Q. From the point $P(a,\ b,\ c)$, let perpendicular $PL$ and $PM$ be drawn to $YOZ$ and $ZOX$ planes, respectively. Then the equation of the plane $OLM$ is
A) $\frac{x}{a} + \frac{y}{b} + \frac{\mathcal{z}}{c} = 0$
B) $\frac{x}{a} + \frac{y}{b} - \frac{\mathcal{z}}{c} = 0$
C) $\frac{x}{a} - \frac{y}{b} - \frac{\mathcal{z}}{c} = 0$
D) $\frac{x}{a} - \frac{y}{b} + \frac{\mathcal{z}}{c} = 0$
Explanation
(b)
Coordinates of $L$ and $M$ are $(0,\ b,\ c)$ and $(a,\ 0,\ c)$, respectively. Therefore, the equation of the plane passing through (0, 0, 0), $(0,\ b,\ c)$ and $(a,\ 0,\ c)$ is
$\left| \begin{matrix}
x - 0 & y - 0 & \mathcal{z} - 0 \\
0 & b & c \\
a & 0 & c
\end{matrix} \right| = 0\ or\ \ \frac{x}{a} + \frac{y}{b} - \frac{\mathcal{z}}{c} = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.