Question:
Let P be a plane passing through the points (2, 1, 0), (4, 1, 1) and (5, 0, 1) and R be any point (2, 1, 6). Then the image of R in the plane P is :
Let P be a plane passing through the points (2, 1, 0), (4, 1, 1) and (5, 0, 1) and R be any point (2, 1, 6). Then the image of R in the plane P is :
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Find the plane through the three points using the cross product of two in-plane vectors, then reflect R using the foot of the perpendicular.
Updated On: Oct 31, 2026
- (4, 3, 2)
- (6, 5, - 2)
- (3, 4, -2)
- (6, 5, 2)
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The Correct Option is B
Solution and Explanation
Let A(2, 1, 0), B(4, 1, 1), C(5, 0, 1). \(\vec{AB}=(2,0,1)\), \(\vec{AC}=(3,-1,1)\).
Normal \(\vec n=\vec{AB}\times\vec{AC}=(0\cdot1-1\cdot(-1),\ 1\cdot3-2\cdot1,\ 2\cdot(-1)-0)=(1,1,-2)\).
Plane: \((x-2)+(y-1)-2z=0\Rightarrow x+y-2z=3\). (Check B: 4 + 1 - 2 = 3; C: 5 + 0 - 2 = 3.)
For R(2, 1, 6): \(\dfrac{x-2}{1}=\dfrac{y-1}{1}=\dfrac{z-6}{-2}=\lambda\) with \(\lambda=-2\cdot\dfrac{(2+1-12-3)}{6}=4\), image \(=(2+4,\ 1+4,\ 6-8)=(6,5,-2)\).
Answer: (B) (6, 5, -2)
Normal \(\vec n=\vec{AB}\times\vec{AC}=(0\cdot1-1\cdot(-1),\ 1\cdot3-2\cdot1,\ 2\cdot(-1)-0)=(1,1,-2)\).
Plane: \((x-2)+(y-1)-2z=0\Rightarrow x+y-2z=3\). (Check B: 4 + 1 - 2 = 3; C: 5 + 0 - 2 = 3.)
For R(2, 1, 6): \(\dfrac{x-2}{1}=\dfrac{y-1}{1}=\dfrac{z-6}{-2}=\lambda\) with \(\lambda=-2\cdot\dfrac{(2+1-12-3)}{6}=4\), image \(=(2+4,\ 1+4,\ 6-8)=(6,5,-2)\).
Answer: (B) (6, 5, -2)
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