Find the minimum distance from the point (2,-2,3) to the plane 6x + 4y - 3z = 2.

Using 11.8

f = distance^2 = (x-2)^2 + (y+2)^2 + (z-3)^2

grad f = < 2x-4, 2y+4, 2z-6 >

g = 6x + 4y - 3z = 2

grad g = <6, 4, -3>

So the 4 equations to solve are:

2x-4 = 6l

2y+4 = 4l

2z-6 = -3l

6x+4y-3z = 2

Solve for x, y, z in terms of l to get

x = 3l + 2, y = 2l - 2, z = -3/2l + 3

and plug this into the last equation to get that l = 14/61

and thus

x = 164/61, y = -94/61, z = 162/61

and the distance is the square root of f at x,y,z

d = 7/sqrt(61).

Using 11.7

Solve for z: z = 2x + 4/3y - 2/3

f = distance^2 = (x-2)^2 + (y+2)^2 + (2x + 4/3y -11/3)^2

grad f = 2< x - 2 + 4x + 8/3y -22/3, y+2 + 8/3x + 16/9y - 44/9>

So we need to solve:

5x + 8/3y = 28/3

8/3x + 25/9y = 26/9

Which has solution x = 164/61 and y = -94/61
and gives the same distance as above.

Using Chapter 9: the formula for the distance from the point (x,y,z) to
the plane ax + by + cz = d, is given by

dist = |ax + by + cz - d|/sqrt(a^2+b^2+c^2) = 7/sqrt(61).

ccollins@math.utk.edu