# Point on plane closest to origin

In Euclidean space, the point on a plane $ax + by + cz = d$ that is closest to the origin has the Cartesian coordinates $(x,y,z)$, where

$\displaystyle x = \frac {ad}{{a^2+b^2+c^2}}$, $\displaystyle y = \frac {bd}{{a^2+b^2+c^2}}$, and $\displaystyle z = \frac {cd}{{a^2+b^2+c^2}}$.

## Restatement using linear algebra

The formula for the closest point may be expressed more succinctly using notation from linear algebra. The expression $ax+by+cz$ in the definition of a plane is a dot product $(a,b,c)\cdot(x,y,z)$, and the expression $a^2+b^2+c^2$ appearing in the solution is the squared norm $|(a,b,c)|^2$. Thus, if $\mathbf{v}=(a,b,c)$ is a given vector, the plane may be described as the set of vectors $\mathbf{w}$ for which $\mathbf{v}\cdot\mathbf{w}=d$ and the closest point on this plane is the vector

$\mathbf{p}=\frac{\mathbf{v}d}{|\mathbf{v}|^2}$.[1][2]

The Euclidean distance from the origin to the plane is the norm of this point,

$\frac{d}{|\mathbf{v}|} = \frac{d}{\sqrt{a^2+b^2+c^2}}$.

## Why this is the closest point

In either the coordinate or vector formulations, one may verify that the given point lies on the given plane by plugging the point into the equation of the plane.

To see that it is the closest point to the origin on the plane, observe that $\mathbf{p}$ is a scalar multiple of the vector $\mathbf{v}$ defining the plane, and is therefore orthogonal to the plane. Thus, if $\mathbf{q}$ is any point on the plane other than $\mathbf{p}$ itself, then the line segments from the origin to $\mathbf{p}$ and from $\mathbf{p}$ to $\mathbf{q}$ form a right triangle, and by the Pythagorean theorem the distance from the origin to $q$ is

$\sqrt{|\mathbf{p}|^2+|\mathbf{p}-\mathbf{q}|^2}$.

Since $|\mathbf{p}-\mathbf{q}|^2$ must be a positive number, this distance is greater than $|\mathbf{p}|$, the distance from the origin to $\mathbf{p}$.[2]

Alternatively, it is possible to rewrite the equation of the plane using dot products with $\mathbf{p}$ in place of the original dot product with $\mathbf{v}$ (because these two vectors are scalar multiples of each other) after which the fact that $\mathbf{p}$ is the closest point becomes an immediate consequence of the Cauchy–Schwarz inequality.[1]

## References

1. ^ a b Strang, Gilbert; Borre, Kai (1997), Linear Algebra, Geodesy, and GPS, SIAM, pp. 22–23, ISBN 9780961408862.
2. ^ a b Shifrin, Ted; Adams, Malcolm (2010), Linear Algebra: A Geometric Approach (2nd ed.), Macmillan, p. 32, ISBN 9781429215213.