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

Answer 1

Answer: (B)

$L:\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\lambda$

$\Rightarrow \frac{a}{I}=\frac{b}{-3}=\frac{c}{-5}$
$\therefore$ Any point on line $L$ is $(\lambda ,-3\lambda ,-5\lambda )$ Equation of line perpendicular to $P_1$ drawn from any point on $L$ is
$\frac{x-\lambda }{1}=\frac{y+3\lambda }{2}=\frac{z+5\lambda }{-1}=\mu$
$\therefore M(\mu +\lambda ,2\mu -3\lambda ,-\mu -5\lambda )$
But M lies on $P_1$,
$\therefore \mu +\lambda +4\mu -6\lambda +\mu +5\lambda +1=0$
$\Rightarrow \mu =\frac{-1}{6}$
$\therefore M\left(\lambda -\frac{1}{6},-3\lambda \frac{-1}{3},-5\lambda +\frac{1}{6}\right)$
For locus of $M$,
$x=\lambda -\frac{1}{6},y=-3\lambda -\frac{1}{3},z=5\lambda +\frac{1}{6}$
$\Rightarrow \frac{x+1/6}{1}=\frac{y+1/3}{-3}=\frac{z-1/6}{-5}=\lambda$
On checking the given point,
we find $\left(\frac{-1}{6},\frac{-1}{3},\frac{1}{6}\right)$ satisfy the above eqn.