Consider the line L given by the equation
(x - 3)/2 = (y - 1)/1 = (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?

Solution

Maths sir · 4 steps

You do not need Q at all: the plane through Q perpendicular to L also passes through the original point.

1. Q is not needed. Let A = (2, 3, -1) and let M be the foot of the perpendicular from A to L. The segment AQ passes through M and is perpendicular to L. So the plane through M perpendicular to L contains A, M and Q, and this is the plane P.
2. Write the plane. The direction of L, (2, 1, 1), is the normal to P. Taking the plane through A with this normal gives the equation below.

   2(x-2) + 1(y-3) + 1(z+1) = 0 ⇒ 2x + y + z = 6

3. Quick check. A point on L is (3+2t, 1+t, 2+t). Putting it in the plane gives 9 + 6t = 6, so t = -1/2 and M = (2, 1/2, 3/2). Then Q = 2M - A = (2, -2, 4), and 2(2) - 2 + 4 = 6, so Q does lie on P.

   Option	Point	2x + y + z	On P?
   (a)	(-1, 1, 2)	-2 + 1 + 2 = 1	no
   (b)	(1, 1, 1)	2 + 1 + 1 = 4	no
   (c)	(1, 1, 2)	2 + 1 + 2 = 5	no
   (d)	(1, 2, 2)	2 + 2 + 2 = 6	yes

4. So. Only (1, 2, 2) satisfies 2x + y + z = 6, so the answer is (d).

The idea to keep. A point and its mirror image in a line lie on the same plane perpendicular to that line. Write that plane through the point you are given and skip finding the image.

Answer from the NTA final answer key. Solution by PrepWiser.