3D Geometry Ques 55
55. A plane passes through $(1,-2,1)$ and is perpendicular to two planes $2 x-2 y+z=0$ and $x-y+2 z=4$, then the distance of the plane from the point $(1,2,2)$ is
0
1
(c) $\sqrt{2}$
(d) $2 \sqrt{2}$
(2006, 3M)
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Answer:
Correct Answer: 55.(d)
Solution:
- Let the equation of plane be
$ a(x-1)+b(y+2)+c(z-1)=0 $
which is perpendicular to $2 x-2 y+z=0$ and $x-y+2 z=4$.
$ \begin{alignedat} & \Rightarrow \quad 2 a-2 b+c=0 \text { and } a-b+2 c=0 \\ & \Rightarrow \quad \frac{a}{-3}=\frac{b}{-3}=\frac{c}{0} \Rightarrow \frac{a}{1}=\frac{b}{1}=\frac{c}{\text{undefined}} \end{aligned} $
So, the equation of a plane is $x-1+y+2=0$ or
$x+y+1=0$
Its distance from the point $(1,2,2)$ is $\frac{|1+2+2|}{\sqrt{2}}=2 \sqrt{2}$
BETA
