Question:
Consider the line L given by the equation
\(\frac{x-3}{2} = \frac{y-1}{1} = \frac{z-2}{1}\).
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?
Consider the line L given by the equation
\(\frac{x-3}{2} = \frac{y-1}{1} = \frac{z-2}{1}\).
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?
\(\frac{x-3}{2} = \frac{y-1}{1} = \frac{z-2}{1}\).
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?
Show Hint
Find the foot of the perpendicular from A(2, 3, -1) to L, double it to get Q, then use the direction ratios (2, 1, 1) of L as the normal of P.
Updated On: Nov 4, 2026
- \((-1, 1, 2)\)
- \((1, 1, 1)\)
- \((1, 1, 2)\)
- \((1, 2, 2)\)
Show Solution
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The Correct Option is D
Solution and Explanation
Points of L: \((3+2t,\ 1+t,\ 2+t)\), direction \((2,1,1)\). Let A = (2, 3, -1).
Foot F of the perpendicular from A: \(\overrightarrow{AF} = (1+2t,\ t-2,\ t+3)\) must satisfy \(2(1+2t) + (t-2) + (t+3) = 0 \Rightarrow 6t + 3 = 0 \Rightarrow t = -\frac{1}{2}\).
So F = (2, 1/2, 3/2). Mirror image Q = 2F - A = (4-2, 1-3, 3+1) = (2, -2, 4).
Plane P passes through Q with normal (2, 1, 1): \(2(x-2) + (y+2) + (z-4) = 0 \Rightarrow 2x + y + z = 6\).
Check options: (-1,1,2): 1; (1,1,1): 4; (1,1,2): 5; (1,2,2): 2 + 2 + 2 = 6. Only \((1,2,2)\) lies on P.
Answer: (4) \((1, 2, 2)\)
Foot F of the perpendicular from A: \(\overrightarrow{AF} = (1+2t,\ t-2,\ t+3)\) must satisfy \(2(1+2t) + (t-2) + (t+3) = 0 \Rightarrow 6t + 3 = 0 \Rightarrow t = -\frac{1}{2}\).
So F = (2, 1/2, 3/2). Mirror image Q = 2F - A = (4-2, 1-3, 3+1) = (2, -2, 4).
Plane P passes through Q with normal (2, 1, 1): \(2(x-2) + (y+2) + (z-4) = 0 \Rightarrow 2x + y + z = 6\).
Check options: (-1,1,2): 1; (1,1,1): 4; (1,1,2): 5; (1,2,2): 2 + 2 + 2 = 6. Only \((1,2,2)\) lies on P.
Answer: (4) \((1, 2, 2)\)
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