Coordinate Geometry in 3 Dimensions

Find the equation of the plane passing through the point $\left(1, -1, -1\right)$ and perpendicular to each of the planes $2 \left(x - 2y\right) + 2 \left(y - z\right) - \left(6 z + x\right) = 0$ and $2 \left(y - x\right)+ 3 \left(y + z\right) + 4 \left(x - z\right) = 0$

The given planes are:

$2 \left(x - 2y\right) + 2 \left(y - z\right) - \left(6 z + x\right) = 0$

i.e. $\;$ $2 x - 4y + 2y - 2z - 6z - x = 0$

i.e. $\;$ $x - 2y - 8z = 0$ $\;\;\; \cdots \; (1a)$

and $\;$ $2 \left(y - x\right)+ 3 \left(y + z\right) + 4 \left(x - z\right) = 0$

i.e. $\;$ $2y - 2x + 3y + 3z + 4x - 4z = 0$

i.e. $\;$ $2x + 5y - z = 0$ $\;\;\; \cdots \; (1b)$

The required plane passes through the point $\left(1, -1, -1\right)$

$\therefore \;$ Let the equation of the required plane be

$a \left(x - 1\right) + b \left(y + 1\right) + c \left(z + 1\right) = 0$ $\;\;\; \cdots \; (2)$

Since plane $(2)$ is perpendicular to each of the planes $(1a)$ and $(1b)$, we have,

$a - 2b -8c = 0$ $\;\;\; \cdots \; (3a)$ and

$2a + 5b -c = 0$ $\;\;\; \cdots \; (3b)$

[Note: for two perpendicular planes $a_1 x + b_1 y + c_1 z + d_1 = 0$ and $a_2 x + b_2 y + c_2 z + d_2 = 0$, we have $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$]

Solving equations $(3a)$ and $(3b)$ we get,

$\dfrac{a}{\left(-2\right) \left(-1\right) - \left(5\right) \left(-8\right)} = \dfrac{b}{\left(1\right) \left(-1\right) - \left(2\right) \left(-8\right)} = \dfrac{c}{\left(1\right) \left(5\right) - \left(2\right) \left(-2\right)} = \lambda$ (say)

i.e. $\;$ $\dfrac{a}{42} = \dfrac{b}{15} = \dfrac{c}{9} = \lambda$

i.e. $\;$ $a = 42 \lambda$, $\;$ $b = 15 \lambda$, $\;$ $c = 9 \lambda$

Substituting the value of $a$, $b$ and $c$ in equation $(2)$, the equation of the required plane is

$42 \lambda \left(x - 1\right) + 15 \lambda \left(y + 1\right) + 9 \lambda \left(z + 1\right) = 0$

i.e. $\;$ $42 x - 42 + 15 y + 15 + 9z + 9 = 0$

i.e. $\;$ $42x + 15y + 9z = 18$

i.e. $\;$ $14x + 5y + 3z = 6$

Coordinate Geometry in 3 Dimensions

Find the points on the line $\dfrac{x + 2}{3} = \dfrac{y + 1}{2} = \dfrac{z - 3}{2}$ at a distance of $3 \sqrt{2}$ units from the point $\left(1, 2, 3\right)$.

Given equation of line is: $\;$ $\dfrac{x + 2}{3} = \dfrac{y + 1}{2} = \dfrac{z - 3}{2} = \lambda$ (say)

$\implies$ $x = 3 \lambda - 2$, $\;$ $y = 2 \lambda - 1$, $\;$ $z = 2 \lambda + 3$

$\therefore \;$ Any point on the given line is: $P \left(3 \lambda -2, 2 \lambda - 1, 2 \lambda + 3\right)$

Given: Distance of $P$ from $Q \left(1, 2, 3\right)$ is $3 \sqrt{2}$ units

i.e. $\sqrt{\left(3 \lambda - 2 - 1\right)^2 + \left(2 \lambda - 1 - 2\right)^2 + \left(2 \lambda + 3 - 3\right)^2} = 3 \sqrt{2}$

i.e. $\;$ $\left(3 \lambda - 3\right)^2 + \left(2 \lambda - 3\right)^2 + 4 \lambda^2 = 18$

i.e. $\;$ $9 \lambda^2 - 18 \lambda + 9 + 4 \lambda^2 - 12 \lambda + 9 + 4 \lambda^2 = 18$

i.e. $\;$ $17 \lambda^2 - 30 \lambda = 0$

i.e. $\;$ $\lambda \left(17 \lambda - 30\right) = 0$

$\implies$ $\lambda = 0$ $\;$ or $\;$ $\lambda = \dfrac{30}{17}$

When $\lambda = 0$, $\;$ $P = \left(-2, -1, 3\right)$

When $\lambda = \dfrac{30}{17}$, $\;$ $P = \left(\dfrac{90}{17} - 2, \; \dfrac{60}{17} - 1, \; \dfrac{60}{17} + 3\right)$ $\;$ i.e. $\;$ $P = \left(\dfrac{56}{17}, \; \dfrac{43}{17}, \; \dfrac{111}{17}\right)$