Question

$P$ is a fixed point $(a,a,a)$ on a line through the origin equally inclined to the axes, then any plane through $P$ perpendicular to $OP$, makes intercepts on the axes, the sum of whose reciprocals is equal to

• A. $a$
• B. $3/2a$
• C. $3a/2$
• D. $a/2$

Answer 1

Answer: (A), (D)

The correct options are
A $a$
D $a/2$

Since, the line is equally inclined to the axes and passes through the origin, its direction ratios are $1,1,1$.
So, its equation is $\frac{x}{1}=\frac{y}{1}=\frac{z}{1}$
A point $P$ on it is given by $(a,a,a)$. So, equation of the plane through $P(a,a,a)$ and perpendicular to $OP$ is
$1(x-a)+1(y-a)+1(z-a)=0$
$(\because OP$ is normal to the plane)
i.e., $x+y+z=3a$
$\Rightarrow \frac{x}{3a}+\frac{y}{3a}+\frac{z}{3a}=1$
Intercepts on axes are $3a,3a$ and $3a$, therefore sum of reciprocals of these intercepts
$=\frac{1}{3a}+\frac{1}{3a}+\frac{1}{3a}=\frac{1}{a}$