Point on plane closest to origin

In Euclidean 3-space we will find the point on an arbitrary plane that is closest to the origin using the method of Lagrange multipliers.

First, let us start with an arbitrary plane, ax + by + cz = d. The distance, L, from the origin to a point (x,y,z) on the plane is given by:

$L = \sqrt{x^2 + y^2 + z^2}.$

Therefore the function that we want to minimize is: $f(x,y,z) = \sqrt{x^2 + y^2 + z^2}.$

Our one constraint on x, y, and z is that the point (x,y,z) must lie on the given plane. Thus, we define g(x,y,z) = ax + by + cz - d.

Next we define a new function with a Lagrange multiplier, λ

$f^* = f(x,y,z) - \lambda g(x,y,z) = \sqrt{x^2 + y^2 + z^2} - \lambda (ax + by + cz - d).$

Take the partial of f ^* with respect to x, y, and z and set each to zero.

$\frac {\partial f^*}{\partial x} = \frac x {\sqrt{x^2 + y^2 + z^2}} - \lambda a = 0$

$\frac {\partial f^*}{\partial y} = \frac y {\sqrt{x^2 + y^2 + z^2}} - \lambda b = 0$

$\frac {\partial f^*}{\partial z} = \frac z {\sqrt{x^2 + y^2 + z^2}} - \lambda c = 0$

Now each partial includes a λ and a $\sqrt{x^2 + y^2 + z^2}$ term.

If we solve each equation for $\lambda \sqrt{x^2 + y^2 + z^2}$ and set them equal to one another

we can find the relation:

$\frac ax = \frac by = \frac cz .$

From this we can obtain y and z as functions of x:

$y = \frac {bx} a$ and $z = \frac {cx} a .$

Substitute these for y and z in the equation of the plane and solve for x to obtain:

$x = \frac {ad}{{a^2+b^2+c^2}} .$

With this x you can solve for y and z:

$y= \frac {bd}{{a^2+b^2+c^2}} ,$

and

$z = \frac {cd}{{a^2+b^2+c^2}} .$

Hence the point on the plane closest to the origin is:

$(x,y,z) = \Big(\frac {ad}{{a^2+b^2+c^2}} , \frac {bd}{{a^2+b^2+c^2}} , \frac {cd}{{a^2+b^2+c^2}}\Big)$

and the distance is given by:

$L = \sqrt{x^2 + y^2 + z^2} = \frac {|d|}{\sqrt{a^2+b^2+c^2}} .$

An applied solution to this problem using linear algebra

This approach is useful in computational geometry and applications of computer graphics.

Given a plane defined by three points p₀, p₁, and p₂.

The normal for this plane is

$n = \frac{(p_1-p_0)\times(p_2-p_0)}{\|(p_1-p_0)\times(p_2-p_0)\|}$

The nearest point p on the plane to the origin is the orthogonal projection of any point on the plane onto the plane's normal

p = nd

Composed of the distance d from the plane to the origin, which is the dot product of n and any point on the plane such as p₀.

$d = n \cdot p_0$