Nick's Calculus Problem: Find Closest Point on Plane

[MarkFL]

Here is the question:

Need help with calculus problem? Optimization.?


Find the point on the plane x - 2y +3z = 6 that is closest to the point (0,1,1).

I can't figure this out. Does anyone know how to do this? Thanks.

I have posted a link there to this thread so the OP can see my work.

 

[MarkFL]

Hello Nick,

Let's let our objective function be the square of the distance from the point on the plane $(x,y,z)$ to the given point $(0,1,1)$:

\(\displaystyle f(x,y,z)=x^2+(y-1)^2+(z-1)^2\)

The point $(x,y,z)$ is constrained to be on the given plane, hence our constraint is:

\(\displaystyle g(x,y,z)=x-2y+3z-6=0\)

Using Lagrange multipliers, we obtain the system:

\(\displaystyle 2x=\lambda\)

\(\displaystyle 2(y-1)=-2\lambda\)

\(\displaystyle 2(z-1)=3\lambda\)

This implies:

\(\displaystyle \lambda=2x=1-y=\frac{2}{3}(z-1)\)

From this, we obtain:

\(\displaystyle y=1-2x,\,z=3x+1\)

Substituting for $y$ and $z$ into the constraint, we find:

\(\displaystyle x-2(1-2x)+3(3x+1)-6=0\)

Solving for $x$ (and using the values for $y$ and $z$ in terms of $x$) we find:

\(\displaystyle x=\frac{5}{14}\implies y=\frac{2}{7},\,z=\frac{29}{14}\)

Thus, the point on the plane $x-2y+3z=6$ that is closest to the point $(0,1,1)$ is:

\(\displaystyle \left(\frac{5}{14},\frac{2}{7},\frac{29}{14} \right)\)