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

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Solution

Let the equation of the plane be

$l x + m y + n z = d$ where $l, m, n, d$ are constants. Now, it is given that the perpendicular distance of the plane from the origin is P units. Hence

$| \frac{l (0) + m (0) + n (0) - d}{\sqrt{l^{2} + m^{2} + n^{2}}} | = P$

Or

$d = P \sqrt{l^{2} + m^{2} + n^{2}}$ .

Hence the equation becomes

$l x + m y + n z = P \sqrt{l^{2} + m^{2} + n^{2}}$ .

Now the intercepts made are a,b,c. Hence the respective points of
intersection of the plane and the co-ordinate axes will be $(a, 0, 0), (0, b, 0), (0, 0, c)$ .

Therefore substitution in the equation of the plane on by one gives us

$l (a) = P \sqrt{l^{2} + m^{2} + n^{2}}$ . Or

$\frac{P}{a} = \frac{l}{\sqrt{l^{2} + m^{2} + n^{2}}}$

Similarly

$\frac{P}{b} = \frac{m}{\sqrt{l^{2} + m^{2} + n^{2}}}$

$\frac{P}{c} = \frac{n}{\sqrt{l^{2} + m^{2} + n^{2}}}$

Hence

$\frac{P^{2}}{a^{2}} + \frac{P^{2}}{b^{2}} + \frac{P^{2}}{c^{2}} = \frac{l^{2}}{l^{2} + m^{2} + n^{2}} + \frac{m^{2}}{l^{2} + m^{2} + n^{2}} + \frac{n^{2}}{l^{2} + m^{2} + n^{2}}$

Hence

$\frac{P^{2}}{a^{2}} + \frac{P^{2}}{b^{2}} + \frac{P^{2}}{c^{2}} = 1$
or
$\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} = \frac{1}{P^{2}}$ .

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