Three Dimensional Geometry Question 244
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
Options:
a
B) $ \frac{3}{2a} $
C) $ \frac{3a}{2} $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
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 $ 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} $ .
BETA
