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