Given a (2,3,1) B (4,1,2) C (6,3,7) d (- 5, - 4,8), find the distance from D to plane ABC
First, find out the normal vector n of the plane ABC, take the vector product of the vector ab {2, - 2,1} and the vector AC {4,0,6} as the normal vector n, that is, n = vector AB, cross product vector AC = - 12i-8j + 8K, get from the plane point equation, - 12 (X-2) - 8 (Y-3) + 8 (Z-1) = 0, that is, the plane ABC equation: 3x + 2y-2z-10 = 0
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- 1. Let a (2,3,1) B (4,1,2) C (6,3,7) d (- 5, - 4,8) find the distance from D to plane ABC, Let a (2,3,1) B (4,1,2) C (6,3,7) d (- 5, - 4,8) find the distance from D to plane ABC?
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