What Is the Closest Point on the Plane x − y + z = 8 to (1, 3, 6)?

[carl123]

Find the point on the plane x − y + z = 8 that is closest to the point (1, 3, 6).

This what I have so far

Z = 8-x+y

d = Sqrt ((x-1)² + (y-3)² + (z-6)²)

f(x,y,z) = d² = (x-1)² + (y-3)² + (z-6)²

Substitute for z

= (x-1)² + (y-3)² + (2-x+y)²

Not sure how to move forward from here

 

[Greg]

Expand $$f(x,y,z)=(x-1)^2+(y-3)^2+(2-x+y)^2$$ then calculate $$\dfrac{\partial f}{\partial x}$$ and $$\dfrac{\partial f}{\partial y}$$, setting the results equal to $$0$$. Solve the resulting system of equations to find $$x$$ and $$y$$ then finish up.

 

[Prove It]

Find the point on the plane x − y + z = 8 that is closest to the point (1, 3, 6).

This what I have so far

Z = 8-x+y

d = Sqrt ((x-1)² + (y-3)² + (z-6)²)

f(x,y,z) = d² = (x-1)² + (y-3)² + (z-6)²

Substitute for z

= (x-1)² + (y-3)² + (2-x+y)²

Not sure how to move forward from here

The closest point on a plane to a point away from the plane is always when the point is perpendicular to the plane. So if we work out the equation of the line that goes through the point (1, 3, 6) which is perpendicular to the plane, then we can use it to find where it intersects the plane. Notice we already have the normal vector of the plane: (1, -1, 1), and so the line is l = (1, -1, 1)t + (1, 3, 6), where t takes on all real values.

So x = t + 1, y = -t + 3, z = t + 6. We can now find where it intersects the plane by substituting into the plane equation.

$\displaystyle \begin{align*} x - y + z &= 8 \\ t + 1 - \left( -t + 3 \right) + t + 6 &= 8 \\ 3t + 4 &= 8 \\ t &= \frac{4}{3} \end{align*}$

and thus $\displaystyle \begin{align*} x = \frac{4}{3} + 1 = \frac{7}{3}, \, y = -\frac{4}{3} + 3 = \frac{5}{3}, \, z = \frac{4}{3} + 6 = \frac{22}{3} \end{align*}$.

The point on the plane x - y + z = 8 that is closest to (1, 3, 6) is $\displaystyle \begin{align*} \left( \frac{7}{3}, \frac{5}{3}, \frac{22}{3} \right) \end{align*}$.