3D Geometry 3 Question 47

####47. 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

(a) 0

(b) 1

(c) $\sqrt{2}$

(d) $2 \sqrt{2}$

(2006, 3M)

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Answer:

Correct Answer: 47. (d)

Solution:

  1. 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{aligned} & \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}{0} . \end{aligned} $

So, the equation of plane is $x-1+y+2=0$ or

$x+y+1=0$

Its distance from the point $(1,2,2)$ is $\frac{|1+2+1|}{\sqrt{2}}=2 \sqrt{2}$



NCERT Chapter Video Solution

Dual Pane