Question

Find the point on the plane ax + by + cz = d at minimum distance from the origin using the method of lagrange multipliers.

Answer 1

The distance between some point $(x,y,z)$ and the origin is given by

$f(x,y,z)=\sqrt{x^2+y^2+z^2}$

so this is the function we're trying to minimize. But notice that $f(x,y,z)$ and $f(x,y,z)^2$ attain their critical points at the same $(x,y,z)$, so we can solve the same problem by minimizing $x^2+y^2+z^2$ instead.

So let's take the Lagrangian to be

$L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(ax+by+cz-d)$

with partial derivatives (set equal to 0)

$L_x=2x+a\lambda=0$
$L_y=2y+b\lambda=0$
$L_z=2z+c\lambda=0$
$L_\lambda=ax+by+cz-d=0$

Now, notice that

$aL_x+bL_y+cL_z=2(ax+by+cz)+(a^2+b^c+c^2)\lambda=0$
$\implies\lambda=-\dfrac{2d}{a^2+b^2+c^2}$

and we can use this to solve for $x,y,z$. We get

$x=\dfrac{ad}{a^2+b^2+c^2}$
$y=\dfrac{bd}{a^2+b^2+c^2}$
$z=\dfrac{cd}{a^2+b^2+c^2}$

At this critical point, we get a minimum distance of

$\sqrt{\left(\dfrac{ad}{a^2+b^2+c^2}\right)^2+\left(\dfrac{bd}{a^2+b^2+c^2}\right)^2+\left(\dfrac{cd}{a^2+b^2+c^2}\right)^2}=\sqrt{\dfrac{d^2}{a^2+b^2+c^2}}$