MHB 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 $(0,1,1)$:

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

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

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

Using Lagrange multipliers, we obtain the system:

$$2x=\lambda$$

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

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

This implies:

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

From this, we obtain:

$$y=1-2x,\,z=3x+1$$

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

$$x-2(1-2x)+3(3x+1)-6=0$$

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

$$x=\frac{5}{14}\implies y=\frac{2}{7},\,z=\frac{29}{14}$$

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

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