In R 3 , let L be a straight line passing through the origin. Suppose that all the points on L are at a constant distance from the two planes P 1 : x + 2 y - z + 1 = 0 and P 2 : 2 x - y + z - 1 = 0 .

Question

In R 3 , let L be a straight line passing through the origin. Suppose that all the points on L are at a constant distance from the two planes P 1 : x + 2 y - z + 1 = 0 and P 2 : 2 x - y + z - 1 = 0 . Let M be the locus of the feet of the perpendiculars drawn from the points on L to the plane P 1 . Which of the following points lie (s) on M ?

TestHub · Accepted Answer

Line L will lie on one of the angle bisector planes and will be parallel to line of intersection of the given planes. Also locus of foot of perpendicular of line L in plane $P_{1}$ will be line M, which will be parallel L.
Foot of perpendicular from (0, 0, 0) on plane $P_{1}$ is $(- \frac{1}{6}, - \frac{1}{3}, \frac{1}{6})$
Hence equation of line M
$\frac{x + \frac{1}{6}}{1} = \frac{y + \frac{1}{3}}{- 3} = \frac{z - \frac{1}{6}}{- 5}$
Where $\hat{i} - 3 \hat{j} - 5 \hat{k}$ is vector parallel to line of intersection of $P_{1}$ and $P_{2}$ .
On checking $(0, - \frac{5}{6} - \frac{2}{3}), (- \frac{1}{6}, - \frac{1}{3}, \frac{1}{6})$ satisfy the above equation.

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