In geometry 3D, let A(1,2,−3)A(1,2,−3)A(1,2,-3) and B(−1,4,1)B(−1,4,1)B(-1,4,1) are two points. In the plane (P):2x−3y+3z−17=0(P):2x-3y+3z-17=0, find a point MM such that the scalar product of two vectors AMAM and BMBM will have the least value. How do I tell Mathematica to do that?

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2 Answers

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ArgMin[{(a – #).(b – #) &[{x, y, z}], 2 x – 3 y + 3 z – 17 == 0}, {x, y, z}]

{29/11, -21/22, 65/22}

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How long Mathematica out put answer? My computer run about 15minutes.

– minthao_2011

Sep 28 ’12 at 15:27

I was assuming you’d made the assignments a = {1, 2, -3}; b = {-1, 4, 1}; first…

– J. M.♦

Sep 28 ’12 at 15:32

Thank you very much.

– minthao_2011

Sep 28 ’12 at 15:37

If I want to find the point MM such that the sum of its distances from the points AA and BB will have the list value, how do I tell Mathematica to do that?

– minthao_2011

Sep 28 ’12 at 16:20

That’s a different question, but here’s a hint: look up EuclideanDistance[].

– J. M.♦

Sep 28 ’12 at 16:23

Let II be midpoint of the segment ABAB. We can prove that, the needing point is projection of the point II on the plane (P)(P). The equation of the line Δ\Delta which passing the point II and perpendicular to the plane (P)(P) is x=2t,y=3−3t,z=−1+3t. x = 2t, \quad y = 3 – 3t, \quad z = -1 + 3t. And then, the coordinates of the point MM is solution of the system of the equations: Equation (P(P and equation Δ\Delta.

Solve[{2 x – 3 y + 3 z – 17 == 0, x == 2 t, y == 3 – 3 t, z == -1 + 3 t}, {x, y, z, t}]