the sum of the products of their corresponding direction ratios is zero:

Substitute the direction ratios of Line 1 and Line 2 into this condition:

$(-3)(3k) + (2k)(1) + (2)(-5) = 0$

$-9k + 2k - 10 = 0$

Combine the terms with k:

$-7k - 10 = 0$

Add 10 to both sides:

$-7k = 10$

Divide by -7:

The value of k is $-\frac{10}{7}$.

Given:

The vector equations of two lines are:

[where $\lambda$ and $\mu$ are scalar parameters]

To Find:

The shortest distance between the two lines.

Solution:

The vector equation of a line is of the form $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\vec{a}$ is the position vector of a point on the line and $\vec{b}$ is a vector parallel to the line.

For Line 1:

For Line 2:

We check if the lines are parallel by comparing their direction vectors $\vec{b_1}$ and $\vec{b_2}$. The direction ratios are $(1, -2, 2)$ and $(3, -2, -2)$. Since the ratios of corresponding components $1/3$, $-2/-2 = 1$, $2/-2 = -1$ are not equal, the vectors $\vec{b_1}$ and $\vec{b_2}$ are not parallel, and thus the lines are skew.

The shortest distance between two skew lines $\vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu \vec{b_2}$ is given by the formula:

First, calculate $\vec{a_2} - \vec{a_1}$:

$\vec{a_2} - \vec{a_1} = (-4\hat{i} - \hat{k}) - (6\hat{i} + 2\hat{j} + 2\hat{k})$

$\vec{a_2} - \vec{a_1} = (-4-6)\hat{i} + (0-2)\hat{j} + (-1-2)\hat{k}$

Next, calculate the cross product $\vec{b_1} \times \vec{b_2}$:

$\vec{b_1} \times \vec{b_2} = \hat{i}[(-2)(-2) - (2)(-2)] - \hat{j}[(1)(-2) - (2)(3)] + \hat{k}[(1)(-2) - (-2)(3)]$

$\vec{b_1} \times \vec{b_2} = \hat{i}[4 + 4] - \hat{j}[-2 - 6] + \hat{k}[-2 + 6]$

Now, calculate the magnitude of the cross product $|\vec{b_1} \times \vec{b_2}|$:

$|\vec{b_1} \times \vec{b_2}| = \sqrt{(8)^2 + (8)^2 + (4)^2}$

$|\vec{b_1} \times \vec{b_2}| = \sqrt{64 + 64 + 16}$

$|\vec{b_1} \times \vec{b_2}| = \sqrt{144}$

Next, calculate the dot product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$:

$(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (-10\hat{i} - 2\hat{j} - 3\hat{k}) \cdot (8\hat{i} + 8\hat{j} + 4\hat{k})$

$(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (-10)(8) + (-2)(8) + (-3)(4)$

$(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = -80 - 16 - 12$

Finally, substitute the values into the shortest distance formula (i):

$d = \frac{|-108|}{|12|}$

$d = \frac{108}{12}$

Simplify the fraction:

The shortest distance between the two lines is 9 units.

Given:

The line passes through the point (1, 2, – 4).

The line is perpendicular to the lines:

To Find:

The vector equation of the line.

Solution:

Let the required line pass through the point A(1, 2, – 4). The position vector of this point is $\vec{a} = \hat{i} + 2\hat{j} - 4\hat{k}$.

The line is perpendicular to the two given lines. The direction ratios of the given lines can be obtained from the denominators of their Cartesian equations.

For Line 1, the direction ratios are $(a_1, b_1, c_1) = (3, -16, 7)$. The direction vector is $\vec{b_1} = 3\hat{i} - 16\hat{j} + 7\hat{k}$.

For Line 2, the direction ratios are $(a_2, b_2, c_2) = (3, 8, -5)$. The direction vector is $\vec{b_2} = 3\hat{i} + 8\hat{j} - 5\hat{k}$.

Since the required line is perpendicular to both Line 1 and Line 2, its direction vector must be perpendicular to both $\vec{b_1}$ and $\vec{b_2}$. A vector perpendicular to two vectors is parallel to their cross product.

Let $\vec{b}$ be the direction vector of the required line. We can take $\vec{b}$ to be proportional to $\vec{b_1} \times \vec{b_2}$.

Calculate the cross product $\vec{b_1} \times \vec{b_2}$:

$\vec{b_1} \times \vec{b_2} = \hat{i}[(-16)(-5) - (7)(8)] - \hat{j}[(3)(-5) - (7)(3)] + \hat{k}[(3)(8) - (-16)(3)]$

$\vec{b_1} \times \vec{b_2} = \hat{i}[80 - 56] - \hat{j}[-15 - 21] + \hat{k}[24 + 48]$

$\vec{b_1} \times \vec{b_2} = \hat{i}[24] - \hat{j}[-36] + \hat{k}[72]$

The direction vector of the required line can be any non-zero scalar multiple of $24\hat{i} + 36\hat{j} + 72\hat{k}$. We can simplify this vector by dividing by the greatest common divisor of the components, which is 12.

Let $\vec{b} = \frac{1}{12} (24\hat{i} + 36\hat{j} + 72\hat{k}) = 2\hat{i} + 3\hat{j} + 6\hat{k}$.

The required line passes through the point with position vector $\vec{a} = \hat{i} + 2\hat{j} - 4\hat{k}$ and is parallel to the vector $\vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}$.

The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{b}$ is given by $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\lambda$ is a scalar parameter.

Substitute the values of $\vec{a}$ and $\vec{b}$:

The vector equation of the line is $\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda (2\hat{i} + 3\hat{j} + 6\hat{k})$.