Question

From the point $P(a,b,c)$, let perpendiculars $PL$ and $PM$ be drawn to $YOZ$ and $ZOX$ planes, respectively. Then the equation of the plane $OLM$ is-

• A. $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0$
• B. $\frac{x}{a}+\frac{y}{b}-\frac{z}{c}=0$
• C. $\frac{x}{a}-\frac{y}{b}-\frac{z}{c}=0$
• D. $\frac{x}{a}-\frac{y}{b}+\frac{z}{c}=0$

Answer 1

The correct option is B

$\frac{x}{a}+\frac{y}{b}-\frac{z}{c}=0$

If perpendicular $PL$ and $PM$ drawn from the point $P(a,b,c)$ to the plane $YOZ$ and $ZOX$ then coordinates of $L$ and $M$ are,
$L=(0,b,c)$ and $M=(a,0,c)$
Now general equation of plane passes through origin is given by,
$x+py+qz=0$
Also this plane passes through $L$ and $M$
$\Rightarrow pb+qc=0$...(1)
and $a+qc=0$ ...(2)
Solving (1) and (2), we get
$p=\frac{a}{b}$ and $q=-\frac{a}{c}$
Hence, equation required plane $OLM$ is,
$\frac{x}{a}+\frac{y}{b}-\frac{z}{c}=0$