| Author |
Comment/Response |
Bill
|
 12/16/12 10:52pm
Hi:
This isn't exactly what you asked for, but perhaps it may be helpful...
Using NMinimize, solve this problem: Find the point P on the plane -2 x + 1 y + 3 z = -3 closest to the point Q(5,0,-7) and the distance between these two points.
NMinimize[{(x - 5)^2 + y^2 + (z + 7)^2, -2 x + 1 y + 3 z == -3}, {x, y, z}]
Out: {56., {x -> 1., y -> 2., z -> -1.}}
distance = Sqrt[56] // N
Out: 7.48331
Hth,
Bill
URL: , |
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