Question

Prove that if a plane has the intercepts a, b, c and is at a distance of P units from the origin, then \(\frac {1}{a^2}+\frac {1}{b^2}+\frac {1}{c^2} =\frac {1}{p^2}\).

Answer 1

The equation of a plane having intercepted a, b, c with x, y and z axes respectively is given by,

\(\frac xa+\frac yb+\frac zc=1\)         ...(1)

The distance (p) of the plane from the origin is given by,

\(P=|\frac {\frac 0a+\frac 0b+\frac 0c-1}{\sqrt {(\frac 1a)^2+(\frac 1b)^2+(\frac 1c)^2}}|\)

⇒ \(p=\frac {1}{\sqrt{\frac {1}{a^2}+\frac {1}{b^2}+\frac {1}{c^2} }}\)

⇒ \(p^2=\frac {1}{{\frac {1}{a^2}+\frac {1}{b^2}+\frac {1}{c^2} }}\)

⇒ \(\frac {1}{p^2}=\frac{1} {a^2}+\frac {1}{b^2}+\frac {1}{c^2} \)