Question

From a point $P(a,b,c)$ perpendiculars $PM$ and $PN$ are drawn to $zx$ and $xy$-planes respectively, $O$ is the origin. An equation of the plane $OMN$ 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 A $\frac{x}{a}-\frac{y}{b}-\frac{z}{c}=0$

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