Multivariable Calculus: Find the coordinates of the point (x, y, z) on the plane z = 2 x + 3 y + 1?

It suffices to mimimize the square of the distance between (x, y, z) = (x, y, 2x+3y+1) and the origin.
==> D = x^2 + y^2 + (2x + 3y + 1)^2.

We find the critical points via first partial derivatives:
D_x = 2x + 4(2x + 3y + 1) = 10x + 12y + 4
D_y = 2y + 6(2x + 3y + 1) = 12x + 20y + 6

Setting these equal to 0 yields (x, y) = (-1/7, -3/14).
(With the second derivative test, this can be checked to give a relative minimum.
Being a sum of squares, this is indeed a global minimum.)

Since z = 2x+3y+1, the desired point is (x, y, z) = (-1/7, 3/14, 19/14).

I hope this helps!

Any element P = (x,y,z) lies on the airplane if it satisfes here A x + B y + C z + D = 0 so 2x + 4y - z + a million = 0 use formulation (A xa + B ya + C za + D) / sqrt(A2 + B2 + C2) subsequently 2(2) + 4(4) - a million(3) + a million/ sqrt (2^2 + 4^2 + -a million^2) simplifies to 4.129483209 good luck!

1

Hi,

I describe a vector approach to this problem:

1. Let the foot of perpendicular from origin O to the plane be F.

2. OF is perpendicular to the plane, so its direction vector is (2, 3, -1).

3. OF has the form (2, 3, -1)t for some real t to be found.

4. F lies on the plane, so it satisfies equation of the plane:

(2t, 3t, -t) . (2, 3, -1) = -1,

from which t can be found.

Thanks.

Cheers,
Wen Shih