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

Answer 1

Toolbox:

• The perpendicular distance of a plane from the origin is $p=\large\frac{d}{\sqrt{A^2+B^2+C^2}}$

Step 1:

Equation of a plane with intercepts $a,b,c$ is $\large\frac{x}{a}+\large\frac{y}{b}+\large\frac{z}{c}=$$1$

The perpendicular distance of a plane is $p=\large\frac{Ax_1+By_1+cz_1+d}{\sqrt{A^2+B^2+C^2}}$

From the origin $(x_1,y_1,z_1)\Rightarrow (0,0,0)$

We can written as

The perpendicular distance of a plane from the origin is $p=\large\frac{d}{\sqrt{A^2+B^2+C^2}}$

$\Rightarrow p=\large\frac{1}{\sqrt{\big(\Large\frac{1}{a}\big)^2+\big(\Large\frac{1}{b}\big)^2+\big(\Large\frac{1}{c}\big)^2}}$

$\Rightarrow p=\large\frac{1}{\sqrt{\big(\Large\frac{1}{a^2}\big)+\big(\Large\frac{1}{b^2}\big)+\big(\Large\frac{1}{c^2}\big)}}$

Step 2:

Squaring on both sides we get

$p^2=\large\frac{1}{\big(\Large\frac{1}{a^2}\big)+\big(\Large\frac{1}{b^2}\big)+\big(\Large\frac{1}{c^2}\big)}$

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

Step 3:

Reciprocating on both sides and simplifying we get

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

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

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

Hence proved.