Question

If a plane has the intercept $a,b,c$ and is at a distance of $p$ units from the origin then prove that
$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{p^2}$

Answer 1

Distance of the point $\left(x_1,y_1,z_1\right)$ from the plane $Ax+By+Cz=D$ is

$\left|\frac{A{x}_1+B{y}_1+C{z}_1-D}{\sqrt{A^2+B^2+C^2}}\right|$

The equation of a plane having intercepts a,b,c on the X-, Y- and Z-axis respectively is

$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$

Comparing with $Ax+By+Cz=D$

$A=\frac{1}{a},B=\frac{1}{b},C=\frac{1}{c},D=1$

$\begin{array}{c}p=\left|\frac{-1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}\right|\\ p=\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}\\ \frac{1}{p}=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}\end{array}$

Squaring both sides

$\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$

Hence proved