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 $L$ on the plane $P_1.$ Which of the following points lie(s) on $M?$

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

Answer 1

Answer: (A), (B)

The correct options are
A $(0,-\frac{5}{6},-\frac{2}{3})$
B $(-\frac{1}{6},-\frac{1}{3},\frac{1}{6})$

Clearly, $L$ is parallel to the planes $P_1$ and $P_2$.
So, the direction ratios of the line $L$
$={\hat{n}}_1\times {\hat{n}}_2=(1,-3,-5)$
So, the equation of the line is $L:\lambda (1,-3,-5)$
Let the feet of the perpendicular from the line $L$ on the plane $P_1$ be $(\alpha ,\beta ,\gamma ).$

$\Rightarrow \frac{\alpha -\lambda }{1}=\frac{\beta +3\lambda }{2}=\frac{\gamma +5\lambda }{-1}=-\frac{(\lambda -6\lambda +5\lambda +1)}{6}$
$\Rightarrow \frac{\alpha +\frac{1}{6}}{1}=\frac{\beta +\frac{2}{6}}{-3}=\frac{\gamma -\frac{1}{6}}{-5}$

So, the locus of $M$ is $\frac{x+\frac{1}{6}}{1}=\frac{y+\frac{1}{3}}{-3}=\frac{z-\frac{1}{6}}{-5}$