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 $P_1:x+2y–z+1=0$ and $P_2:2x–y+z–1=0$. 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 $M?$

• A. $(-\frac{5}{6},0,\frac{1}{6})$
• B. $(-\frac{1}{6},-\frac{1}{3},\frac{1}{6})$
• C. $(0,-\frac{5}{6},-\frac{2}{3})$
• D. $(-\frac{1}{3},0,\frac{2}{3})$

Answer 1

Answer: (B), (C)

$(0,-\frac{5}{6},-\frac{2}{3})$

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

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 P1 will be line M, which will be parallel L.

Foot of perpendicular from (0, 0, 0) on plane P1 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.