# Thread: Vector calculus - planes

1. ## Vector calculus - planes

A plane has equation x- 2y +z = 4. Show that the square, say D^2, of the distance from the point P = (-1,3,2) to a point (x,y,z) on the plane is given by
D^2 = 2x^2 - 2x + 14 + 5y^2 + 2y - 4xy.
By finding the minimum of D^2 determine the shortest distance from P to the plane. (You should verify that the distance is indeed a minimum.) Let Q be the point on the plane clasest to P. Show that the line QP is perpendicular to the plane at Q.

2. Originally Posted by njr008
A plane has equation x- 2y +z = 4. Show that the square, say D^2, of the distance from the point P = (-1,3,2) to a point (x,y,z) on the plane is given by
D^2 = 2x^2 - 2x + 14 + 5y^2 + 2y - 4xy.
By finding the minimum of D^2 determine the shortest distance from P to the plane. (You should verify that the distance is indeed a minimum.) Let Q be the point on the plane clasest to P. Show that the line QP is perpendicular to the plane at Q.
$D^2 = (x - [-1])^2 + (y - 3)^2 + (z - 2)^2$.

Now substitute z = 4 - x + 2y. Expand. Simplify. Spadework for you to do.

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Let $f(x, y) = 2x^2 - 2x + 14 + 5y^2 + 2y - 4xy$ and find extrema in the usual way:

Solve simultaneously:

$\frac{\partial f}{\partial x} = 0$ .... (1)

$\frac{\partial f}{\partial y} = 0$ .... (2)