Question

Use lagrange multipliers to find the point on the plane x − 2y + 3z = 6 that is closest to the point (0, 2, 5).

Answer 1

Lagrangian:

$L(x,y,z,\lambda)=x^2+(y-2)^2+(z-5)^2+\lambda(x-2y+3z-6)$

where the function we want to minimize is actually $\sqrt{x^2+(y-2)^2+(z-5)^2}$, but it's easy to see that $\sqrt{f(\mathbf x)}$ and $f(\mathbf x)$ have critical points at the same vector $\mathbf x$.

Derivatives of the Lagrangian set equal to zero:

$L_x=2x+\lambda=0\implies x=-\dfrac\lambda2$
$L_y=2(y-2)-2\lambda=0\implies y=2+\lambda$
$L_z=2(z-5)+3\lambda=0\implies z=5-\dfrac{3\lambda}2$
$L_\lambda=x-2y+3z-6=0$

Substituting the first three equations into the fourth gives

$-\dfrac\lambda2-2(2+\lambda)+3\left(5-\dfrac{3\lambda}2\right)=6$
$11-7\lambda=6\implies \lambda=\dfrac57$

Solving for $x,y,z$, we get a single critical point at $\left(-\dfrac5{14},\dfrac{19}7,\dfrac{55}{14}\right)$, which in turn gives the least distance between the plane and (0, 2, 5) of $\dfrac5{\sqrt{14}}$.