Point on plane closest to origin

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Here we will find the point on an arbitrary plane that is closest to the origin using 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, g = ax + by + cz - d.

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

f^* = f(x,y,z) - \lambda g = \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 \frac \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 derive y and z as functions of x:

y = \frac {bx} a


z = \frac {cx} a

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


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

From x you can solve for y and z:

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


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

And 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}}