# How to find the distance of a plane to the origin

Given the equation of a plane: ax+by+cz=D, or in vector notation $\mathbf{r}\cdot \left( \begin{array}{c} a\\ b\\ c\\ \end{array}\right)=D$, how do we find the (shortest) distance of a plane to the origin?

(When a question asks for the distance of a plane to the origin, by definition it means the shortest distance.)

One way to derive the formula is this:

## Derivation

Let X be the point on the plane nearest to the origin.

$\overrightarrow{OX}$ must be perpendicular to the plane, i.e. parallel to the normal vector $\mathbf{n}=\left(\begin{array}{c}a\\b\\c\\\end{array}\right)$.

Furthermore, X lies on the plane, hence we have $\boxed{\overrightarrow{OX}\cdot\mathbf{n}=D}$

Using the formula for dot product, we can get $|\overrightarrow{OX}\cdot\mathbf{n}|=|\overrightarrow{OX}||\mathbf{n}|\cos \theta=D$

Since $\overrightarrow{OX}$ is parallel to $\mathbf{n}$, $\theta$ is either 0 or 180 degrees, hence $\cos \theta$ is either 1 or -1.

Thus, we have $|\overrightarrow{OX}||\mathbf{n}|=|D|$.

The shortest distance from the point X to the origin is then $\displaystyle|\overrightarrow{OX}|=\frac{|D|}{|\mathbf{n}|}=\frac{|D|}{\sqrt{a^2+b^2+c^2}}$

Ans: Shortest distance from point to plane is $\displaystyle\boxed{\frac{|D|}{\sqrt{a^2+b^2+c^2}}}$

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