Vector proof relationship between distance and intercepts

Question

If $a$, $b$ and $c$ are the $x$-intercept, $y$-intercept, and $z$-intercept of a plane, respectively, and $d$ is the distance from the origin to the plane show that $$\frac{1}{d^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$$

Answer 1

The following will use $\mathbf x$ for vector $\overrightarrow{OX}$, and $|\mathbf{x}|^2 = \mathbf{x} \cdot \mathbf{x} = x^2$ for magnitudes.

Let $\mathbf{i} = \frac{\mathbf{a}}{|\mathbf{a}|}$, $\mathbf{j} = \frac{\mathbf{b}}{|\mathbf{b}|}$, $\mathbf{k} = \frac{\mathbf{c}}{|\mathbf{c}|}$ be the unit vectors along the axes. Let $D$ be the foot of the perpendicular from the origin $O$ onto the plane $ABC$ with components $d = u \,\mathbf{i} + v \,\mathbf{j} + w \,\mathbf{k}$.

• $OD \perp AB$ $\implies$ $0 = \mathbf d \cdot(\mathbf b - \mathbf a)= v \,b - u \,a$. Then by symmetry in $a,b,c$ it follows that $u\,a=v\,b=w\,c=\lambda$ for some $\lambda \in \mathbb{R}\;$ so $\;\mathbf d = \lambda(\frac{1}{a}\,\mathbf i + \frac{1}{b}\,\mathbf j+\frac{1}{c}\,\mathbf k)$.

• $OD \perp DA$ $\implies$ $0 = \mathbf d \cdot(\mathbf d - \mathbf a)= d^2 - \lambda \,\frac{1}{a}\,a = \lambda^2(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}) - \lambda$ from which it follows that $\lambda = 1 \,\Big/ \left(\frac{1}{a^2} + \frac{1}{c^2} + \frac{1}{c^2}\right)$.

Therefore $\mathbf d$ is given by the following expresion, from which the magnitude calculates easily: $$\mathbf d = \frac{1}{\frac{1}{a^2} + \frac{1}{c^2} + \frac{1}{c^2}}\left(\frac{1}{a}\,\mathbf i + \frac{1}{b}\,\mathbf j+\frac{1}{c}\,\mathbf k\right)$$