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Consider the line L given by the equation \frac{\mathrm{x}-3}{2}=\frac{\mathrm{y}-1}{1}=\frac{\mathrm{z}-2}{1}. Let Q be the mirror image of the point (2,3,-1) with respect to L. Let a plane P be such that it passes through Q, and the line L is perpendicular to P. Then which of the following points is on the plane P ?
 
Option: 1 (-1,1,2)
 
Option: 2 (1,1,1)
Option: 3 (1,1,2)
 
Option: 4 (1,2,2)

Answers (1)

best_answer

Let Q be the image of A(2,3,-1) in given line

So A Q \perp line L

Also given that plane P is perpendicular to line L

\therefore A also lies on plane P.
 

For Plane P.
A(2,3,-1) lies on the plane and L \perp P

\Rightarrow \vec{n}=2 i+j+k \text {. }

Plane\: is \: 2(x-2)+1(y-3)+1(z+1)=0

             \Rightarrow 2 x+y+z-6=0

Checking the given options, (D) (1,2,2) satisfies it.

Hence option (4) is correct answer.

Posted by

Kuldeep Maurya

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