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The plane — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsThe plane5 Marks
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 $x+2 y+2 z=5$
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Answer

We know that, equation of plane passing through the point $\left(x_1, y_1, z_1\right)$ is given by,
$a\left(x-x_1\right)+b\left(y-y_1\right)+c\left(z-z_1\right)=0$
Here, the required plane is pasing through $(-1,1,1)$
$a(x+1)+b(y-1)+c(z-1)=0 \ldots \text { (i) }$
It is also passing through $(-1,1,1)$, so it must satisfy the equation (i),
$a(1+1)+b(1-1)+c(1-1)=0$
$2 a-2 b=0 \ldots \text { (ii) }$
We know that, plane $a_1 x+b_1 y+c_1 z+d_1=0$ and $a_2 x+b_2 y+c_2 z+d_2=0$ are perpendicular if $a_1 a_2+b_1 b_2+c_1 c_2=0 \ldots$ (iii)
Given that, plane (i) is perpendicular to plane
$x+2 y+2 z=5 \ldots \text { (iv) }$
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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