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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Author: mathtuition88

https://mathtuition88.com/

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