Distance between plane and elliptic paraboloid

Question

Find a point on the plane $z = x + y - 2$ and a point on the surface $z = x^2 + y^2$ such that the distance between them is minimized.

I know what is happening. But I just don't know where to start. I am self-teaching vector calculus. Any help would be appreciated! Thanks in advance.

Answer 1

Here's a hint: Find the point on the surface where the normal vector is parallel to the normal vector of the plane.

Answer 2

The gradient of the paraboloid is

$ g = (2 x, 2 y , -1) $

And the normal to the plane is

$ n = (1, 1, -1) $

And we want to have

$ g = k n $

for $k \ne 0 $.

Thus

$ 2 x = k$

$ 2 y = k $

$ -1 = - k $

Hence, $ k = 1 $ and $ x = y = \dfrac{1}{2} $ , i.e. the point of the paraboloid closest to the plane is $ R = ( \dfrac{1}{2}, \dfrac{1}{2}, \dfrac{1}{2} ) $

The projection of $R$ onto the plane is

$ Q = R_0 + P (R - R_0) $

where $R_0$ is a point on the plane, for example, $ R_0 = (0,0,-2) $

and

$ P = I - \dfrac{ n n^T }{ n^T n } = \begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && 1 \end{bmatrix} - \dfrac{1}{3} \begin{bmatrix} 1 && 1 && - 1 \\ 1 && 1 && -1 \\ -1 && -1 && 1 \end{bmatrix} = \dfrac{1}{3} \begin{bmatrix} 2 && -1 && 1 \\ -1 && 2 && 1 \\ 1 && 1 && 2 \end{bmatrix} $

Therefore,

$ Q = \begin{bmatrix} 0 \\ 0 \\ -2 \end{bmatrix} + \dfrac{1}{3} \begin{bmatrix} 2 && -1 && 1 \\ -1 && 2 && 1 \\ 1 && 1 && 2 \end{bmatrix} \begin{bmatrix} \frac{1}{2} \\ \frac{1}{2} \\ \frac{5}{2} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ -2 \end{bmatrix} + \dfrac{1}{6} \begin{bmatrix} 6 \\ 6 \\ 12 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$

Therefore, $Q= (1, 1, 0) $ is the point on the plane that is closest to the paraboloid.

Answer 3

I suggest you to use the extension of Lagrange method to find minimum of function $$f(x_1,y_1,z_1,x_2,y_2,z_2) = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$$ express the distance of two point $P_1(x_1,y_1,z_1)$ and $P_2(x_2,y_2,z_2)$ subject to 2 constrain are

$$g_1(x_1,y_1,z_1,x_2,y_2,z_2) = x_1 + y_1 -2 = 0$$ and $$g_2(x_1,y_1,z_1,x_2,y_2,z_2) = x_2^2 + y_2^2 - z_2 = 0$$ Lagrange extension