Question

There is a point P(a, a, a) on the line passing through the origin and equally inclined with axes. The equation of the plane perpendicular to OP and passing through P cuts the intercepts on axes, the sum of whose reciprocals is –

• A. a
• B. 3/2a
• C. 3a/2
• D. 1/a

Answer 1

Answer: (D)

1/a

Answer 2

OP = $\sqrt{a^{2} + a^{2} + a^{2}} = a\sqrt{3}$

D.R.’s of OP = $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$

Equation of Reqd. plane is lx +my + nz = p

Ž x + y + z = 3a Ž $\frac{x}{3a} + \frac{y}{3a} + \frac{z}{3a}$ = 1

Intercept on axis = 3a, 3a, 3a

Sum of their reciprocals =$\frac{1}{3a} + \frac{1}{3a} + \frac{1}{3a}$ = $\frac{1}{a}$.