Consider the line L given by the equation
${{x - 3} \over 2} = {{y - 1} \over 1} = {{z - 2} \over 1}$.
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
Plane p is ${ \bot ^r}$ to line ${{x - 3} \over 2} = {{y - 1} \over 1} = {{z - 2} \over 1}$ & passes through pt. (2, 3) equation of plane p <br><br>2(x $-$ 2) + 1(y $-$ 3) + 1 (z + 1) = 0<br><br>2x + y + z $-$ 6 = 0<br><br>Point (1, 2, 2) satisfies above equation
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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