Chapter 12 Introduction To Three Dimensional Geometry

To locate a point $P(x, y, z)$ in three-dimensional space, we use a coordinate system defined by three mutually perpendicular axes: the x-axis, the y-axis, and the z-axis. These axes intersect at a point called the origin $(0, 0, 0)$.

The coordinates $(x, y, z)$ of a point represent the directed distances from the origin along the respective axes or parallel to them.

(i) Locating the point $(2, 3, 4)$:

We start from the origin $(0, 0, 0)$.

1. Move 2 units along the positive x-axis.

2. From this point, move 3 units parallel to the positive y-axis.

3. From there, move 4 units parallel to the positive z-axis.

The final position reached is the location of the point $(2, 3, 4)$.

(ii) Locating the point $(-2, -2, 3)$:

We start from the origin $(0, 0, 0)$.

1. Move 2 units along the negative x-axis (since the x-coordinate is $-2$).

2. From this point, move 2 units parallel to the negative y-axis (since the y-coordinate is $-2$).

3. From there, move 3 units parallel to the positive z-axis (since the z-coordinate is $3$).

The final position reached is the location of the point $(-2, -2, 3)$.

In a three-dimensional coordinate system, a plane is a flat, two-dimensional surface that extends infinitely. Simple linear equations involving only one variable, like $x = c$, $y = c$, or $z = c$ (where $c$ is a constant), represent planes that are parallel to one of the coordinate planes.

(i) Sketching the plane $x = 1$:

The equation $x = 1$ represents the set of all points $(x, y, z)$ in space where the x-coordinate is always equal to 1, while the y and z coordinates can take any real value.

This means that every point on this plane is at a constant distance of 1 unit from the yz-plane, measured along the x-axis (in the positive direction).

Therefore, the plane $x = 1$ is parallel to the yz-plane and passes through the point $(1, 0, 0)$ on the x-axis. It is also perpendicular to the x-axis.

(ii) Sketching the plane $y = 3$:

The equation $y = 3$ represents the set of all points $(x, y, z)$ in space where the y-coordinate is always equal to 3, while the x and z coordinates can take any real value.

This means that every point on this plane is at a constant distance of 3 units from the xz-plane, measured along the y-axis (in the positive direction).

Therefore, the plane $y = 3$ is parallel to the xz-plane and passes through the point $(0, 3, 0)$ on the y-axis. It is also perpendicular to the y-axis.

(iii) Sketching the plane $z = 4$:

The equation $z = 4$ represents the set of all points $(x, y, z)$ in space where the z-coordinate is always equal to 4, while the x and y coordinates can take any real value.

This means that every point on this plane is at a constant distance of 4 units from the xy-plane, measured along the z-axis (in the positive direction).

Therefore, the plane $z = 4$ is parallel to the xy-plane and passes through the point $(0, 0, 4)$ on the z-axis. It is also perpendicular to the z-axis.

Let the given point be $P(3, 4, 5)$. We need to find the coordinates of the feet of the perpendiculars drawn from P onto the x-axis, y-axis, and z-axis.

Finding the coordinates of L (foot of perpendicular on x-axis):

The x-axis is the set of points where the y and z coordinates are both zero. Thus, any point on the x-axis has coordinates of the form $(x, 0, 0)$.

The foot of the perpendicular from $P(3, 4, 5)$ to the x-axis is the point on the x-axis that has the same x-coordinate as P. This is because the perpendicular line segment from P to the x-axis will be parallel to the yz-plane.

Therefore, the coordinates of L are obtained by setting the y and z coordinates of P to zero.

The coordinates of L are $(3, 0, 0)$.

Finding the coordinates of M (foot of perpendicular on y-axis):

The y-axis is the set of points where the x and z coordinates are both zero. Thus, any point on the y-axis has coordinates of the form $(0, y, 0)$.

The foot of the perpendicular from $P(3, 4, 5)$ to the y-axis is the point on the y-axis that has the same y-coordinate as P.

Therefore, the coordinates of M are obtained by setting the x and z coordinates of P to zero.

The coordinates of M are $(0, 4, 0)$.

Finding the coordinates of N (foot of perpendicular on z-axis):

The z-axis is the set of points where the x and y coordinates are both zero. Thus, any point on the z-axis has coordinates of the form $(0, 0, z)$.

The foot of the perpendicular from $P(3, 4, 5)$ to the z-axis is the point on the z-axis that has the same z-coordinate as P.

Therefore, the coordinates of N are obtained by setting the x and y coordinates of P to zero.

The coordinates of N are $(0, 0, 5)$.

Let the given point be $P(3, 4, 5)$. We need to find the coordinates of the feet of the perpendiculars drawn from P onto the xy-plane, yz-plane, and zx-plane.

Finding the coordinates of L (foot of perpendicular on xy-plane):

The xy-plane is the set of all points $(x, y, z)$ where the z-coordinate is zero. Any point on the xy-plane has coordinates of the form $(x, y, 0)$.

The foot of the perpendicular from $P(3, 4, 5)$ to the xy-plane is the point on the xy-plane that has the same x and y coordinates as P. This is because the perpendicular line segment from P to the xy-plane is parallel to the z-axis.

Therefore, the coordinates of L are obtained by setting the z-coordinate of P to zero.

The coordinates of L are $(3, 4, 0)$.

Finding the coordinates of M (foot of perpendicular on yz-plane):

The yz-plane is the set of all points $(x, y, z)$ where the x-coordinate is zero. Any point on the yz-plane has coordinates of the form $(0, y, z)$.

The foot of the perpendicular from $P(3, 4, 5)$ to the yz-plane is the point on the yz-plane that has the same y and z coordinates as P. This is because the perpendicular line segment from P to the yz-plane is parallel to the x-axis.

Therefore, the coordinates of M are obtained by setting the x-coordinate of P to zero.

The coordinates of M are $(0, 4, 5)$.

Finding the coordinates of N (foot of perpendicular on zx-plane):

The zx-plane is the set of all points $(x, y, z)$ where the y-coordinate is zero. Any point on the zx-plane has coordinates of the form $(x, 0, z)$.

The foot of the perpendicular from $P(3, 4, 5)$ to the zx-plane is the point on the zx-plane that has the same x and z coordinates as P. This is because the perpendicular line segment from P to the zx-plane is parallel to the y-axis.

Therefore, the coordinates of N are obtained by setting the y-coordinate of P to zero.

The coordinates of N are $(3, 0, 5)$.

Given the point $P(3, 4, 5)$.

From Example 4, we know the coordinates of the feet of the perpendiculars drawn from P on the xy, yz, and zx-planes are:

L (on xy-plane) has coordinates $(3, 4, 0)$.

M (on yz-plane) has coordinates $(0, 4, 5)$.

N (on zx-plane) has coordinates $(3, 0, 5)$.

We need to find the distance of points L, M, and N from the point P.

The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in 3D space is given by the distance formula:

$\text{Distance} = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$

Distance PL:

Point P is $(3, 4, 5)$ and point L is $(3, 4, 0)$.

$PL = \sqrt{(3-3)^2 + (4-4)^2 + (0-5)^2}$

$PL = \sqrt{(0)^2 + (0)^2 + (-5)^2}$

$PL = \sqrt{0 + 0 + 25}$

$PL = \sqrt{25}$

$PL = 5$ units.

Distance PM:

Point P is $(3, 4, 5)$ and point M is $(0, 4, 5)$.

$PM = \sqrt{(0-3)^2 + (4-4)^2 + (5-5)^2}$

$PM = \sqrt{(-3)^2 + (0)^2 + (0)^2}$

$PM = \sqrt{9 + 0 + 0}$

$PM = \sqrt{9}$

$PM = 3$ units.

Distance PN:

Point P is $(3, 4, 5)$ and point N is $(3, 0, 5)$.

$PN = \sqrt{(3-3)^2 + (0-4)^2 + (5-5)^2}$

$PN = \sqrt{(0)^2 + (-4)^2 + (0)^2}$

$PN = \sqrt{0 + 16 + 0}$

$PN = \sqrt{16}$

$PN = 4$ units.

Thus, the distances are $PL = 5$, $PM = 3$, and $PN = 4$. Note that these distances are the absolute values of the coordinates of P that are zeroed out in L, M, and N respectively when projecting onto the planes.

Given points are $P(2, 4, 6)$, $Q(-2, -2, -2)$, and $R(6, 10, 14)$.

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

Calculate distances:

$PQ = \sqrt{(-2 - 2)^2 + (-2 - 4)^2 + (-2 - 6)^2} = \sqrt{(-4)^2 + (-6)^2 + (-8)^2} = \sqrt{16 + 36 + 64} = \sqrt{116} = 2\sqrt{29}$

$QR = \sqrt{(6 - (-2))^2 + (10 - (-2))^2 + (14 - (-2))^2} = \sqrt{8^2 + 12^2 + 16^2} = \sqrt{64 + 144 + 256} = \sqrt{464} = 4\sqrt{29}$

$PR = \sqrt{(6 - 2)^2 + (10 - 4)^2 + (14 - 6)^2} = \sqrt{4^2 + 6^2 + 8^2} = \sqrt{16 + 36 + 64} = \sqrt{116} = 2\sqrt{29}$

Check for collinearity:

The distances are $PQ = 2\sqrt{29}$, $QR = 4\sqrt{29}$, and $PR = 2\sqrt{29}$.

Sum of the two smaller distances: $PQ + PR = 2\sqrt{29} + 2\sqrt{29} = 4\sqrt{29}$.

This sum is equal to the largest distance $QR = 4\sqrt{29}$.

Since $PQ + PR = QR$, the points P, Q, and R are collinear. The point P lies between Q and R.

Let the coordinates of the point equidistant from the four given points be $P(x, y, z)$.

The given points are $O(0, 0, 0)$, $A(l, 0, 0)$, $B(0, m, 0)$, and $C(0, 0, n)$.

According to the problem, the distance from P to each of these points is equal. Thus, $PO = PA = PB = PC$.

Squaring the distances, we have $PO^2 = PA^2 = PB^2 = PC^2$.

Using the distance formula, the square of the distance from a point $(x_1, y_1, z_1)$ to $(x_2, y_2, z_2)$ is $(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2$.

Let's calculate the squared distances from $P(x, y, z)$ to each of the given points:

$PO^2 = (x-0)^2 + (y-0)^2 + (z-0)^2 = x^2 + y^2 + z^2$

$PA^2 = (x-l)^2 + (y-0)^2 + (z-0)^2 = (x-l)^2 + y^2 + z^2$

$PB^2 = (x-0)^2 + (y-m)^2 + (z-0)^2 = x^2 + (y-m)^2 + z^2$

$PC^2 = (x-0)^2 + (y-0)^2 + (z-n)^2 = x^2 + y^2 + (z-n)^2$

Now, we equate the squared distances $PO^2$ with $PA^2$, $PB^2$, and $PC^2$ to find the values of $x, y, z$.

Equating $PO^2$ and $PA^2$:

$x^2 + y^2 + z^2 = (x-l)^2 + y^2 + z^2$

$x^2 = x^2 - 2lx + l^2$

$0 = -2lx + l^2$

$2lx = l^2$

Assuming $l \neq 0$, we divide by $2l$ to get:

$x = \frac{l^2}{2l} = \frac{l}{2}$

Equating $PO^2$ and $PB^2$:

$x^2 + y^2 + z^2 = x^2 + (y-m)^2 + z^2$

$y^2 = (y-m)^2$

$y^2 = y^2 - 2my + m^2$

$0 = -2my + m^2$

$2my = m^2$

Assuming $m \neq 0$, we divide by $2m$ to get:

$y = \frac{m^2}{2m} = \frac{m}{2}$

Equating $PO^2$ and $PC^2$:

$x^2 + y^2 + z^2 = x^2 + y^2 + (z-n)^2$

$z^2 = (z-n)^2$

$z^2 = z^2 - 2nz + n^2$

$0 = -2nz + n^2$

$2nz = n^2$

Assuming $n \neq 0$, we divide by $2n$ to get:

$z = \frac{n^2}{2n} = \frac{n}{2}$

Assuming $l, m, n$ are non-zero (which ensures A, B, C are distinct from O and the points are non-coplanar), the coordinates of the point equidistant from the four given points O, A, B, and C are $\left(\frac{l}{2}, \frac{m}{2}, \frac{n}{2}\right)$.

This point is the center of the circumsphere passing through the four points.

Let the point on the x-axis equidistant from A and B be $P(x, 0, 0)$.

The given points are $A(3, 2, 2)$ and $B(5, 5, 4)$.

Since P is equidistant from A and B, we have $PA = PB$.

Squaring both sides, we get $PA^2 = PB^2$.

Using the distance formula squared, $(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2$:

Calculate $PA^2$:

$PA^2 = (x - 3)^2 + (0 - 2)^2 + (0 - 2)^2$

$PA^2 = (x - 3)^2 + (-2)^2 + (-2)^2$

$PA^2 = (x^2 - 6x + 9) + 4 + 4$

$PA^2 = x^2 - 6x + 17$

Calculate $PB^2$:

$PB^2 = (x - 5)^2 + (0 - 5)^2 + (0 - 4)^2$

$PB^2 = (x - 5)^2 + (-5)^2 + (-4)^2$

$PB^2 = (x^2 - 10x + 25) + 25 + 16$

$PB^2 = x^2 - 10x + 66$

Equating $PA^2$ and $PB^2$:

$x^2 - 6x + 17 = x^2 - 10x + 66$

Subtracting $x^2$ from both sides:

$-6x + 17 = -10x + 66$

Adding $10x$ to both sides and subtracting 17 from both sides:

$10x - 6x = 66 - 17$

$4x = 49$

$x = \frac{49}{4}$

Thus, the x-coordinate of the point is $\frac{49}{4}$. Since the point is on the x-axis, its y and z coordinates are 0.

The coordinates of the required point are $\left(\frac{49}{4}, 0, 0\right)$.

Let the point on the y-axis be $P$. Any point on the y-axis has coordinates of the form $(0, y, 0)$. So, let $P(0, y, 0)$.

Let the given point be $A(1, 2, 3)$.

The distance between P and A is given as $\sqrt{10}$.

Using the distance formula between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in 3D space:

$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$

The distance PA is $\sqrt{(1-0)^2 + (2-y)^2 + (3-0)^2}$.

We are given $PA = \sqrt{10}$.

So, $\sqrt{(1-0)^2 + (2-y)^2 + (3-0)^2} = \sqrt{10}$

Squaring both sides:

$(1-0)^2 + (2-y)^2 + (3-0)^2 = 10$

$1^2 + (2-y)^2 + 3^2 = 10$

$1 + (4 - 4y + y^2) + 9 = 10$

$y^2 - 4y + 4 + 1 + 9 = 10$

$y^2 - 4y + 14 = 10$

Subtracting 10 from both sides:

$y^2 - 4y + 14 - 10 = 0$

$y^2 - 4y + 4 = 0$

We solve the quadratic equation for $y$. This equation can be factored as a perfect square:

$(y - 2)^2 = 0$

Taking the square root of both sides:

$y - 2 = 0$

$y = 2$

Since $y=2$, the coordinates of the point on the y-axis are $(0, y, 0) = (0, 2, 0)$.

The point on the y-axis equidistant from the given point is $(0, 2, 0)$.

Let the two given points be $P_1(2, 3, 5)$ and $P_2(5, 9, 7)$.

When planes are drawn through these two points parallel to the coordinate planes, they form a rectangular parallelepiped (also known as a cuboid). The two given points $P_1$ and $P_2$ are opposite vertices of this parallelepiped.

The edges of the parallelepiped are parallel to the x, y, and z axes. The lengths of the edges are determined by the absolute differences of the coordinates of the two opposite vertices.

Length of the edges:

The length of the edge parallel to the x-axis is the absolute difference of the x-coordinates:

Length along x-axis = $|x_2 - x_1| = |5 - 2| = |3| = 3$ units.

The length of the edge parallel to the y-axis is the absolute difference of the y-coordinates:

Length along y-axis = $|y_2 - y_1| = |9 - 3| = |6| = 6$ units.

The length of the edge parallel to the z-axis is the absolute difference of the z-coordinates:

Length along z-axis = $|z_2 - z_1| = |7 - 5| = |2| = 2$ units.

The lengths of the edges of the parallelepiped are 3, 6, and 2 units.

Length of the diagonal:

The length of the diagonal of a rectangular parallelepiped with edge lengths $a, b, c$ is given by $\sqrt{a^2 + b^2 + c^2}$.

Alternatively, the diagonal connecting the two given opposite vertices $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$ is simply the distance between these two points, which is given by the distance formula:

Diagonal length = $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$

Using $P_1(2, 3, 5)$ and $P_2(5, 9, 7)$:

Diagonal length = $\sqrt{(5-2)^2 + (9-3)^2 + (7-5)^2}$

Diagonal length = $\sqrt{3^2 + 6^2 + 2^2}$

Diagonal length = $\sqrt{9 + 36 + 4}$

Diagonal length = $\sqrt{49}$

Diagonal length = $7$ units.

The lengths of the edges of the parallelepiped are 3, 6, and 2 units.

The length of the diagonal of the parallelepiped is 7 units.

Let the given points be $A(0, 7, 10)$, $B(-1, 6, 6)$, and $C(-4, 9, 6)$.

To determine the type of triangle formed by these points, we will calculate the squares of the distances between each pair of points using the distance formula squared: $d^2 = (x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2$.

Calculate $AB^2$:

$A(0, 7, 10)$, $B(-1, 6, 6)$

$AB^2 = (-1 - 0)^2 + (6 - 7)^2 + (6 - 10)^2$

$AB^2 = (-1)^2 + (-1)^2 + (-4)^2$

$AB^2 = 1 + 1 + 16$

$AB^2 = 18$

Calculate $BC^2$:

$B(-1, 6, 6)$, $C(-4, 9, 6)$

$BC^2 = (-4 - (-1))^2 + (9 - 6)^2 + (6 - 6)^2$

$BC^2 = (-4 + 1)^2 + (3)^2 + (0)^2$

$BC^2 = (-3)^2 + 3^2 + 0^2$

$BC^2 = 9 + 9 + 0$

$BC^2 = 18$

Calculate $AC^2$:

$A(0, 7, 10)$, $C(-4, 9, 6)$

$AC^2 = (-4 - 0)^2 + (9 - 7)^2 + (6 - 10)^2$

$AC^2 = (-4)^2 + (2)^2 + (-4)^2$

$AC^2 = 16 + 4 + 16$

$AC^2 = 36$

We have $AB^2 = 18$, $BC^2 = 18$, and $AC^2 = 36$.

Since $AB^2 = BC^2 = 18$, it follows that $AB = BC = \sqrt{18}$.

This means that two sides of the triangle ABC have equal length. Therefore, the triangle is an isosceles triangle.

Now, let's check if it is a right-angled triangle using the Pythagorean theorem. The theorem states that in a right-angled triangle, the square of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides.

In our case, the squared distances are 18, 18, and 36. The largest squared distance is $AC^2 = 36$. Let's check if the sum of the other two squared distances equals this value:

$AB^2 + BC^2 = 18 + 18 = 36$

We observe that $AB^2 + BC^2 = AC^2$ ($18 + 18 = 36$).

This confirms that the triangle satisfies the Pythagorean theorem. Therefore, the triangle is a right-angled triangle, with the right angle at the vertex B (opposite to side AC).

Since the triangle ABC is both isosceles (because $AB = BC$) and right-angled (because $AB^2 + BC^2 = AC^2$), the points (0, 7, 10), (–1, 6, 6) and (– 4, 9, 6) form a right-angled isosceles triangle.

Let the given points be $A(5, -1, 1)$, $B(7, -4, 7)$, $C(1, -6, 10)$, and $D(-1, -3, 4)$.

To show that these points form a rhombus, we need to prove that all four sides of the quadrilateral ABCD have equal length.

We will calculate the square of the distance between consecutive points using the squared distance formula between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$, which is $d^2 = (x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2$.

Calculate $AB^2$:

Using $A(5, -1, 1)$ and $B(7, -4, 7)$:

$AB^2 = (7 - 5)^2 + (-4 - (-1))^2 + (7 - 1)^2$

$AB^2 = (2)^2 + (-4 + 1)^2 + (6)^2$

$AB^2 = (2)^2 + (-3)^2 + (6)^2$

$AB^2 = 4 + 9 + 36$

$AB^2 = 49$

Calculate $BC^2$:

Using $B(7, -4, 7)$ and $C(1, -6, 10)$:

$BC^2 = (1 - 7)^2 + (-6 - (-4))^2 + (10 - 7)^2$

$BC^2 = (-6)^2 + (-6 + 4)^2 + (3)^2$

$BC^2 = (-6)^2 + (-2)^2 + (3)^2$

$BC^2 = 36 + 4 + 9$

$BC^2 = 49$

Calculate $CD^2$:

Using $C(1, -6, 10)$ and $D(-1, -3, 4)$:

$CD^2 = (-1 - 1)^2 + (-3 - (-6))^2 + (4 - 10)^2$

$CD^2 = (-2)^2 + (-3 + 6)^2 + (-6)^2$

$CD^2 = (-2)^2 + (3)^2 + (-6)^2$

$CD^2 = 4 + 9 + 36$

$CD^2 = 49$

Calculate $DA^2$:

Using $D(-1, -3, 4)$ and $A(5, -1, 1)$:

$DA^2 = (5 - (-1))^2 + (-1 - (-3))^2 + (1 - 4)^2$

$DA^2 = (5 + 1)^2 + (-1 + 3)^2 + (-3)^2$

$DA^2 = (6)^2 + (2)^2 + (-3)^2$

$DA^2 = 36 + 4 + 9$

$DA^2 = 49$

We have $AB^2 = 49$, $BC^2 = 49$, $CD^2 = 49$, and $DA^2 = 49$.

This means $AB = \sqrt{49} = 7$, $BC = \sqrt{49} = 7$, $CD = \sqrt{49} = 7$, and $DA = \sqrt{49} = 7$.

Since all four sides are equal in length ($AB = BC = CD = DA = 7$), the quadrilateral ABCD is a rhombus.

Let the two points be $P_1(2, 4, 5)$ and $P_2(3, 5, -4)$.

The xz-plane is defined by the equation $y = 0$.

Let the point where the line segment joining $P_1$ and $P_2$ is divided by the xz-plane be $R$.

Since R lies on the xz-plane, its y-coordinate is 0. Let $R$ divide the line segment $P_1P_2$ in the ratio $\lambda : 1$.

Using the section formula, the coordinates of the point R that divides the segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in the ratio $\lambda : 1$ are given by:

$R = \left( \frac{1 \cdot x_1 + \lambda \cdot x_2}{1+\lambda}, \frac{1 \cdot y_1 + \lambda \cdot y_2}{1+\lambda}, \frac{1 \cdot z_1 + \lambda \cdot z_2}{1+\lambda} \right)$

Substituting the coordinates of $P_1(2, 4, 5)$ and $P_2(3, 5, -4)$, the coordinates of R are:

$R = \left( \frac{1(2) + \lambda(3)}{1+\lambda}, \frac{1(4) + \lambda(5)}{1+\lambda}, \frac{1(5) + \lambda(-4)}{1+\lambda} \right)$

$R = \left( \frac{2 + 3\lambda}{1+\lambda}, \frac{4 + 5\lambda}{1+\lambda}, \frac{5 - 4\lambda}{1+\lambda} \right)$

Since R lies on the xz-plane, its y-coordinate is 0.

Therefore, we set the y-coordinate of R equal to 0:

$\frac{4 + 5\lambda}{1+\lambda} = 0$

For this fraction to be zero, the numerator must be zero (assuming the denominator is non-zero):

$4 + 5\lambda = 0$

$5\lambda = -4$

$\lambda = -\frac{4}{5}$

The ratio is $\lambda : 1 = -\frac{4}{5} : 1$. The negative value of $\lambda$ indicates that the point R divides the line segment $P_1P_2$ externally.

The magnitude of the ratio is $|-\frac{4}{5}| : 1 = \frac{4}{5} : 1$. Multiplying by 5, the ratio is $4 : 5$.

Thus, the xz-plane divides the line segment joining the points (2, 4, 5) and (3, 5, – 4) externally in the ratio 4 : 5.

Let the given points be $A(-2, 0, 6)$ and $B(10, -6, -12)$.

We are looking for a point P that is five-sixth of the way from A to B. This means that the distance from A to P is $\frac{5}{6}$ of the distance from A to B.

If P lies on the line segment AB such that AP = $\frac{5}{6}$ AB, then the remaining part PB must be $AB - AP = AB - \frac{5}{6} AB = \frac{1}{6} AB$.

Therefore, the point P divides the line segment AB internally in the ratio AP : PB = $\frac{5}{6} AB : \frac{1}{6} AB = 5 : 1$.

Let the ratio be $m : n = 5 : 1$.

Using the section formula for internal division, the coordinates of the point $P(x, y, z)$ that divides the line segment joining $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ in the ratio $m : n$ are given by:

$x = \frac{m x_2 + n x_1}{m+n}$

$y = \frac{m y_2 + n y_1}{m+n}$

$z = \frac{m z_2 + n z_1}{m+n}$

Substituting the coordinates of $A(-2, 0, 6)$ and $B(10, -6, -12)$ and the ratio $m : n = 5 : 1$:

Calculate the x-coordinate of P:

$x = \frac{5(10) + 1(-2)}{5+1} = \frac{50 - 2}{6} = \frac{48}{6} = 8$

Calculate the y-coordinate of P:

$y = \frac{5(-6) + 1(0)}{5+1} = \frac{-30 + 0}{6} = \frac{-30}{6} = -5$

Calculate the z-coordinate of P:

$z = \frac{5(-12) + 1(6)}{5+1} = \frac{-60 + 6}{6} = \frac{-54}{6} = -9$

Thus, the coordinates of the point P are $(8, -5, -9)$.

We are given a rectangular parallelepiped (cuboid) with one vertex at the origin $O(0, 0, 0)$ and another vertex at $P(3, 5, 6)$. The cuboid is in the first octant, and its edges lie along the coordinate axes. This means the faces of the parallelepiped are parallel to the coordinate planes (xy, yz, zx planes).

The given vertex $P(3, 5, 6)$ is the vertex diagonally opposite to the origin. Its coordinates $(3, 5, 6)$ represent the lengths of the edges along the positive x, y, and z axes, respectively, originating from the origin.

Vertices of the Parallelepiped:

A rectangular parallelepiped has 8 vertices. Since one vertex is at the origin and the edges lie along the axes, the vertices will have coordinates $(x, y, z)$ where each coordinate is either 0 or the corresponding coordinate of the opposite vertex $(3, 5, 6)$.

The vertices are:

1. The origin: $O(0, 0, 0)$

2. The vertex on the x-axis: $(3, 0, 0)$

3. The vertex on the y-axis: $(0, 5, 0)$

4. The vertex on the z-axis: $(0, 0, 6)$

5. The vertex on the xy-plane (but not on an axis): $(3, 5, 0)$

6. The vertex on the yz-plane (but not on an axis): $(0, 5, 6)$

7. The vertex on the zx-plane (but not on an axis): $(3, 0, 6)$

8. The given vertex opposite to the origin: $(3, 5, 6)$

Edges of the Parallelepiped:

A rectangular parallelepiped has 12 edges, with 4 edges parallel to each coordinate axis. The lengths of the edges are determined by the dimensions of the parallelepiped, which are given by the coordinates of the vertex opposite to the origin.

The lengths of the edges are:

- Length along the x-direction: 3 units.

- Length along the y-direction: 5 units.

- Length along the z-direction: 6 units.

The edges can be described by the line segments connecting the vertices:

Edges parallel to the x-axis (length 3):

- From $(0, 0, 0)$ to $(3, 0, 0)$

- From $(0, 5, 0)$ to $(3, 5, 0)$

- From $(0, 0, 6)$ to $(3, 0, 6)$

- From $(0, 5, 6)$ to $(3, 5, 6)$

Edges parallel to the y-axis (length 5):

- From $(0, 0, 0)$ to $(0, 5, 0)$

- From $(3, 0, 0)$ to $(3, 5, 0)$

- From $(0, 0, 6)$ to $(0, 5, 6)$

- From $(3, 0, 6)$ to $(3, 5, 6)$

Edges parallel to the z-axis (length 6):

- From $(0, 0, 0)$ to $(0, 0, 6)$

- From $(3, 0, 0)$ to $(3, 0, 6)$

- From $(0, 5, 0)$ to $(0, 5, 6)$

- From $(3, 5, 0)$ to $(3, 5, 6)$

Let the given points be $A(3, 2, 0)$, $B(5, 3, 2)$, and $C(-9, 6, -3)$.

AD is the angle bisector of $\angle BAC$ and meets BC at D. According to the Angle Bisector Theorem, the angle bisector of a vertex angle divides the opposite side in the ratio of the lengths of the other two sides of the triangle.

For $\triangle ABC$, the angle bisector AD on side BC means that point D divides the side BC in the ratio $BD : DC = AB : AC$.

First, we need to calculate the lengths of the sides AB and AC using the distance formula in 3D: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.

Calculate the length of AB:

Using $A(3, 2, 0)$ and $B(5, 3, 2)$:

$AB = \sqrt{(5 - 3)^2 + (3 - 2)^2 + (2 - 0)^2}$

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

$AB = \sqrt{4 + 1 + 4}$

$AB = \sqrt{9}$

$AB = 3$

Calculate the length of AC:

Using $A(3, 2, 0)$ and $C(-9, 6, -3)$:

$AC = \sqrt{(-9 - 3)^2 + (6 - 2)^2 + (-3 - 0)^2}$

$AC = \sqrt{(-12)^2 + (4)^2 + (-3)^2}$

$AC = \sqrt{144 + 16 + 9}$

$AC = \sqrt{169}$

$AC = 13$

According to the Angle Bisector Theorem, D divides BC in the ratio $BD : DC = AB : AC$.

So, the ratio is $3 : 13$.

Let D divide BC internally in the ratio $m : n = 3 : 13$.

We will use the section formula for internal division to find the coordinates of D. If a point $D(x, y, z)$ divides the line segment joining $B(x_1, y_1, z_1)$ and $C(x_2, y_2, z_2)$ in the ratio $m : n$, its coordinates are given by:

$x = \frac{m x_2 + n x_1}{m+n}$

$y = \frac{m y_2 + n y_1}{m+n}$

$z = \frac{m z_2 + n z_1}{m+n}$

Substituting the coordinates of $B(5, 3, 2)$ and $C(-9, 6, -3)$ and the ratio $m : n = 3 : 13$:

Calculate the x-coordinate of D:

$x = \frac{3(-9) + 13(5)}{3+13} = \frac{-27 + 65}{16} = \frac{38}{16} = \frac{19}{8}$

Calculate the y-coordinate of D:

$y = \frac{3(6) + 13(3)}{3+13} = \frac{18 + 39}{16} = \frac{57}{16}$

Calculate the z-coordinate of D:

$z = \frac{3(-3) + 13(2)}{3+13} = \frac{-9 + 26}{16} = \frac{17}{16}$

Thus, the coordinates of the point D are $\left(\frac{19}{8}, \frac{57}{16}, \frac{17}{16}\right)$.

Let the required point in the yz-plane be $P$. Since the point lies in the yz-plane, its x-coordinate is 0. Thus, the coordinates of P are of the form $(0, y, z)$.

The given points are $A(2, 0, 3)$, $B(0, 3, 2)$, and $C(0, 0, 1)$.

According to the problem, the point P is equidistant from A, B, and C. This means $PA = PB = PC$.

Squaring the distances, we have $PA^2 = PB^2 = PC^2$.

Using the distance formula squared, $d^2 = (x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2$, we calculate the squared distances from $P(0, y, z)$ to A, B, and C.

Calculate $PA^2$:

$PA^2 = (0 - 2)^2 + (y - 0)^2 + (z - 3)^2$

$PA^2 = (-2)^2 + y^2 + (z^2 - 6z + 9)$

$PA^2 = 4 + y^2 + z^2 - 6z + 9$

$PA^2 = y^2 + z^2 - 6z + 13$

Calculate $PB^2$:

$PB^2 = (0 - 0)^2 + (y - 3)^2 + (z - 2)^2$

$PB^2 = 0^2 + (y^2 - 6y + 9) + (z^2 - 4z + 4)$

$PB^2 = y^2 - 6y + z^2 - 4z + 13$

Calculate $PC^2$:

$PC^2 = (0 - 0)^2 + (y - 0)^2 + (z - 1)^2$

$PC^2 = 0^2 + y^2 + (z^2 - 2z + 1)$

$PC^2 = y^2 + z^2 - 2z + 1$

Now, we equate the squared distances to form equations.

Equating $PA^2$ and $PB^2$:

$y^2 + z^2 - 6z + 13 = y^2 - 6y + z^2 - 4z + 13$

Subtract $y^2 + z^2 + 13$ from both sides:

$-6z = -6y - 4z$

Adding $6y$ and $6z$ to both sides:

$6y = 6z - 4z$

$6y = 2z$

Divide by 2:

Equating $PB^2$ and $PC^2$:

$y^2 - 6y + z^2 - 4z + 13 = y^2 + z^2 - 2z + 1$

Subtract $y^2 + z^2$ from both sides:

$-6y - 4z + 13 = -2z + 1$

Add $2z$ and subtract $13$ from both sides:

$-6y - 4z + 2z = 1 - 13$

$-6y - 2z = -12$

Divide by -2:

Now we solve the system of linear equations using equations (i) and (ii).

Substitute the value of $z$ from equation (i) into equation (ii):

$3y + (3y) = 6$

$6y = 6$

Divide by 6:

$y = 1$

Substitute the value of $y=1$ back into equation (i):

$z = 3(1)$

$z = 3$

The coordinates of the point P are $(0, y, z)$. Since $y=1$ and $z=3$, the coordinates are $(0, 1, 3)$.

The point in the yz-plane equidistant from the three given points is (0, 1, 3).

Let the given point be $P(3, 4, 5)$.

The foot of the perpendicular from a point $(x, y, z)$ to the y-axis is the point on the y-axis with coordinates $(0, y, 0)$.

So, the foot of the perpendicular from $P(3, 4, 5)$ on the y-axis is the point $M(0, 4, 0)$.

The question asks for the length of the perpendicular segment PM, which is the distance from point P to its foot on the y-axis.

Using the distance formula between $P(3, 4, 5)$ and $M(0, 4, 0)$:

$PM = \sqrt{(0 - 3)^2 + (4 - 4)^2 + (0 - 5)^2}$

$PM = \sqrt{(-3)^2 + (0)^2 + (-5)^2}$

$PM = \sqrt{9 + 0 + 25}$

$PM = \sqrt{34}$

The length of the foot of the perpendicular (meaning the length of the perpendicular segment) is $\sqrt{34}$.

Comparing this with the given options, we find that $\sqrt{34}$ corresponds to option (B).

The correct answer is (B) $\sqrt{34}$.

Let the given point be $P(6, 7, 8)$.

The xy-plane is defined by the equation $z = 0$.

The perpendicular distance from a point $(x_1, y_1, z_1)$ to the plane $Ax + By + Cz + D = 0$ is given by $\frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$.

For the xy-plane, the equation is $z = 0$, which can be written as $0 \cdot x + 0 \cdot y + 1 \cdot z + 0 = 0$. So, $A=0, B=0, C=1, D=0$.

The perpendicular distance from $P(6, 7, 8)$ to the xy-plane ($z=0$) is:

Distance $= \frac{|0(6) + 0(7) + 1(8) + 0|}{\sqrt{0^2 + 0^2 + 1^2}}$

Distance $= \frac{|8|}{\sqrt{1}}$

Distance $= \frac{8}{1} = 8$

Alternatively, the perpendicular distance of a point $(x, y, z)$ from the xy-plane is simply the absolute value of its z-coordinate.

For the point $P(6, 7, 8)$, the perpendicular distance from the xy-plane is $|z| = |8| = 8$.

Comparing this with the given options, we find that 8 corresponds to option (A).

The correct answer is (A) 8.

Let the given point be $P(6, 7, 8)$.

The xy-plane is the set of all points $(x, y, z)$ where the z-coordinate is 0.

The foot of the perpendicular drawn from a point $(x_1, y_1, z_1)$ to the xy-plane is the point on the xy-plane that has the same x and y coordinates as the original point, and its z-coordinate is 0.

This is because the perpendicular line segment from P to the xy-plane is parallel to the z-axis.

For the point $P(6, 7, 8)$, the foot of the perpendicular L on the xy-plane will have the x-coordinate of P, the y-coordinate of P, and a z-coordinate of 0.

Therefore, the coordinates of point L are $(6, 7, 0)$.

Comparing this with the given options, we find that $(6, 7, 0)$ corresponds to option (B).

The correct answer is (B) (6, 7, 0).

Let the given point be $P(6, 7, 8)$.

The x-axis is the set of all points $(x, y, z)$ where the y and z coordinates are both 0.

The foot of the perpendicular drawn from a point $(x_1, y_1, z_1)$ to the x-axis is the point on the x-axis that has the same x-coordinate as the original point, and its y and z coordinates are 0.

This is because the perpendicular line segment from P to the x-axis is parallel to the yz-plane.

For the point $P(6, 7, 8)$, the foot of the perpendicular L on the x-axis will have the x-coordinate of P, and the y and z coordinates equal to 0.

Therefore, the coordinates of point L are $(6, 0, 0)$.

Comparing this with the given options, we find that $(6, 0, 0)$ corresponds to option (A).

The correct answer is (A) (6, 0, 0).

The locus of a point is the set of all points that satisfy a given condition or set of conditions.

We are given the conditions $y = 0$ and $z = 0$.

In a three-dimensional coordinate system, the y-axis is defined by $x=0$ and $z=0$. The z-axis is defined by $x=0$ and $y=0$. The x-axis is defined by $y=0$ and $z=0$.

The set of all points $(x, y, z)$ such that $y=0$ and $z=0$ means that the y-coordinate is zero and the z-coordinate is zero, while the x-coordinate can be any real number.

Points of the form $(x, 0, 0)$ lie on the x-axis.

For example, $(1, 0, 0)$, $(-5, 0, 0)$, $(\sqrt{2}, 0, 0)$ are all points on the x-axis.

Therefore, the locus of a point for which $y=0$ and $z=0$ is the x-axis.

Comparing this with the given options, we find that the equation of the x-axis corresponds to option (A).

The correct answer is (A) equation of x-axis.

Let the given point be $P(3, 4, 5)$.

The xz-plane is the set of all points $(x, y, z)$ where the y-coordinate is 0. The equation of the xz-plane is $y = 0$.

The foot of the perpendicular drawn from a point $(x_1, y_1, z_1)$ to the xz-plane is the point on the xz-plane that has the same x and z coordinates as the original point, and its y-coordinate is 0.

This is because the line segment representing the perpendicular from P to the xz-plane is parallel to the y-axis.

For the point $P(3, 4, 5)$, the foot of the perpendicular L on the xz-plane will have the x-coordinate of P, the z-coordinate of P, and the y-coordinate equal to 0.

Therefore, the coordinates of point L are $(3, 0, 5)$.

Comparing this with the given options, we find that $(3, 0, 5)$ corresponds to option (C).

The correct answer is (C) (3, 0, 5).

A plane parallel to the xy-plane has the equation $z = k$, where $k$ is a constant. This means that every point on such a plane has the same z-coordinate.

If a line is parallel to the xy-plane, all points on the line are at a constant distance from the xy-plane. The distance of a point $(x, y, z)$ from the xy-plane is $|z|$. For this distance to be constant for all points on the line, the absolute value of the z-coordinate must be constant.

If the line is above or below the xy-plane, the z-coordinate itself must be constant. If the line is on the xy-plane, the z-coordinate is 0 for all points.

Therefore, for a line parallel to the xy-plane, all the points on the line have equal z-coordinates.

A line is parallel to xy-plane if all the points on the line have equal z-coordinates.

The equation $x = b$ describes the set of all points $(x, y, z)$ in three-dimensional space where the x-coordinate is a constant value $b$, and the y and z coordinates can take any real values.

Consider the coordinate planes:

• The xy-plane has the equation $z = 0$.
• The yz-plane has the equation $x = 0$.
• The zx-plane (or xz-plane) has the equation $y = 0$.

The plane $x = b$ is perpendicular to the x-axis. Any plane perpendicular to the x-axis is parallel to the plane formed by the y and z axes, which is the yz-plane.

Every point on the plane $x=b$ is at a constant directed distance $b$ from the yz-plane.

The equation x = b represents a plane parallel to yz-plane.

Let the given point be $P(3, 5, 6)$.

The y-axis is the set of points $(x, y, z)$ where $x=0$ and $z=0$. Any point on the y-axis can be represented as $(0, y, 0)$.

The foot of the perpendicular from point $P(3, 5, 6)$ to the y-axis is the point on the y-axis with the same y-coordinate as P and x and z coordinates equal to 0. This point is $M(0, 5, 0)$.

The perpendicular distance of P from the y-axis is the distance between P and M.

Using the distance formula between $P(3, 5, 6)$ and $M(0, 5, 0)$:

Distance $= \sqrt{(0 - 3)^2 + (5 - 5)^2 + (0 - 6)^2}$

Distance $= \sqrt{(-3)^2 + (0)^2 + (-6)^2}$

Distance $= \sqrt{9 + 0 + 36}$

Distance $= \sqrt{45}$

Distance $= \sqrt{9 \times 5} = 3\sqrt{5}$

Alternatively, the perpendicular distance of a point $(x, y, z)$ from the y-axis is the distance from $(x, y, z)$ to $(0, y, 0)$, which is $\sqrt{(x-0)^2 + (y-y)^2 + (z-0)^2} = \sqrt{x^2 + 0^2 + z^2} = \sqrt{x^2 + z^2}$.

For the point $P(3, 5, 6)$, the perpendicular distance from the y-axis is $\sqrt{3^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}$.

Perpendicular distance of the point P (3, 5, 6) from y-axis is $3\sqrt{5}$.

Let the given point be $P(3, 4, 5)$.

The zx-plane (or xz-plane) is the plane containing the x-axis and the z-axis. The characteristic property of any point in the zx-plane is that its y-coordinate is always zero.

The foot of the perpendicular from a point to a plane is the point on the plane that is closest to the given point. For a coordinate plane like the zx-plane, the foot of the perpendicular from a point $(x_1, y_1, z_1)$ is found by setting the coordinate that is zero on the plane to zero, while keeping the other coordinates.

Since the zx-plane has $y=0$, the foot of the perpendicular from $P(3, 4, 5)$ onto the zx-plane will have the same x and z coordinates as P, and its y-coordinate will be 0.

Therefore, the coordinates of the foot of the perpendicular L are $(3, 0, 5)$.

L is the foot of perpendicular drawn from the point P (3, 4, 5) on zx-planes. The coordinates of L are (3, 0, 5).

Let the given point be $P(a, b, c)$.

The z-axis is the set of points $(x, y, z)$ where $x=0$ and $y=0$. Any point on the z-axis can be represented as $(0, 0, z)$.

The foot of the perpendicular drawn from a point $(x_1, y_1, z_1)$ to the z-axis is the point on the z-axis with the same z-coordinate as the original point and x and y coordinates equal to 0.

So, the foot of the perpendicular from $P(a, b, c)$ on the z-axis is the point $N(0, 0, c)$.

The question asks for the length of the foot of the perpendicular drawn from the point P on the z-axis. This refers to the length of the perpendicular segment PN, which is the distance between P and N.

Using the distance formula between $P(a, b, c)$ and $N(0, 0, c)$:

Distance $= \sqrt{(0 - a)^2 + (0 - b)^2 + (c - c)^2}$

Distance $= \sqrt{(-a)^2 + (-b)^2 + (0)^2}$

Distance $= \sqrt{a^2 + b^2 + 0}$

Distance $= \sqrt{a^2 + b^2}$

The length of the foot of perpendicular drawn from the point P (a, b, c) on z-axis is $\sqrt{a^2 + b^2}$.

The y-axis is a straight line and the z-axis is a straight line. These two lines intersect at the origin $(0, 0, 0)$.

In geometry, two intersecting lines uniquely determine a plane.

The plane containing the y-axis and the z-axis is the set of all points $(x, y, z)$ where the x-coordinate is 0. This plane is indeed called the yz-plane, and its equation is $x=0$.

Therefore, the y-axis and z-axis together determine the yz-plane.

The statement is True.

The eight octants in a three-dimensional coordinate system are determined by the signs of the x, y, and z coordinates.

Let's list the sign conventions for the coordinates $(x, y, z)$ in each octant:

• I Octant: $(+, +, +)$ (where $x>0, y>0, z>0$)
• II Octant: $(-, +, +)$ (where $x<0, y>0, z>0$)
• III Octant: $(-, -, +)$ (where $x<0, y<0, z>0$)
• IV Octant: $(+, -, +)$ (where $x>0, y<0, z>0$)
• V Octant: $(+, +, -)$ (where $x>0, y>0, z<0$)
• VI Octant: $(-, +, -)$ (where $x<0, y>0, z<0$)
• VII Octant: $(-, -, -)$ (where $x<0, y<0, z<0$)
• VIII Octant: $(+, -, -)$ (where $x>0, y<0, z<0$)

The given point is $(4, 5, -6)$. Let's examine the signs of its coordinates:

• x-coordinate is $4$, which is positive ($+$).
• y-coordinate is $5$, which is positive ($+$).
• z-coordinate is $-6$, which is negative ($-$).

The signs of the coordinates are $(+, +, -)$.

Looking at the list of octants, the $(+, +, -)$ sign combination corresponds to the V Octant.

The statement claims that the point $(4, 5, -6)$ lies in the VI^th octant, but based on the signs of the coordinates, it lies in the V Octant.

The statement is False.

The xy-plane is defined by the equation $z = 0$. This plane contains all points $(x, y, 0)$ where $x$ and $y$ can be any real numbers.

The xz-plane (or zx-plane) is defined by the equation $y = 0$. This plane contains all points $(x, 0, z)$ where $x$ and $z$ can be any real numbers.

The intersection of the xy-plane and the xz-plane is the set of points that lie on both planes simultaneously. This means the coordinates of the points in the intersection must satisfy both equations:

$z = 0$

$y = 0$

So, a point $(x, y, z)$ is in the intersection if and only if $y=0$ and $z=0$. The x-coordinate can be any real number.

The set of all points $(x, 0, 0)$ where $x \in \mathbb{R}$ is precisely the definition of the x-axis.

Therefore, the x-axis is the intersection of the xy-plane and the xz-plane.

The statement is True.

In a standard three-dimensional Cartesian coordinate system, the three coordinate planes (xy-plane, yz-plane, and zx-plane) are mutually perpendicular to each other. Their intersection is the origin.

These three planes divide the entire space into regions. Each region is characterized by the signs of the x, y, and z coordinates.

For example, the region where $x>0, y>0, z>0$ is one region (the first octant). The region where $x<0, y>0, z>0$ is another region (the second octant).

Since each coordinate can be positive ($>0$) or negative ($<0$) relative to the plane where that coordinate is zero, there are $2 \times 2 \times 2 = 8$ possible combinations of signs for the three coordinates (excluding points on the planes or axes).

Each of these 8 combinations of signs corresponds to one of the 8 octants into which the space is divided by the three mutually perpendicular coordinate planes.

The statement is True.

The equation of a plane is given by $z = 6$. This means that for any point $(x, y, z)$ on this plane, the z-coordinate is always 6, regardless of the values of x and y.

A plane parallel to the xy-plane has an equation of the form $z = k$, where $k$ is a constant. Since the given equation is $z = 6$, it fits this form, so the plane is indeed parallel to the xy-plane.

The z-intercept of a plane is the z-coordinate of the point where the plane intersects the z-axis. The z-axis is defined by the equations $x=0$ and $y=0$.

To find the z-intercept, we set $x=0$ and $y=0$ in the equation of the plane. In this case, the equation is already $z = 6$. So, any point on the plane with $x=0$ and $y=0$ has a z-coordinate of 6. The point of intersection with the z-axis is $(0, 0, 6)$.

The z-intercept is the z-coordinate of this intersection point, which is 6 units.

The statement that the plane $z=6$ is parallel to the xy-plane and has a z-intercept of 6 units is correct.

The statement is True.

The yz-plane is defined as the plane that contains both the y-axis and the z-axis.

Any point lying on the y-axis has coordinates of the form $(0, y, 0)$.

Any point lying on the z-axis has coordinates of the form $(0, 0, z)$.

Any linear combination of points on the y-axis and z-axis, or any point that can be reached by starting at the origin and moving only parallel to the y-axis and/or the z-axis, will have an x-coordinate of 0.

Conversely, any point $(x, y, z)$ where $x=0$ has the form $(0, y, z)$. Such a point lies in the plane that contains the y-axis and is parallel to the z-axis, or equivalently, contains the z-axis and is parallel to the y-axis. This is precisely the yz-plane.

Therefore, the set of all points $(x, y, z)$ satisfying the equation $x=0$ is the yz-plane.

The statement is True.

The x-axis is a straight line in the three-dimensional coordinate system.

Points on the x-axis are characterized by having their y-coordinate and z-coordinate equal to zero.

A point on the x-axis can be represented in the general form $(x, 0, 0)$, where $x$ is the x-coordinate of the point.

If the x-coordinate of a point on the x-axis is given as $x_0$, then its y and z coordinates must be 0.

Therefore, the coordinates of a point on the x-axis with x-coordinate $x_0$ are indeed $(x_0, 0, 0)$.

The statement is True.

The equation $x = x_0$, where $x_0$ is a constant, represents the set of all points $(x, y, z)$ in three-dimensional space such that the x-coordinate is always equal to $x_0$. The values of y and z can be any real numbers.

The yz-plane is defined by the equation $x=0$.

A plane parallel to the yz-plane is a plane where the distance from any point on the plane to the yz-plane is constant. The distance of a point $(x, y, z)$ from the yz-plane (the plane $x=0$) is $|x|$.

If the equation of a plane is $x = x_0$, then for any point on this plane, the x-coordinate is fixed at $x_0$. The distance from any point $(x_0, y, z)$ on this plane to the yz-plane is $|x_0|$, which is a constant.

Thus, the plane $x = x_0$ is parallel to the yz-plane (unless $x_0=0$, in which case it is the yz-plane itself).

The statement is True.

Let's solve each part of Column C1.

(a) Finding the third vertex:

Let the two given vertices be $A(3, -5, 7)$ and $B(-1, 7, -6)$. Let the third vertex be $C(x, y, z)$.

The centroid of the triangle ABC is given as the origin $G(0, 0, 0)$.

The coordinates of the centroid are $G\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.

Substituting the given values:

$\frac{3 + (-1) + x}{3} = 0 \implies 2 + x = 0 \implies x = -2$

$\frac{-5 + 7 + y}{3} = 0 \implies 2 + y = 0 \implies y = -2$

$\frac{7 + (-6) + z}{3} = 0 \implies 1 + z = 0 \implies z = -1$

The third vertex is $(-2, -2, -1)$. This matches C2(ii).

(b) Finding the centroid from midpoints:

Let the midpoints of the sides of the triangle be $D(1, 2, -3)$, $E(3, 0, 1)$, and $F(-1, 1, -4)$.

The centroid of a triangle formed by vertices is the same as the centroid of the triangle formed by the midpoints of its sides.

Centroid of triangle DEF = $\left(\frac{1 + 3 + (-1)}{3}, \frac{2 + 0 + 1}{3}, \frac{-3 + 1 + (-4)}{3}\right)$

= $\left(\frac{3}{3}, \frac{3}{3}, \frac{-6}{3}\right)$

= $(1, 1, -2)$. This matches C2(iv).

(c) Determining the shape of the quadrilateral:

Let the points be $A(3, -1, -1)$, $B(5, -4, 0)$, $C(2, 3, -2)$, $D(0, 6, -3)$.

Calculate the square of the lengths of the sides:

$AB^2 = (5-3)^2 + (-4-(-1))^2 + (0-(-1))^2 = 2^2 + (-3)^2 + 1^2 = 4 + 9 + 1 = 14$

$BC^2 = (2-5)^2 + (3-(-4))^2 + (-2-0)^2 = (-3)^2 + 7^2 + (-2)^2 = 9 + 49 + 4 = 62$

$CD^2 = (0-2)^2 + (6-3)^2 + (-3-(-2))^2 = (-2)^2 + 3^2 + (-1)^2 = 4 + 9 + 1 = 14$

$DA^2 = (3-0)^2 + (-1-6)^2 + (-1-(-3))^2 = 3^2 + (-7)^2 + 2^2 = 9 + 49 + 4 = 62$

Since $AB^2 = CD^2$ and $BC^2 = DA^2$, the opposite sides are equal ($AB=CD=\sqrt{14}$, $BC=DA=\sqrt{62}$). This means the quadrilateral is a parallelogram.

Calculate the square of the lengths of the diagonals:

$AC^2 = (2-3)^2 + (3-(-1))^2 + (-2-(-1))^2 = (-1)^2 + 4^2 + (-1)^2 = 1 + 16 + 1 = 18$

$BD^2 = (0-5)^2 + (6-(-4))^2 + (-3-0)^2 = (-5)^2 + 10^2 + (-3)^2 = 25 + 100 + 9 = 134$

Since $AC^2 \neq BD^2$, the diagonals are not equal. Thus, it is a parallelogram but not a rectangle or a square.

The points are the vertices of a Parallelogram. This matches C2(i).

(d) Checking for collinearity:

Let the points be $A(1, -1, 3)$, $B(2, -4, 5)$, and $C(5, -13, 11)$.

Calculate the distances between the points:

$AB = \sqrt{(2-1)^2 + (-4-(-1))^2 + (5-3)^2} = \sqrt{1^2 + (-3)^2 + 2^2} = \sqrt{1 + 9 + 4} = \sqrt{14}$

$BC = \sqrt{(5-2)^2 + (-13-(-4))^2 + (11-5)^2} = \sqrt{3^2 + (-9)^2 + 6^2} = \sqrt{9 + 81 + 36} = \sqrt{126} = 3\sqrt{14}$

$AC = \sqrt{(5-1)^2 + (-13-(-1))^2 + (11-3)^2} = \sqrt{4^2 + (-12)^2 + 8^2} = \sqrt{16 + 144 + 64} = \sqrt{224} = 4\sqrt{14}$

Check if the sum of the two smaller distances equals the largest distance:

$AB + BC = \sqrt{14} + 3\sqrt{14} = 4\sqrt{14}$

Since $AB + BC = AC$ ($4\sqrt{14} = 4\sqrt{14}$), the points A, B, and C are Collinear.

This matches C2(v).

(e) Determining the type of triangle:

Let the points be $A(2, 4, 3)$, $B(4, 1, 9)$, and $C(10, -1, 6)$.

Calculate the square of the lengths of the sides:

$AB^2 = (4-2)^2 + (1-4)^2 + (9-3)^2 = 2^2 + (-3)^2 + 6^2 = 4 + 9 + 36 = 49$

$BC^2 = (10-4)^2 + (-1-1)^2 + (6-9)^2 = 6^2 + (-2)^2 + (-3)^2 = 36 + 4 + 9 = 49$

$AC^2 = (10-2)^2 + (-1-4)^2 + (6-3)^2 = 8^2 + (-5)^2 + 3^2 = 64 + 25 + 9 = 98$

Since $AB^2 = BC^2 = 49$, we have $AB = BC = \sqrt{49} = 7$. The triangle is isosceles.

Check for a right angle using the Pythagorean theorem. The largest side squared is $AC^2 = 98$.

$AB^2 + BC^2 = 49 + 49 = 98$

Since $AB^2 + BC^2 = AC^2$, the triangle satisfies the Pythagorean theorem and is right-angled at B.

Therefore, the points are the vertices of an Isosceles right-angled triangle.

This matches C2(iii).

Matching Summary:

(a) - (ii)

(b) - (iv)

(c) - (i)

(d) - (v)

(e) - (iii)

To locate a point $P(x, y, z)$ in three-dimensional space, we start from the origin $(0, 0, 0)$ and move along the coordinate axes or parallel to them. We move $x$ units along the x-axis, then $y$ units parallel to the y-axis, and finally $z$ units parallel to the z-axis. The sign of the coordinate determines the direction of movement along the respective axis (positive or negative direction).

(i) (1, – 1, 3)

Start from the origin $(0, 0, 0)$. Move 1 unit along the positive x-axis. From there, move 1 unit parallel to the negative y-axis. Finally, move 3 units parallel to the positive z-axis. The endpoint is the location of the point $(1, -1, 3)$.

(ii) (– 1, 2, 4)

Start from the origin $(0, 0, 0)$. Move 1 unit along the negative x-axis. From there, move 2 units parallel to the positive y-axis. Finally, move 4 units parallel to the positive z-axis. The endpoint is the location of the point $(-1, 2, 4)$.

(iii) (– 2, – 4, –7)

Start from the origin $(0, 0, 0)$. Move 2 units along the negative x-axis. From there, move 4 units parallel to the negative y-axis. Finally, move 7 units parallel to the negative z-axis. The endpoint is the location of the point $(-2, -4, -7)$.

(iv) (– 4, 2, – 5)

Start from the origin $(0, 0, 0)$. Move 4 units along the negative x-axis. From there, move 2 units parallel to the positive y-axis. Finally, move 5 units parallel to the negative z-axis. The endpoint is the location of the point $(-4, 2, -5)$.

The three coordinate planes divide the space into eight regions called octants. The octant is determined by the signs of the coordinates $(x, y, z)$ of a point.

(i) (1, 2, 3): The coordinates are $(+, +, +)$. This point lies in Octant I.

(ii) (4, – 2, 3): The coordinates are $(+, -, +)$. This point lies in Octant IV.

(iii) (4, –2, –5): The coordinates are $(+, -, -)$. This point lies in Octant VIII.

(iv) (4, 2, –5): The coordinates are $(+, +, -)$. This point lies in Octant V.

(v) (– 4, 2, 5): The coordinates are $(-, +, +)$. This point lies in Octant II.

(vi) (–3, –1, 6): The coordinates are $(-, -, +)$. This point lies in Octant III.

(vii) (2, – 4, – 7): The coordinates are $(+, -, -)$. This point lies in Octant VIII.

(viii) (– 4, 2, – 5): The coordinates are $(-, +, -)$. This point lies in Octant VI.

Let the coordinates of the point P be $(x, y, z)$.

The foot of the perpendicular from P to the x-axis lies on the x-axis, so its y and z coordinates are 0. The x-coordinate is the same as that of P. Thus, the coordinates of A are $(x, 0, 0)$.

The foot of the perpendicular from P to the y-axis lies on the y-axis, so its x and z coordinates are 0. The y-coordinate is the same as that of P. Thus, the coordinates of B are $(0, y, 0)$.

The foot of the perpendicular from P to the z-axis lies on the z-axis, so its x and y coordinates are 0. The z-coordinate is the same as that of P. Thus, the coordinates of C are $(0, 0, z)$.

(i) For the point P (3, 4, 2):

The coordinates of A (foot of perpendicular on x-axis) are (3, 0, 0).

The coordinates of B (foot of perpendicular on y-axis) are (0, 4, 0).

The coordinates of C (foot of perpendicular on z-axis) are (0, 0, 2).

(ii) For the point P (–5, 3, 7):

The coordinates of A (foot of perpendicular on x-axis) are (–5, 0, 0).

The coordinates of B (foot of perpendicular on y-axis) are (0, 3, 0).

The coordinates of C (foot of perpendicular on z-axis) are (0, 0, 7).

(iii) For the point P (4, – 3, – 5):

The coordinates of A (foot of perpendicular on x-axis) are (4, 0, 0).

The coordinates of B (foot of perpendicular on y-axis) are (0, –3, 0).

The coordinates of C (foot of perpendicular on z-axis) are (0, 0, –5).

Let the coordinates of the point P be $(x, y, z)$.

The foot of the perpendicular from P to the xy-plane lies on the xy-plane, meaning its z-coordinate is 0. The x and y coordinates remain the same as P. Thus, the coordinates of A on the xy-plane are $(x, y, 0)$.

The foot of the perpendicular from P to the yz-plane lies on the yz-plane, meaning its x-coordinate is 0. The y and z coordinates remain the same as P. Thus, the coordinates of B on the yz-plane are $(0, y, z)$.

The foot of the perpendicular from P to the zx-plane lies on the zx-plane, meaning its y-coordinate is 0. The x and z coordinates remain the same as P. Thus, the coordinates of C on the zx-plane are $(x, 0, z)$.

(i) For the point P (3, 4, 5):

The coordinates of A (foot of perpendicular on xy-plane) are (3, 4, 0).

The coordinates of B (foot of perpendicular on yz-plane) are (0, 4, 5).

The coordinates of C (foot of perpendicular on zx-plane) are (3, 0, 5).

(ii) For the point P (–5, 3, 7):

The coordinates of A (foot of perpendicular on xy-plane) are (–5, 3, 0).

The coordinates of B (foot of perpendicular on yz-plane) are (0, 3, 7).

The coordinates of C (foot of perpendicular on zx-plane) are (–5, 0, 7).

(iii) For the point P (4, – 3, – 5):

The coordinates of A (foot of perpendicular on xy-plane) are (4, – 3, 0).

The coordinates of B (foot of perpendicular on yz-plane) are (0, – 3, – 5).

The coordinates of C (foot of perpendicular on zx-plane) are (4, 0, – 5).

Let the two points be $P_1 = (2, 0, 0)$ and $P_2 = (-3, 0, 0)$.

We can find the distance between these two points using the distance formula in three dimensions.

The distance $d$ between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by:

$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$

For the given points $P_1(2, 0, 0)$ and $P_2(-3, 0, 0)$:

$x_1 = 2$, $y_1 = 0$, $z_1 = 0$

$x_2 = -3$, $y_2 = 0$, $z_2 = 0$

Substitute these values into the distance formula:

$d = \sqrt{(-3 - 2)^2 + (0 - 0)^2 + (0 - 0)^2}$

$d = \sqrt{(-5)^2 + 0^2 + 0^2}$

$d = \sqrt{25 + 0 + 0}$

$d = \sqrt{25}$

$d = 5$

Alternatively, since both points lie on the x-axis (y and z coordinates are zero), the distance is simply the absolute difference of their x-coordinates:

$d = |x_2 - x_1| = |-3 - 2| = |-5| = 5$

The distance between the points (2, 0, 0) and (–3, 0, 0) is 5 units.

Let the origin be $O = (0, 0, 0)$ and the given point be $P = (6, 6, 7)$.

We need to find the distance between $O$ and $P$.

The distance $d$ between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by the formula:

$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$

Here, $(x_1, y_1, z_1) = (0, 0, 0)$ and $(x_2, y_2, z_2) = (6, 6, 7)$.

Substituting these values into the distance formula:

$d = \sqrt{(6 - 0)^2 + (6 - 0)^2 + (7 - 0)^2}$

$d = \sqrt{6^2 + 6^2 + 7^2}$

$d = \sqrt{36 + 36 + 49}$

$d = \sqrt{72 + 49}$

$d = \sqrt{121}$

$d = 11$

The distance from the origin to the point (6, 6, 7) is 11 units.

Let the origin be $O = (0, 0, 0)$ and the given point be $P = (x, y, \sqrt{1 - x^2 - y^2})$.

We need to find the distance between the origin $O$ and the point $P$ using the distance formula in three dimensions.

The distance $d$ between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by:

$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$

Here, $(x_1, y_1, z_1) = (0, 0, 0)$ and $(x_2, y_2, z_2) = (x, y, \sqrt{1 - x^2 - y^2})$.

Substitute these values into the distance formula:

$d = \sqrt{(x - 0)^2 + (y - 0)^2 + (\sqrt{1 - x^2 - y^2} - 0)^2}$

$d = \sqrt{x^2 + y^2 + (\sqrt{1 - x^2 - y^2})^2}$

Since the square of a square root is the number itself (assuming the number is non-negative, which $1 - x^2 - y^2$ must be for $\sqrt{1 - x^2 - y^2}$ to be a real number), we have:

$d = \sqrt{x^2 + y^2 + (1 - x^2 - y^2)}$

We are given the condition $x^2 + y^2 = 1$. Substitute this into the expression for $d$:

$d = \sqrt{1 + (1 - (x^2 + y^2))}$

$d = \sqrt{1 + (1 - 1)}$

$d = \sqrt{1 + 0}$

$d = \sqrt{1}$

$d = 1$

Thus, the distance from the origin to the point $(x, y, \sqrt{1 - x^2 - y^2})$ is 1 unit, provided that $x^2 + y^2 = 1$.

Let the given points be A(1, – 1, 3), B(2, – 4, 5), and C(5, – 13, 11).

To show that three points A, B, and C are collinear, we can calculate the distances between all pairs of points (AB, BC, AC) and check if the sum of the lengths of two segments is equal to the length of the third segment.

We use the distance formula between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.

Calculate the distance AB:

A(1, –1, 3) and B(2, –4, 5)

$AB = \sqrt{(2-1)^2 + (-4 - (-1))^2 + (5-3)^2}$

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

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

$AB = \sqrt{1 + 9 + 4}$

$AB = \sqrt{14}$

Calculate the distance BC:

B(2, –4, 5) and C(5, –13, 11)

$BC = \sqrt{(5-2)^2 + (-13 - (-4))^2 + (11-5)^2}$

$BC = \sqrt{(3)^2 + (-13+4)^2 + (6)^2}$

$BC = \sqrt{3^2 + (-9)^2 + 6^2}$

$BC = \sqrt{9 + 81 + 36}$

$BC = \sqrt{126}$

$BC = \sqrt{9 \times 14}$

$BC = 3\sqrt{14}$

Calculate the distance AC:

A(1, –1, 3) and C(5, –13, 11)

$AC = \sqrt{(5-1)^2 + (-13 - (-1))^2 + (11-3)^2}$

$AC = \sqrt{(4)^2 + (-13+1)^2 + (8)^2}$

$AC = \sqrt{4^2 + (-12)^2 + 8^2}$

$AC = \sqrt{16 + 144 + 64}$

$AC = \sqrt{224}$

$AC = \sqrt{16 \times 14}$

$AC = 4\sqrt{14}$

Now, let's check if the sum of two distances equals the third distance.

Consider $AB + BC = \sqrt{14} + 3\sqrt{14} = (1+3)\sqrt{14} = 4\sqrt{14}$.

We see that $AB + BC = 4\sqrt{14}$ and $AC = 4\sqrt{14}$.

Therefore, $AB + BC = AC$.

Since the sum of the distances between two pairs of points equals the distance between the third pair, the points A, B, and C lie on the same line. Hence, the points are collinear.

Given:

Three consecutive vertices of a parallelogram ABCD are A (6, – 2, 4), B (2, 4, – 8), and C (–2, 2, 4).

To Find:

The coordinates of the fourth vertex D.

Solution:

Let the coordinates of the fourth vertex D be $(x, y, z)$.

In a parallelogram, the diagonals bisect each other. This means that the midpoint of diagonal AC is the same as the midpoint of diagonal BD.

The formula for the midpoint of a line segment joining points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right)$.

Let's find the midpoint of the diagonal AC.

Point A is (6, – 2, 4) and Point C is (–2, 2, 4).

Midpoint of AC = $\left(\frac{6 + (-2)}{2}, \frac{-2 + 2}{2}, \frac{4 + 4}{2}\right)$

Midpoint of AC = $\left(\frac{4}{2}, \frac{0}{2}, \frac{8}{2}\right)$

Midpoint of AC = $(2, 0, 4)$

Now, let's find the midpoint of the diagonal BD.

Point B is (2, 4, – 8) and Point D is $(x, y, z)$.

Midpoint of BD = $\left(\frac{2 + x}{2}, \frac{4 + y}{2}, \frac{-8 + z}{2}\right)$

Since the midpoints of AC and BD are the same, we can equate their corresponding coordinates:

$\frac{2 + x}{2} = 2$

$2 + x = 4$

$x = 4 - 2$

$x = 2$

$\frac{4 + y}{2} = 0$

$4 + y = 0$

$y = -4$

$\frac{-8 + z}{2} = 4$

$-8 + z = 8$

$z = 8 + 8$

$z = 16$

Thus, the coordinates of the fourth vertex D are (2, – 4, 16).

Given:

The vertices of triangle ABC are A (0, 4, 1), B (2, 3, – 1), and C (4, 5, 0).

Solution:

To show that triangle ABC is right-angled, we can calculate the square of the lengths of its sides and check if the Pythagorean theorem holds, i.e., if the sum of the squares of two sides is equal to the square of the third side.

The square of the distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by the formula: $d^2 = (x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2$.

Calculate the square of the length of side AB using points A(0, 4, 1) and B(2, 3, – 1):

$AB^2 = (2 - 0)^2 + (3 - 4)^2 + (-1 - 1)^2$

$AB^2 = (2)^2 + (-1)^2 + (-2)^2$

$AB^2 = 4 + 1 + 4$

$AB^2 = 9$

Calculate the square of the length of side BC using points B(2, 3, – 1) and C(4, 5, 0):

$BC^2 = (4 - 2)^2 + (5 - 3)^2 + (0 - (-1))^2$

$BC^2 = (2)^2 + (2)^2 + (0 + 1)^2$

$BC^2 = 2^2 + 2^2 + 1^2$

$BC^2 = 4 + 4 + 1$

$BC^2 = 9$

Calculate the square of the length of side AC using points A(0, 4, 1) and C(4, 5, 0):

$AC^2 = (4 - 0)^2 + (5 - 4)^2 + (0 - 1)^2$

$AC^2 = (4)^2 + (1)^2 + (-1)^2$

$AC^2 = 16 + 1 + 1$

$AC^2 = 18$

Now, we check if the Pythagorean theorem holds for these side lengths:

We have $AB^2 = 9$, $BC^2 = 9$, and $AC^2 = 18$.

Let's check if the sum of the squares of the two smaller sides equals the square of the largest side:

$AB^2 + BC^2 = 9 + 9 = 18$

We see that $AB^2 + BC^2 = 18$ and $AC^2 = 18$.

Thus, $AB^2 + BC^2 = AC^2$.

Since the sum of the squares of two sides (AB and BC) is equal to the square of the third side (AC), the triangle ABC is a right-angled triangle by the converse of the Pythagorean theorem. The right angle is at the vertex opposite to the side AC, which is vertex B.

Given:

The centroid of the triangle is the origin, which is G (0, 0, 0).

Two vertices of the triangle are A (2, 4, 6) and B (0, –2, –5).

To Find:

The coordinates of the third vertex of the triangle.

Solution:

Let the coordinates of the third vertex be C $(x, y, z)$.

The coordinates of the centroid of a triangle with vertices $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, and $(x_3, y_3, z_3)$ are given by the formula:

Centroid $(G) = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3}\right)$

In this case, the centroid is G (0, 0, 0), and the vertices are A (2, 4, 6), B (0, –2, –5), and C $(x, y, z)$.

Using the centroid formula, we have:

$\left(\frac{2 + 0 + x}{3}, \frac{4 + (-2) + y}{3}, \frac{6 + (-5) + z}{3}\right) = (0, 0, 0)$

Equating the corresponding coordinates, we get:

$\frac{2 + x}{3} = 0$

$2 + x = 0 \times 3$

$2 + x = 0$

$x = -2$

$\frac{4 - 2 + y}{3} = 0$

$\frac{2 + y}{3} = 0$

$2 + y = 0 \times 3$

$2 + y = 0$

$y = -2$

$\frac{6 - 5 + z}{3} = 0$

$\frac{1 + z}{3} = 0$

$1 + z = 0 \times 3$

$1 + z = 0$

$z = -1$

Therefore, the coordinates of the third vertex C are (–2, –2, –1).

Given:

The midpoints of the sides of a triangle are D (1, 2, – 3), E (3, 0, 1), and F (– 1, 1, – 4).

To Find:

The centroid of the triangle.

Solution:

Let the vertices of the triangle be A, B, and C. Let D be the midpoint of AB, E be the midpoint of BC, and F be the midpoint of CA.

A key property regarding the centroid of a triangle states that the centroid of the triangle formed by joining the midpoints of the sides of the original triangle is the same as the centroid of the original triangle.

Let G be the centroid of the triangle ABC. G is also the centroid of the triangle DEF.

The coordinates of the centroid of a triangle with vertices $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, and $(x_3, y_3, z_3)$ are given by the formula:

$G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3}\right)$

Using the coordinates of the midpoints D(1, 2, –3), E(3, 0, 1), and F(–1, 1, –4) as the vertices of the triangle DEF, the centroid G is:

$G = \left(\frac{1 + 3 + (-1)}{3}, \frac{2 + 0 + 1}{3}, \frac{-3 + 1 + (-4)}{3}\right)$

$G = \left(\frac{1 + 3 - 1}{3}, \frac{2 + 1}{3}, \frac{-3 + 1 - 4}{3}\right)$

$G = \left(\frac{3}{3}, \frac{3}{3}, \frac{-6}{3}\right)$

$G = (1, 1, -2)$

The centroid of the triangle is (1, 1, –2).

Given:

The midpoints of the sides of a triangle are D (5, 7, 11), E (0, 8, 5), and F (2, 3, – 1).

To Find:

The coordinates of the vertices of the triangle.

Solution:

Let the vertices of the triangle be A $(x_1, y_1, z_1)$, B $(x_2, y_2, z_2)$, and C $(x_3, y_3, z_3)$.

Let D (5, 7, 11) be the midpoint of AB, E (0, 8, 5) be the midpoint of BC, and F (2, 3, – 1) be the midpoint of CA.

Using the midpoint formula, the coordinates of the midpoint of a segment with endpoints $(x_a, y_a, z_a)$ and $(x_b, y_b, z_b)$ are $\left(\frac{x_a + x_b}{2}, \frac{y_a + y_b}{2}, \frac{z_a + z_b}{2}\right)$.

For midpoint D (5, 7, 11) of AB:

$\frac{x_1 + x_2}{2} = 5 \implies x_1 + x_2 = 10$

$\frac{y_1 + y_2}{2} = 7 \implies y_1 + y_2 = 14$

$\frac{z_1 + z_2}{2} = 11 \implies z_1 + z_2 = 22$

For midpoint E (0, 8, 5) of BC:

$\frac{x_2 + x_3}{2} = 0 \implies x_2 + x_3 = 0$

$\frac{y_2 + y_3}{2} = 8 \implies y_2 + y_3 = 16$

$\frac{z_2 + z_3}{2} = 5 \implies z_2 + z_3 = 10$

For midpoint F (2, 3, – 1) of CA:

$\frac{x_3 + x_1}{2} = 2 \implies x_3 + x_1 = 4$

$\frac{y_3 + y_1}{2} = 3 \implies y_3 + y_1 = 6$

$\frac{z_3 + z_1}{2} = -1 \implies z_3 + z_1 = -2$

Now, we have three systems of linear equations. Let's solve for the x-coordinates first:

$x_1 + x_2 = 10$ ... (1)

$x_2 + x_3 = 0$ ... (2)

$x_3 + x_1 = 4$ ... (3)

Adding equations (1), (2), and (3):

$(x_1 + x_2) + (x_2 + x_3) + (x_3 + x_1) = 10 + 0 + 4$

$2x_1 + 2x_2 + 2x_3 = 14$

$2(x_1 + x_2 + x_3) = 14$

$x_1 + x_2 + x_3 = 7$ ... (4)

Substitute equation (1) into equation (4):

$10 + x_3 = 7$

$x_3 = 7 - 10 = -3$

Substitute equation (2) into equation (4):

$x_1 + 0 = 7$

$x_1 = 7$

Substitute equation (3) into equation (4):

$x_2 + 4 = 7$

$x_2 = 7 - 4 = 3$

So, the x-coordinates of the vertices are $x_1=7$, $x_2=3$, and $x_3=-3$.

Similarly, for the y-coordinates:

$y_1 + y_2 = 14$

$y_2 + y_3 = 16$

$y_3 + y_1 = 6$

Adding these gives $2(y_1 + y_2 + y_3) = 14 + 16 + 6 = 36$, so $y_1 + y_2 + y_3 = 18$.

$(y_1+y_2)+y_3 = 18 \implies 14+y_3 = 18 \implies y_3 = 4$

$y_1+(y_2+y_3) = 18 \implies y_1+16 = 18 \implies y_1 = 2$

$(y_3+y_1)+y_2 = 18 \implies 6+y_2 = 18 \implies y_2 = 12$

So, the y-coordinates are $y_1=2$, $y_2=12$, and $y_3=4$.

For the z-coordinates:

$z_1 + z_2 = 22$

$z_2 + z_3 = 10$

$z_3 + z_1 = -2$

Adding these gives $2(z_1 + z_2 + z_3) = 22 + 10 + (-2) = 30$, so $z_1 + z_2 + z_3 = 15$.

$(z_1+z_2)+z_3 = 15 \implies 22+z_3 = 15 \implies z_3 = -7$

$z_1+(z_2+z_3) = 15 \implies z_1+10 = 15 \implies z_1 = 5$

$(z_3+z_1)+z_2 = 15 \implies -2+y_2 = 15 \implies z_2 = 17$

So, the z-coordinates are $z_1=5$, $z_2=17$, and $z_3=-7$.

Combining the coordinates, the vertices of the triangle are:

A = $(x_1, y_1, z_1)$ = (7, 2, 5)

B = $(x_2, y_2, z_2)$ = (3, 12, 17)

C = $(x_3, y_3, z_3)$ = (–3, 4, –7)

Given:

Three vertices of a parallelogram ABCD are A (1, 2, 3), B (– 1, – 2, – 1), and C (2, 3, 2).

To Find:

The coordinates of the fourth vertex D.

Solution:

Let the coordinates of the fourth vertex D be $(x, y, z)$.

In a parallelogram, the diagonals bisect each other. This means that the midpoint of diagonal AC is the same as the midpoint of diagonal BD.

The formula for the midpoint of a line segment joining points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right)$.

Let's find the midpoint of the diagonal AC.

Point A is (1, 2, 3) and Point C is (2, 3, 2).

Midpoint of AC = $\left(\frac{1 + 2}{2}, \frac{2 + 3}{2}, \frac{3 + 2}{2}\right)$

Midpoint of AC = $\left(\frac{3}{2}, \frac{5}{2}, \frac{5}{2}\right)$

Now, let's find the midpoint of the diagonal BD.

Point B is (– 1, – 2, – 1) and Point D is $(x, y, z)$.

Midpoint of BD = $\left(\frac{-1 + x}{2}, \frac{-2 + y}{2}, \frac{-1 + z}{2}\right)$

Since the midpoints of AC and BD are the same, we can equate their corresponding coordinates:

From equation (i):

$-1 + x = 3$

$x = 3 + 1$

$x = 4$

From equation (ii):

$-2 + y = 5$

$y = 5 + 2$

$y = 7$

From equation (iii):

$-1 + z = 5$

$z = 5 + 1$

$z = 6$

Thus, the coordinates of the fourth vertex D are (4, 7, 6).

Given:

The endpoints of the line segment are A (2, 1, – 3) and B (5, – 8, 3).

To Find:

The coordinates of the points that trisect the line segment AB.

Solution:

Let the points that trisect the line segment AB be P and Q. Trisection means the points divide the segment into three equal parts.

This implies that the ratio AP : PB = 1 : 2 and AQ : QB = 2 : 1.

We will use the section formula for a point dividing a line segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in the ratio $m:n$ internally. The coordinates of the point are given by:

$\left(\frac{nx_1 + mx_2}{m+n}, \frac{ny_1 + my_2}{m+n}, \frac{nz_1 + mz_2}{m+n}\right)$

Finding the coordinates of point P:

Point P divides AB in the ratio 1:2. Here $(x_1, y_1, z_1) = (2, 1, -3)$, $(x_2, y_2, z_2) = (5, -8, 3)$, $m=1$, and $n=2$.

$x_P = \frac{2(2) + 1(5)}{1+2} = \frac{4 + 5}{3} = \frac{9}{3} = 3$

$y_P = \frac{2(1) + 1(-8)}{1+2} = \frac{2 - 8}{3} = \frac{-6}{3} = -2$

$z_P = \frac{2(-3) + 1(3)}{1+2} = \frac{-6 + 3}{3} = \frac{-3}{3} = -1$

So, the coordinates of point P are (3, –2, –1).

Finding the coordinates of point Q:

Point Q divides AB in the ratio 2:1. Here $(x_1, y_1, z_1) = (2, 1, -3)$, $(x_2, y_2, z_2) = (5, -8, 3)$, $m=2$, and $n=1$.

$x_Q = \frac{1(2) + 2(5)}{2+1} = \frac{2 + 10}{3} = \frac{12}{3} = 4$

$y_Q = \frac{1(1) + 2(-8)}{2+1} = \frac{1 - 16}{3} = \frac{-15}{3} = -5$

$z_Q = \frac{1(-3) + 2(3)}{2+1} = \frac{-3 + 6}{3} = \frac{3}{3} = 1$

So, the coordinates of point Q are (4, –5, 1).

The two points that trisect the line segment joining A (2, 1, – 3) and B (5, – 8, 3) are (3, –2, –1) and (4, –5, 1).

Alternate Solution:

Once the first point of trisection P (3, –2, –1) is found, the second point of trisection Q is the midpoint of the line segment PB.

Using the midpoint formula for points P(3, –2, –1) and B(5, –8, 3):

Midpoint of PB = $\left(\frac{3 + 5}{2}, \frac{-2 + (-8)}{2}, \frac{-1 + 3}{2}\right)$

= $\left(\frac{8}{2}, \frac{-10}{2}, \frac{2}{2}\right)$

= $(4, -5, 1)$

This confirms the coordinates of the second point of trisection Q as (4, –5, 1).

Given:

The vertices of triangle ABC are A (a, 1, 3), B (– 2, b, – 5), and C (4, 7, c).

The centroid of the triangle is the origin, which is G (0, 0, 0).

To Find:

The values of a, b, and c.

Solution:

The coordinates of the centroid of a triangle with vertices $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, and $(x_3, y_3, z_3)$ are given by the formula:

$G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3}\right)$

In this case, $(x_1, y_1, z_1) = (a, 1, 3)$, $(x_2, y_2, z_2) = (-2, b, -5)$, $(x_3, y_3, z_3) = (4, 7, c)$, and the centroid is $(0, 0, 0)$.

Using the centroid formula, we have:

$\left(\frac{a + (-2) + 4}{3}, \frac{1 + b + 7}{3}, \frac{3 + (-5) + c}{3}\right) = (0, 0, 0)$

$\left(\frac{a - 2 + 4}{3}, \frac{1 + 7 + b}{3}, \frac{3 - 5 + c}{3}\right) = (0, 0, 0)$

$\left(\frac{a + 2}{3}, \frac{8 + b}{3}, \frac{-2 + c}{3}\right) = (0, 0, 0)$

Equating the corresponding coordinates to the coordinates of the origin (0, 0, 0):

From equation (i):

$a + 2 = 0 \times 3$

$a + 2 = 0$

$a = -2$

From equation (ii):

$8 + b = 0 \times 3$

$8 + b = 0$

$b = -8$

From equation (iii):

$-2 + c = 0 \times 3$

$-2 + c = 0$

$c = 2$

The values are a = –2, b = –8, and c = 2.

Given:

The vertices of triangle ABC are A (2, 2, – 3), B (5, 6, 9), and C (2, 7, 9).

AD is the internal bisector of angle A and meets BC at point D.

To Find:

The coordinates of point D.

Solution:

According to the Angle Bisector Theorem, the internal bisector of an angle of a triangle divides the opposite side in the ratio of the lengths of the sides containing the angle.

In triangle ABC, the angle bisector AD divides the side BC in the ratio AB : AC. That is, BD : DC = AB : AC.

First, we calculate the lengths of the sides AB and AC using the distance formula between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.

Calculate the length of AB:

$A = (2, 2, -3)$, $B = (5, 6, 9)$

$AB = \sqrt{(5-2)^2 + (6-2)^2 + (9-(-3))^2}$

$AB = \sqrt{(3)^2 + (4)^2 + (12)^2}$

$AB = \sqrt{9 + 16 + 144}$

$AB = \sqrt{169}$

$AB = 13$

Calculate the length of AC:

$A = (2, 2, -3)$, $C = (2, 7, 9)$

$AC = \sqrt{(2-2)^2 + (7-2)^2 + (9-(-3))^2}$

$AC = \sqrt{(0)^2 + (5)^2 + (12)^2}$

$AC = \sqrt{0 + 25 + 144}$

$AC = \sqrt{169}$

$AC = 13$

The ratio in which D divides BC is AB : AC = $13 : 13 = 1 : 1$.

This means that point D is the midpoint of the line segment BC.

We use the section formula for internal division to find the coordinates of point D. Since the ratio is 1:1, we can use the midpoint formula.

The midpoint of a segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right)$.

Using points B (5, 6, 9) and C (2, 7, 9):

The coordinates of D $(x_D, y_D, z_D)$ are:

$x_D = \frac{5 + 2}{2} = \frac{7}{2}$

$y_D = \frac{6 + 7}{2} = \frac{13}{2}$

$z_D = \frac{9 + 9}{2} = \frac{18}{2} = 9$

The coordinates of point D are $\left(\frac{7}{2}, \frac{13}{2}, 9\right)$.

Given:

Three points A (2, 3, 4), B (–1, 2, – 3), and C (– 4, 1, – 10).

To Show:

The points A, B, and C are collinear.

To Find:

The ratio in which C divides the line segment AB.

Solution:

To check for collinearity and find the ratio of division using the section formula, let us assume that the point C divides the line segment joining A (2, 3, 4) and B (–1, 2, – 3) in the ratio $\lambda : 1$.

Using the section formula for a point dividing a line segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in the ratio $\lambda : 1$, the coordinates of the point are given by:

$\left(\frac{1 \cdot x_1 + \lambda \cdot x_2}{1+\lambda}, \frac{1 \cdot y_1 + \lambda \cdot y_2}{1+\lambda}, \frac{1 \cdot z_1 + \lambda \cdot z_2}{1+\lambda}\right)$

Substituting the coordinates of A (2, 3, 4), B (–1, 2, – 3), and the assumed ratio $\lambda : 1$, the coordinates of the dividing point are:

$\left(\frac{2 - \lambda}{1+\lambda}, \frac{3 + 2\lambda}{1+\lambda}, \frac{4 - 3\lambda}{1+\lambda}\right)$

Since we are assuming that point C (– 4, 1, – 10) is this dividing point, we equate its coordinates with the calculated coordinates:

$(-4, 1, -10) = \left(\frac{2 - \lambda}{1+\lambda}, \frac{3 + 2\lambda}{1+\lambda}, \frac{4 - 3\lambda}{1+\lambda}\right)$

Equating the corresponding coordinates, we get three equations:

From equation (1):

$-4(1+\lambda) = 2 - \lambda$

$-4 - 4\lambda = 2 - \lambda$

$-4 - 2 = 4\lambda - \lambda$

$-6 = 3\lambda$

$\lambda = \frac{-6}{3}$

$\lambda = -2$

From equation (2):

$1(1+\lambda) = 3 + 2\lambda$

$1 + \lambda = 3 + 2\lambda$

$1 - 3 = 2\lambda - \lambda$

$-2 = \lambda$

$\lambda = -2$

From equation (3):

$-10(1+\lambda) = 4 - 3\lambda$

$-10 - 10\lambda = 4 - 3\lambda$

$-10 - 4 = 10\lambda - 3\lambda$

$-14 = 7\lambda$

$\lambda = \frac{-14}{7}$

$\lambda = -2$

Since the value of $\lambda$ is the same ($\lambda = -2$) from all three equations, our assumption that C divides AB in the ratio $\lambda : 1$ is consistent. This implies that the three points A, B, and C lie on the same line. Therefore, the points A, B, and C are collinear.

The ratio in which C divides AB is $\lambda : 1 = -2 : 1$.

A negative ratio indicates that the division is external. Specifically, C divides AB externally in the ratio $2 : 1$.

This means that the distance from A to C is 2 times the distance from B to C, and C lies outside the segment AB on the line passing through A and B.

The points A, B, and C are collinear, and C divides AB externally in the ratio 2 : 1.

Given:

The midpoints of the sides of a triangle are D (1, 5, – 1), E (0, 4, – 2), and F (2, 3, 4).

To Find:

1. The coordinates of the vertices of the triangle.

2. The centroid of the triangle.

Solution:

Let the vertices of the triangle be A $(x_1, y_1, z_1)$, B $(x_2, y_2, z_2)$, and C $(x_3, y_3, z_3)$.

Let D (1, 5, – 1) be the midpoint of AB, E (0, 4, – 2) be the midpoint of BC, and F (2, 3, 4) be the midpoint of CA.

Using the midpoint formula, the coordinates of the midpoint of a segment with endpoints $(x_a, y_a, z_a)$ and $(x_b, y_b, z_b)$ are $\left(\frac{x_a + x_b}{2}, \frac{y_a + y_b}{2}, \frac{z_a + z_b}{2}\right)$.

For midpoint D (1, 5, – 1) of AB:

For midpoint E (0, 4, – 2) of BC:

For midpoint F (2, 3, 4) of CA:

Adding equations (1), (4), and (7) for the x-coordinates:

$(x_1 + x_2) + (x_2 + x_3) + (x_3 + x_1) = 2 + 0 + 4$

$2x_1 + 2x_2 + 2x_3 = 6$

$x_1 + x_2 + x_3 = 3$ ... (10)

Subtracting (1) from (10): $(x_1 + x_2 + x_3) - (x_1 + x_2) = 3 - 2 \implies x_3 = 1$

Subtracting (4) from (10): $(x_1 + x_2 + x_3) - (x_2 + x_3) = 3 - 0 \implies x_1 = 3$

Subtracting (7) from (10): $(x_1 + x_2 + x_3) - (x_3 + x_1) = 3 - 4 \implies x_2 = -1$

So, $x_1=3, x_2=-1, x_3=1$.

Adding equations (2), (5), and (8) for the y-coordinates:

$(y_1 + y_2) + (y_2 + y_3) + (y_3 + y_1) = 10 + 8 + 6$

$2y_1 + 2y_2 + 2y_3 = 24$

$y_1 + y_2 + y_3 = 12$ ... (11)

Subtracting (2) from (11): $(y_1 + y_2 + y_3) - (y_1 + y_2) = 12 - 10 \implies y_3 = 2$

Subtracting (5) from (11): $(y_1 + y_2 + y_3) - (y_2 + y_3) = 12 - 8 \implies y_1 = 4$

Subtracting (8) from (11): $(y_1 + y_2 + y_3) - (y_3 + y_1) = 12 - 6 \implies y_2 = 6$

So, $y_1=4, y_2=6, y_3=2$.

Adding equations (3), (6), and (9) for the z-coordinates:

$(z_1 + z_2) + (z_2 + z_3) + (z_3 + z_1) = -2 + (-4) + 8$

$2z_1 + 2z_2 + 2z_3 = 2$

$z_1 + z_2 + z_3 = 1$ ... (12)

Subtracting (3) from (12): $(z_1 + z_2 + z_3) - (z_1 + z_2) = 1 - (-2) \implies z_3 = 3$

Subtracting (6) from (12): $(z_1 + z_2 + z_3) - (z_2 + z_3) = 1 - (-4) \implies z_1 = 5$

Subtracting (9) from (12): $(z_1 + z_2 + z_3) - (z_3 + z_1) = 1 - 8 \implies z_2 = -7$

So, $z_1=5, z_2=-7, z_3=3$.

The coordinates of the vertices are:

A = $(x_1, y_1, z_1)$ = (3, 4, 5)

B = $(x_2, y_2, z_2)$ = (–1, 6, –7)

C = $(x_3, y_3, z_3)$ = (1, 2, 3)

Now, we find the centroid of the triangle ABC using the vertex coordinates A (3, 4, 5), B (–1, 6, –7), and C (1, 2, 3).

The coordinates of the centroid $G = (x_G, y_G, z_G)$ are:

$x_G = \frac{x_1 + x_2 + x_3}{3} = \frac{3 + (-1) + 1}{3} = \frac{3 - 1 + 1}{3} = \frac{3}{3} = 1$

$y_G = \frac{y_1 + y_2 + y_3}{3} = \frac{4 + 6 + 2}{3} = \frac{12}{3} = 4$

$z_G = \frac{z_1 + z_2 + z_3}{3} = \frac{5 + (-7) + 3}{3} = \frac{5 - 7 + 3}{3} = \frac{1}{3}$

The centroid of the triangle is $\left(1, 4, \frac{1}{3}\right)$.

Alternate Method for Centroid:

The centroid of a triangle is the same as the centroid of the triangle formed by joining the midpoints of its sides.

Using the midpoint coordinates D (1, 5, – 1), E (0, 4, – 2), and F (2, 3, 4) to find the centroid:

$x_G = \frac{1 + 0 + 2}{3} = \frac{3}{3} = 1$

$y_G = \frac{5 + 4 + 3}{3} = \frac{12}{3} = 4$

$z_G = \frac{-1 + (-2) + 4}{3} = \frac{-3 + 4}{3} = \frac{1}{3}$

This gives the same centroid coordinates, $\left(1, 4, \frac{1}{3}\right)$.

Given:

Three points A (0, – 1, – 7), B (2, 1, – 9), and C (6, 5, – 13).

To Prove:

The points A, B, and C are collinear.

To Find:

The ratio in which point A divides the line segment BC.

Proof/Solution:

To prove collinearity and find the ratio of division using the section formula, let us assume that the first point A divides the line segment joining the other two points B (2, 1, – 9) and C (6, 5, – 13) in the ratio $\lambda : 1$.

Using the section formula for a point dividing a line segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in the ratio $\lambda : 1$, the coordinates of the dividing point are given by:

$\left(\frac{1 \cdot x_1 + \lambda \cdot x_2}{1+\lambda}, \frac{1 \cdot y_1 + \lambda \cdot y_2}{1+\lambda}, \frac{1 \cdot z_1 + \lambda \cdot z_2}{1+\lambda}\right)$

Substituting the coordinates of B (2, 1, – 9) and C (6, 5, – 13), and the assumed ratio $\lambda : 1$, the coordinates of the dividing point are:

$\left(\frac{2 + 6\lambda}{1+\lambda}, \frac{1 + 5\lambda}{1+\lambda}, \frac{-9 - 13\lambda}{1+\lambda}\right)$

Since point A (0, – 1, – 7) is this dividing point, we equate its coordinates with the calculated coordinates:

$(0, -1, -7) = \left(\frac{2 + 6\lambda}{1+\lambda}, \frac{1 + 5\lambda}{1+\lambda}, \frac{-9 - 13\lambda}{1+\lambda}\right)$

Equating the corresponding coordinates, we get three equations:

From equation (1):

$0 = 2 + 6\lambda$

$6\lambda = -2$

$\lambda = \frac{-2}{6} = -\frac{1}{3}$

From equation (2):

$-1(1+\lambda) = 1 + 5\lambda$

$-1 - \lambda = 1 + 5\lambda$

$-1 - 1 = 5\lambda + \lambda$

$-2 = 6\lambda$

$\lambda = \frac{-2}{6} = -\frac{1}{3}$

From equation (3):

$-7(1+\lambda) = -9 - 13\lambda$

$-7 - 7\lambda = -9 - 13\lambda$

$-7 + 9 = -13\lambda + 7\lambda$

$2 = -6\lambda$

$\lambda = \frac{2}{-6} = -\frac{1}{3}$

Since the value of $\lambda$ is consistent ($\lambda = -1/3$) from all three equations, the points A, B, and C lie on the same line. This proves that the points are collinear.

The ratio in which A divides BC is $\lambda : 1 = -\frac{1}{3} : 1$.

To express this as a ratio of integers, we can multiply both parts by 3: $(-1/3) \times 3 : 1 \times 3 = -1 : 3$.

The negative sign indicates that the division is external. Therefore, the first point A divides the join of the other two points B and C externally in the ratio 1 : 3.

Given:

A cube with edge length = 2 units.

One vertex coincides with the origin (0, 0, 0).

The three edges passing through the origin coincide with the positive direction of the x, y, and z axes.

To Find:

The coordinates of all the vertices of the cube.

Solution:

Let the origin be O(0, 0, 0). Since the edges passing through the origin lie along the positive x, y, and z axes and the edge length is 2 units, the vertices along these axes will have coordinates where one coordinate is 2 and the other two are 0.

The vertices are:

1. The vertex at the origin: (0, 0, 0)

2. The vertex along the positive x-axis (at a distance of 2 from the origin): (2, 0, 0)

3. The vertex along the positive y-axis (at a distance of 2 from the origin): (0, 2, 0)

4. The vertex along the positive z-axis (at a distance of 2 from the origin): (0, 0, 2)

The remaining vertices can be found by moving parallel to the axes from the origin or the vertices on the axes.

5. The vertex in the xy-plane (moving 2 units in positive x and 2 units in positive y from the origin): (2, 2, 0)

6. The vertex in the yz-plane (moving 2 units in positive y and 2 units in positive z from the origin): (0, 2, 2)

7. The vertex in the xz-plane (moving 2 units in positive x and 2 units in positive z from the origin): (2, 0, 2)

8. The vertex opposite to the origin (moving 2 units in positive x, 2 units in positive y, and 2 units in positive z from the origin): (2, 2, 2)

The coordinates of the eight vertices of the cube are:

(0, 0, 0)

(2, 0, 0)

(0, 2, 0)

(0, 0, 2)

(2, 2, 0)

(0, 2, 2)

(2, 0, 2)

(2, 2, 2)

Given: The point P is $(3, 4, 5)$.

To Find: The distance of point P from the yz-plane.

Solution:

The coordinates of the given point P are $(3, 4, 5)$. Let the point be $(x, y, z)$.

So, we have $x=3$, $y=4$, and $z=5$.

The distance of a point $(x, y, z)$ from the yz-plane is given by the absolute value of its x-coordinate, i.e., $|x|$.

Distance from yz-plane $= |3|$

Distance from yz-plane $= 3$ units.

Therefore, the distance of the point P(3, 4, 5) from the yz-plane is 3 units.

The correct option is (A).

Given: The point P is $(3, 4, 5)$.

To Find: The distance of the point P(3, 4, 5) from the y-axis. (Interpreting the question's phrasing "length of foot of perpendicular drawn from the point P (3, 4, 5) on y-axis" as the distance from the point to the axis, which is the length of the perpendicular segment).

Solution:

Let the point P be $(x_1, y_1, z_1) = (3, 4, 5)$.

The y-axis is the line where $x=0$ and $z=0$.

The foot of the perpendicular from a point $(x_1, y_1, z_1)$ onto the y-axis is the point on the y-axis with coordinates $(0, y_1, 0)$.

For the point P(3, 4, 5), the foot of the perpendicular on the y-axis is the point F(0, 4, 0).

The distance of the point P from the y-axis is the distance between P(3, 4, 5) and F(0, 4, 0).

Using the distance formula between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in 3D space, which is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.

Distance PF $= \sqrt{(0-3)^2 + (4-4)^2 + (0-5)^2}$

Distance PF $= \sqrt{(-3)^2 + (0)^2 + (-5)^2}$

Distance PF $= \sqrt{9 + 0 + 25}$

Distance PF $= \sqrt{34}$ units.

Alternatively, the distance of a point $(x, y, z)$ from the y-axis is given by the formula $\sqrt{x^2 + z^2}$.

For point P(3, 4, 5), the distance from the y-axis is $\sqrt{3^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34}$ units.

Therefore, the distance of the point P(3, 4, 5) from the y-axis is $\sqrt{34}$ units.

The correct option is (B).

Given: The point P is $(3, 4, 5)$ and the origin O is $(0, 0, 0)$.

To Find: The distance of the point P(3, 4, 5) from the origin (0, 0, 0).

Solution:

Let the point P be $(x_2, y_2, z_2) = (3, 4, 5)$ and the origin O be $(x_1, y_1, z_1) = (0, 0, 0)$.

The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in 3D space is given by the formula:

Distance $= \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$

Distance PO $= \sqrt{(3-0)^2 + (4-0)^2 + (5-0)^2}$

Distance PO $= \sqrt{(3)^2 + (4)^2 + (5)^2}$

Distance PO $= \sqrt{9 + 16 + 25}$

Distance PO $= \sqrt{50}$ units.

Therefore, the distance of the point (3, 4, 5) from the origin (0, 0, 0) is $\sqrt{50}$ units.

The correct option is (A).

Given:

Two points are $(x_1, y_1, z_1) = (a, 0, 1)$ and $(x_2, y_2, z_2) = (0, 1, 2)$.

The distance between the points is given as $\sqrt{27}$.

To Find:

The value of $a$.

Solution:

The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in 3D space is given by the formula:

Distance $= \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$

Substituting the given points and distance into the formula:

$\sqrt{27} = \sqrt{(0-a)^2 + (1-0)^2 + (2-1)^2}$

$\sqrt{27} = \sqrt{(-a)^2 + (1)^2 + (1)^2}$

$\sqrt{27} = \sqrt{a^2 + 1 + 1}$

$\sqrt{27} = \sqrt{a^2 + 2}$

Squaring both sides of the equation:

$(\sqrt{27})^2 = (\sqrt{a^2 + 2})^2$

$27 = a^2 + 2$

Subtract 2 from both sides:

$27 - 2 = a^2$

$25 = a^2$

Taking the square root of both sides:

$a = \pm\sqrt{25}$

$a = \pm 5$

Thus, the value of $a$ can be $5$ or $-5$.

The correct option is (B).

Given: The question asks for the intersection of two planes that form the x-axis.

To Identify: The two planes whose intersection is the x-axis.

Solution:

In a three-dimensional Cartesian coordinate system, the coordinate planes are:

The xy-plane, where the z-coordinate of every point is 0. The equation of the xy-plane is $z=0$.

The yz-plane, where the x-coordinate of every point is 0. The equation of the yz-plane is $x=0$.

The xz-plane, where the y-coordinate of every point is 0. The equation of the xz-plane is $y=0$.

The x-axis is the set of points where both the y-coordinate and the z-coordinate are 0.

Points on the x-axis have coordinates of the form $(x, 0, 0)$.

This means that for any point on the x-axis, $y=0$ and $z=0$.

The plane defined by $y=0$ is the xz-plane.

The plane defined by $z=0$ is the xy-plane.

The intersection of the xy-plane ($z=0$) and the xz-plane ($y=0$) is the set of points that satisfy both $y=0$ and $z=0$, which is precisely the definition of the x-axis.

Let's check the other options:

(B) yz-plane ($x=0$) and zx-plane ($y=0$). Their intersection is the z-axis (where $x=0$ and $y=0$).

(C) xy-plane ($z=0$) and yz-plane ($x=0$). Their intersection is the y-axis (where $x=0$ and $z=0$).

Thus, the x-axis is the intersection of the xy and xz planes.

The correct option is (A).

Given: The question asks for the equations representing the y-axis in a 3D coordinate system.

To Identify: The equations that define the y-axis.

Solution:

In a three-dimensional Cartesian coordinate system, the y-axis is the line that passes through the origin and is perpendicular to both the x-axis and the z-axis.

Any point on the y-axis has coordinates of the form $(0, y, 0)$.

This means that for any point on the y-axis, the x-coordinate is always 0, and the z-coordinate is always 0.

Therefore, the equations that define the y-axis are $x=0$ and $z=0$.

Let's examine the given options:

(A) $x=0, y=0$: This represents the z-axis.

(B) $y=0, z=0$: This represents the x-axis.

(C) $z=0, x=0$: This represents the y-axis.

Thus, the equation of the y-axis is $x=0, z=0$.

The correct option is (C).

Given: The point is P($-2, -3, -4$).

Solution:

In a three-dimensional coordinate system, the three coordinate planes (xy-plane, yz-plane, and xz-plane) divide the space into eight regions called octants.

The sign of the coordinates ($x$, $y$, $z$) determines which octant a point lies in.

The signs for each octant are as follows:

First Octant: $(+, +, +)$ ($x>0, y>0, z>0$)

Second Octant: $(-, +, +)$ ($x<0, y>0, z>0$)

Third Octant: $(-, -, +)$ ($x<0, y<0, z>0$)

Fourth Octant: $(+, -, +)$ ($x>0, y<0, z>0$)

Fifth Octant: $(+, +, -)$ ($x>0, y>0, z<0$)

Sixth Octant: $(-, +, -)$ ($x<0, y>0, z<0$)

Seventh Octant: $(-, -, -)$ ($x<0, y<0, z<0$)

Eighth Octant: $(+, -, -)$ ($x>0, y<0, z<0$)

For the given point P($-2, -3, -4$), the coordinates are:

$x = -2$ (negative)

$y = -3$ (negative)

$z = -4$ (negative)

The signs of the coordinates are $(-, -, -)$.

According to the octant definitions, the point with $(-, -, -)$ coordinates lies in the Seventh Octant.

The correct option is (B).

Given: A plane is parallel to the yz-plane.

To Identify: The axis to which the plane is perpendicular.

Solution:

In a three-dimensional coordinate system, the yz-plane is the plane where the x-coordinate is always zero. Its equation is $x=0$.

A plane that is parallel to the yz-plane will have an equation of the form $x=k$, where $k$ is a constant.

The normal vector to the plane $x=k$ is in the direction of the x-axis, which is $(1, 0, 0)$.

A plane is perpendicular to a line if its normal vector is parallel to the direction vector of the line.

The direction vector of the x-axis is $(1, 0, 0)$.

Since the normal vector of the plane ($1, 0, 0$) is parallel to the direction vector of the x-axis ($1, 0, 0$), the plane is perpendicular to the x-axis.

Geometrically, the yz-plane is "vertical" and contains the y and z axes. The x-axis is the line perpendicular to this plane, passing through the origin. Any plane parallel to the yz-plane maintains this orientation and is therefore also perpendicular to the x-axis.

The correct option is (A).

Given: The conditions for a point are $y = 0$ and $z = 0$.

To Identify: The locus of a point satisfying these conditions.

Solution:

The locus of a point is the set of all points that satisfy a given condition or set of conditions.

In a three-dimensional Cartesian coordinate system, the conditions $y = 0$ and $z = 0$ define a specific line.

The equation $y=0$ represents the xz-plane.

The equation $z=0$ represents the xy-plane.

The locus of points satisfying both $y=0$ and $z=0$ is the intersection of the xz-plane and the xy-plane.

Consider a point $(x, y, z)$. If $y=0$ and $z=0$, the coordinates of the point must be of the form $(x, 0, 0)$.

Points of the form $(x, 0, 0)$, where $x$ can be any real number, lie on the x-axis.

Therefore, the locus of a point for which $y = 0$ and $z = 0$ is the x-axis.

The equations $y=0, z=0$ together represent the equation of the x-axis in 3D space.

The correct option is (A).

Given: The condition for the locus of a point in 3D space is $x = 0$.

To Identify: The geometric representation of the locus of a point for which $x = 0$.

Solution:

The locus of a point is the set of all points that satisfy a given condition.

In a three-dimensional Cartesian coordinate system, a point is represented by $(x, y, z)$.

The condition $x = 0$ means that the x-coordinate of any point on the locus must be zero.

The coordinates of points satisfying $x = 0$ are of the form $(0, y, z)$, where $y$ and $z$ can take any real values.

Let's consider the coordinate planes in 3D space:

The xy-plane is defined by the equation $z = 0$.

The yz-plane is defined by the equation $x = 0$.

The xz-plane is defined by the equation $y = 0$.

Comparing the given condition $x=0$ with the definitions of the coordinate planes, we see that the locus of points for which $x = 0$ is the yz-plane.

The yz-plane contains the y-axis (where $x=0, z=0$) and the z-axis (where $x=0, y=0$) and all points between them where the x-coordinate is zero.

The correct option is (B).

Given:

Two points P$_1$(5, 8, 10) and P$_2$(3, 6, 8).

A parallelopiped is formed by planes parallel to the coordinate planes passing through these two points. These points represent opposite vertices of the parallelopiped.

To Find:

The length of the diagonal of the parallelopiped.

Solution:

When a parallelopiped is formed by planes through two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ parallel to the coordinate planes, the lengths of its edges ($l, m, n$) are given by the absolute difference of the corresponding coordinates of the two points.

Let $(x_1, y_1, z_1) = (5, 8, 10)$ and $(x_2, y_2, z_2) = (3, 6, 8)$.

Length of the edge along the x-direction, $l = |x_2 - x_1| = |3 - 5| = |-2| = 2$.

Length of the edge along the y-direction, $m = |y_2 - y_1| = |6 - 8| = |-2| = 2$.

Length of the edge along the z-direction, $n = |z_2 - z_1| = |8 - 10| = |-2| = 2$.

The dimensions of the parallelopiped are 2, 2, and 2.

The length of the diagonal of a parallelopiped with edge lengths $l, m, n$ is given by the formula $\sqrt{l^2 + m^2 + n^2}$.

Diagonal length $= \sqrt{2^2 + 2^2 + 2^2}$

Diagonal length $= \sqrt{4 + 4 + 4}$

Diagonal length $= \sqrt{12}$

Simplifying the square root:

$\sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3}$ units.

Therefore, the length of the diagonal of the parallelopiped is $2\sqrt{3}$.

The correct option is (A).

Given: The point P is $(3, 4, 5)$.

To Find: The coordinates of the foot of the perpendicular drawn from point P on the xy-plane.

Solution:

Let the coordinates of the point P be $(x_1, y_1, z_1) = (3, 4, 5)$.

The xy-plane is the plane where the z-coordinate of every point is zero. The equation of the xy-plane is $z=0$.

When a perpendicular is drawn from a point $(x_1, y_1, z_1)$ to the xy-plane, the foot of the perpendicular will have the same x and y coordinates as the original point, but its z-coordinate will be 0 (since it lies on the xy-plane).

So, the coordinates of the foot of the perpendicular, L, from the point P$(3, 4, 5)$ to the xy-plane are $(3, 4, 0)$.

Now, let's check the given options:

(A) $(3, 0, 0)$ - This point lies on the x-axis (where $y=0$ and $z=0$).

(B) $(0, 4, 5)$ - This point lies on the yz-plane (where $x=0$).

(C) $(3, 0, 5)$ - This point lies on the xz-plane (where $y=0$).

Our calculated coordinates for L are $(3, 4, 0)$, which does not match any of the options (A), (B), or (C).

Therefore, the correct answer is none of these.

The correct option is (D).

Given: The point P is $(3, 4, 5)$.

To Find: The coordinates of the foot of the perpendicular drawn from point P on the x-axis.

Solution:

Let the coordinates of the point P be $(x_1, y_1, z_1) = (3, 4, 5)$.

The x-axis is the line where the y-coordinate is zero and the z-coordinate is zero. The equations of the x-axis are $y=0$ and $z=0$.

When a perpendicular is drawn from a point $(x_1, y_1, z_1)$ to the x-axis, the foot of the perpendicular will have the same x-coordinate as the original point, but its y and z coordinates will be 0 (since it lies on the x-axis).

So, the coordinates of the foot of the perpendicular, L, from the point P$(3, 4, 5)$ to the x-axis are $(3, 0, 0)$.

Let's verify the options:

(A) $(3, 0, 0)$ - This point satisfies $y=0$ and $z=0$, so it is on the x-axis. It also has the same x-coordinate as P.

(B) $(0, 4, 0)$ - This point is on the y-axis.

(C) $(0, 0, 5)$ - This point is on the z-axis.

The coordinates of L are indeed $(3, 0, 0)$.

The correct option is (A).

The three axes OX, OY, OZ determine the coordinate planes.

In a three-dimensional Cartesian coordinate system, the three axes OX (x-axis), OY (y-axis), and OZ (z-axis) are mutually perpendicular and intersect at the origin O ($0, 0, 0$).

These axes define three principal planes, known as the coordinate planes:

1. The plane containing the x-axis (OX) and the y-axis (OY) is called the XY-plane. The equation of the XY-plane is $z=0$.

2. The plane containing the y-axis (OY) and the z-axis (OZ) is called the YZ-plane. The equation of the YZ-plane is $x=0$.

3. The plane containing the z-axis (OZ) and the x-axis (OX) is called the ZX-plane (or XZ-plane). The equation of the ZX-plane is $y=0$.

These three coordinate planes divide the entire space into eight regions, called octants.

The three planes determine a rectangular parallelopiped which has three pairs of rectangular faces.

A rectangular parallelopiped is a three-dimensional figure bounded by six rectangular faces.

These six faces consist of three pairs of opposite faces. The faces within each pair are parallel and congruent rectangles.

For example, in a typical rectangular parallelopiped:

1. The top and bottom faces form one pair.

2. The front and back faces form a second pair.

3. The left and right faces form a third pair.

Therefore, there are three pairs of rectangular faces.

The coordinates of a point are the perpendicular distance from the point on the respectives axes.

In a three-dimensional coordinate system, the coordinates $(x, y, z)$ of a point P are defined by its distances from the coordinate planes.

Specifically:

The x-coordinate ($x$) is the perpendicular distance of the point P from the YZ-plane (where $x=0$).

The y-coordinate ($y$) is the perpendicular distance of the point P from the ZX-plane (where $y=0$).

The z-coordinate ($z$) is the perpendicular distance of the point P from the XY-plane (where $z=0$).

These distances are taken with a sign, which depends on the side of the coordinate plane the point lies on relative to the origin.

The three coordinate planes divide the space into eight parts.

The three coordinate planes are the XY-plane ($z=0$), the YZ-plane ($x=0$), and the ZX-plane ($y=0$).

These three mutually perpendicular planes intersect at the origin ($0, 0, 0$).

They divide the entire three-dimensional space into eight regions, just as two perpendicular lines divide a plane into four quadrants.

These eight regions are called octants.

Each octant is defined by the sign of the x, y, and z coordinates.

If a point P lies in yz-plane, then the coordinates of a point on yz-plane is of the form $(0, y, z)$.

The yz-plane is defined as the plane where the x-coordinate of any point is zero.

Any point lying in the yz-plane is at a perpendicular distance of 0 from the yz-plane itself.

The coordinates of a point $(x, y, z)$ represent its distances from the coordinate planes.

For a point to be in the yz-plane, its distance from the yz-plane (which is the x-coordinate) must be zero.

Therefore, if a point lies in the yz-plane, its x-coordinate is $0$, while its y and z coordinates can be any real numbers.

Thus, the form of the coordinates is $(0, y, z)$.

The equation of yz-plane is $x=0$.

The yz-plane is one of the three principal coordinate planes in a three-dimensional Cartesian coordinate system.

It is defined as the plane that contains the y-axis and the z-axis.

Any point $(x, y, z)$ that lies on the yz-plane must have its distance from the yz-plane equal to zero. The distance of a point $(x, y, z)$ from the yz-plane is given by the absolute value of its x-coordinate, $|x|$.

For a point to be on the yz-plane, its x-coordinate must be $0$.

Therefore, the equation that describes all points lying on the yz-plane is $x=0$.

If the point P lies on z-axis, then coordinates of P are of the form $(0, 0, z)$.

A point that lies on the z-axis is located along the intersection of the ZX-plane ($y=0$) and the YZ-plane ($x=0$).

For a point to be on the z-axis, its perpendicular distance from both the YZ-plane (given by the x-coordinate) and the ZX-plane (given by the y-coordinate) must be zero.

Therefore, the x-coordinate and the y-coordinate of any point on the z-axis are both $0$.

The z-coordinate can be any real number, representing the position of the point along the z-axis relative to the origin.

Thus, the coordinates of a point P on the z-axis are of the form $(0, 0, z)$, where $z$ is a real number.

The equations of z-axis are $x=0, y=0$.

In a three-dimensional coordinate system, a line is typically defined by the intersection of two planes.

The z-axis is the line where the YZ-plane and the ZX-plane intersect.

The YZ-plane is defined by the equation $x=0$, because any point in this plane has an x-coordinate of zero.

The ZX-plane is defined by the equation $y=0$, because any point in this plane has a y-coordinate of zero.

Therefore, the points that lie on the z-axis are precisely those points that satisfy both conditions $x=0$ and $y=0$.

Thus, the equations of the z-axis are $x=0$ and $y=0$. The z-coordinate ($z$) can be any real number.

A line is parallel to xy-plane if all the points on the line have equal z-coordinate.

The xy-plane is defined by the equation $z=0$. A plane parallel to the xy-plane has an equation of the form $z=k$, where $k$ is a constant.

If a line is parallel to the xy-plane, it means the line lies entirely within a plane that is parallel to the xy-plane.

For all points $(x, y, z)$ lying on such a line, their distance from the xy-plane is constant.

The distance of a point $(x, y, z)$ from the xy-plane is given by $|z|$.

If this distance is constant and non-zero, or if the line lies within the xy-plane ($z=0$), the z-coordinate of all points on the line must be the same constant value.

Thus, for a line to be parallel to the xy-plane, the z-coordinate of every point on the line must be equal.

A line is parallel to x-axis if all the points on the line have equal y and z coordinates.

A line parallel to the x-axis means that the line extends infinitely in the positive and negative x directions, but its position relative to the YZ-plane and the ZX-plane remains fixed.

The y-coordinate of a point $(x, y, z)$ represents its perpendicular distance from the ZX-plane ($y=0$).

The z-coordinate of a point $(x, y, z)$ represents its perpendicular distance from the XY-plane ($z=0$).

If a line is parallel to the x-axis, all points $(x, y_0, z_0)$ on the line must have constant y and z values. The x-coordinate is free to change as you move along the line.

Thus, for a line to be parallel to the x-axis, the y-coordinate and the z-coordinate of every point on the line must be equal to constant values ($y_0$ and $z_0$ respectively).

x = a represent a plane parallel to yz-plane.

In a three-dimensional Cartesian coordinate system, the equation of a plane can be represented in various forms.

The equation $x=a$, where $a$ is a constant, means that the x-coordinate of every point on the surface is fixed at the value $a$. The y and z coordinates can take any real values.

The YZ-plane is defined by the equation $x=0$. Any point in the YZ-plane has its x-coordinate equal to 0.

A plane defined by $x=a$ is a set of points that are all at the same perpendicular distance (with sign) from the YZ-plane.

Since the distance from the YZ-plane is constant ($a$), the plane $x=a$ is parallel to the YZ-plane.

Similarly, $y=b$ represents a plane parallel to the ZX-plane, and $z=c$ represents a plane parallel to the XY-plane.

The plane parallel to yz - plane is perpendicular to x-axis.

The yz-plane is defined by the equation $x=0$. This plane contains the y-axis and the z-axis.

The x-axis is perpendicular to both the y-axis and the z-axis.

Therefore, the x-axis is perpendicular to the yz-plane.

A plane parallel to the yz-plane has an equation of the form $x=a$, where $a$ is a constant.

Any plane parallel to a given plane is perpendicular to the same lines or axes that the original plane is perpendicular to.

Since the yz-plane is perpendicular to the x-axis, any plane parallel to the yz-plane (like $x=a$) is also perpendicular to the x-axis.

The length of the longest piece of a string that can be stretched straight in a rectangular room whose dimensions are 10, 13 and 8 units are $\mathbf{\sqrt{333}}$ units.

The longest piece of a string that can be stretched straight in a rectangular room is along the space diagonal of the room.

Let the dimensions of the rectangular room be length $l$, width $w$, and height $h$.

Given dimensions are 10, 13, and 8 units. We can set $l=10$, $w=13$, and $h=8$ (the assignment order does not affect the diagonal length).

The length of the space diagonal ($d$) of a rectangular parallelopiped is given by the formula:

$d = \sqrt{l^2 + w^2 + h^2}$

Substituting the given dimensions:

$d = \sqrt{10^2 + 13^2 + 8^2}$

$d = \sqrt{100 + 169 + 64}$

$d = \sqrt{333}$

The length of the longest string is $\sqrt{333}$ units.

If the distance between the points (a, 2, 1) and (1, –1, 1) is 5, then a is equal to 5 or -3.

Let the two points be $P(a, 2, 1)$ and $Q(1, -1, 1)$.

The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in three dimensions is given by the formula:

$D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Given that the distance between P and Q is 5, we have:

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

$5 = \sqrt{(1 - a)^2 + (-3)^2 + (0)^2}$

$5 = \sqrt{(1 - a)^2 + 9 + 0}$

$5 = \sqrt{(1 - a)^2 + 9}$

Squaring both sides of the equation:

$5^2 = (1 - a)^2 + 9$

$25 = (1 - a)^2 + 9$

Subtract 9 from both sides:

$(1 - a)^2 = 25 - 9$

$(1 - a)^2 = 16$

Taking the square root of both sides:

$1 - a = \pm \sqrt{16}$

$1 - a = \pm 4$

This gives two possible cases:

Case 1: $1 - a = 4$

$-a = 4 - 1$

$-a = 3$

$a = -3$

Case 2: $1 - a = -4$

$-a = -4 - 1$

$-a = -5$

$a = 5$

Thus, the possible values for a are 5 or -3.

If the mid-points of the sides of a triangle AB; BC; CA are D (1, 2, – 3), E (3, 0, 1) and F (–1, 1, – 4), then the centriod of the triangle ABC is $(1, 1, -2)$.

Let the vertices of the triangle ABC be $A(x_A, y_A, z_A)$, $B(x_B, y_B, z_B)$, and $C(x_C, y_C, z_C)$.

Let the given midpoints be $D(1, 2, -3)$, $E(3, 0, 1)$, and $F(-1, 1, -4)$.

We know that the centroid of a triangle formed by joining the midpoints of the sides of a given triangle is the same as the centroid of the original triangle.

So, the centroid of triangle ABC is the same as the centroid of the triangle DEF whose vertices are the midpoints D, E, and F.

The coordinates of the centroid G of a triangle with vertices $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, and $(x_3, y_3, z_3)$ are given by:

$G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3}\right)$

Using the coordinates of the midpoints D(1, 2, -3), E(3, 0, 1), and F(-1, 1, -4), the centroid of triangle DEF is:

$x_G = \frac{1 + 3 + (-1)}{3} = \frac{3}{3} = 1$

$y_G = \frac{2 + 0 + 1}{3} = \frac{3}{3} = 1$

$z_G = \frac{-3 + 1 + (-4)}{3} = \frac{-6}{3} = -2$

The centroid of triangle DEF is $(1, 1, -2)$.

Since the centroid of triangle ABC is the same as the centroid of triangle DEF, the centroid of triangle ABC is $(1, 1, -2)$.

Here are the correct matches between Column $C_1$ and Column $C_2$:

(a) In xy-plane $\to$ (iii) z-coordinate is zero

Explanation: In the $xy$-plane, every point has its $z$-coordinate equal to zero.

(b) Point (2, 3, 4) lies in the $\to$ (i) Ist octant

Explanation: The first octant of the 3D coordinate system is defined by all points $(x, y, z)$ where $x > 0$, $y > 0$, and $z > 0$. Since 2, 3, and 4 are all positive, the point (2, 3, 4) lies in the first octant.

(c) Locus of the points having x coordinate 0 is $\to$ (ii) yz-plane

Explanation: The set of all points $(x, y, z)$ such that $x = 0$ forms the $yz$-plane.

(d) A line is parallel to x-axis if and only $\to$ (vi) if all the points on the line have equal y and z-coordinates.

Explanation: A line parallel to the $x$-axis has a direction vector proportional to $(1, 0, 0)$. This means as you move along the line, only the $x$-coordinate changes, while the $y$ and $z$ coordinates remain constant for all points on that line.

(e) If x = 0, y = 0 taken together will represent the $\to$ (iv) z-axis

Explanation: The equation $x = 0$ represents the $yz$-plane, and the equation $y = 0$ represents the $xz$-plane. The intersection of these two planes is the $z$-axis, where both the $x$ and $y$ coordinates are zero.

(f) z = c represent the plane $\to$ (v) plane parallel to xy-plane

Explanation: The equation $z = c$ (where $c$ is a constant) represents a plane where the $z$-coordinate of every point is $c$. This plane is parallel to the $xy$-plane (which is given by $z = 0$) and is located at a distance of $|c|$ units from the $xy$-plane.

(g) Planes x = a, y = b represent the line $\to$ (viii) parallel to z - axis.

Explanation: The intersection of the plane $x = a$ (parallel to the $yz$-plane) and the plane $y = b$ (parallel to the $xz$-plane) is a line where $x$ is always $a$ and $y$ is always $b$, but $z$ can take any real value. This line is parallel to the $z$-axis.

(h) Coordinates of a point are the distances from the origin to the feet of perpendiculars $\to$ (vii) from the point on the respective

Explanation: The coordinates $(x, y, z)$ of a point in 3D space represent the directed distances from the origin to the feet of the perpendiculars dropped from the point onto the respective coordinate axes (or planes, depending on the interpretation, but the options suggest axes/related concepts). The statement in (vii) is incomplete but is the intended match given the choices.

(i) A ball is the solid region in the space enclosed by a $\to$ (x) sphere

Explanation: A sphere is the set of all points in 3D space that are a fixed distance from a central point. A ball is the solid region containing all points within or on a sphere.

(j) Region in the plane enclosed by a circle is known as a $\to$ (ix) disc

Explanation: A circle is a curve. The region in the plane bounded by a circle (including the circle itself) is called a closed disk or disc. The region excluding the circle is an open disk.