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. $\frac{3}{2a}$
• C. $\frac{3a}{2}$
• D. $\frac{1}{a}$

Answer 1

The correct option is D $\frac{1}{a}$

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
[$\because$ OPis normal to the plane]
i.e. x+y+z =3a or $\frac{x}{3a}+\frac{y}{3a}+\frac{z}{3a}=1$
Intercepts on axes are 3a, 3a, 3a. Therefore sum of reciprocals of these intercepts
$=\frac{1}{3a}+\frac{1}{3a}+\frac{1}{3a}=\frac{3}{3a}=\frac{1}{a}$.