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Question

Find the equation of the plane passing through the points whose coordinates are (-1, 1, 1) and (1, -1, 1) and perpendicular to the plane x + 2y + 2z = 5.

Vidyadip · Accepted Answer

We know that, equation of plane passing through the point $(x_1, y_1, z_1)$ is given by,
$a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$
Here, the required plane is pasing through$ (-1, 1, 1)$
$a(x + 1) + b(y - 1) + c(z - 1) = 0 ....(i)$
It is also passing through (-1, 1, 1), so it must satisfy the equation $(i),$
$a(1 + 1) + b(1 - 1) + c(1 - 1) = 0$
$2a - 2b = 0 ....(ii)$
We know that, plane $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ are perpendicular if
$a_1a_2 + b_1b_2 + c_1c_2= 0 ....(iii)$
Given that, plane (i) is perpendicular to plane
$x + 2y + 2z = 5 ....(iv)$
Using plane (i), (iv) in equation $(iii),$
$a_1a_2 + b_1b_2 + c_1c_2= 0$
$(a)(1) + (b)(2) + (c)(2) = 0$
$a + 2b + 2c = 0 ....(v)$
Solving (ii) and (v) by cross-multiplication,
$\frac{\text{a}}{(-2)(2)-(2)(0)}=\frac{\text{b}}{(1)(0)-(2)(2)}=\frac{\text{c}}{(2)(2)-(1)(-2)}$
$\frac{\text{a}}{-4-0}=\frac{\text{b}}{0-4}=\frac{\text{c}}{4+2}$
$\frac{\text{a}}{-4}=\frac{\text{b}}{-4}=\frac{\text{c}}{6}=\lambda(\text{say})$
$\text{a}=-4\lambda,\text{b}=-4\lambda,\text{c}=6\lambda$
Put a, b, c in equation (i),
$\text{a}(\text{x}+1)+\text{b}(\text{y}-1)+\text{c}(\text{z}-1)=0$
$(-4\lambda)(\text{x}+1)+(-4\lambda)(\text{y}-1)+(6\lambda)(\text{z}-1)=0$
$-4\lambda\text{x}+4\lambda-4\lambda\text{y}+4\lambda+6\lambda\text{z}-6\lambda=0$
$-4\lambda\text{x}-4\lambda\text{y}+6\lambda\text{z}-6\lambda=0$
Dividing by $(-2\lambda),$ we get
$2x + 2y - 3z + 3 = 0$
The equation of required plane is,
$2x + 2y - 3z + 3 = 0$

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