Nick's question at Yahoo! Answers regarding finding the point on a plane closest to a given point

MarkFL

Staff member
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

Staff member
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:

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