Question: Let P be a plane with equation ax + by + cz = d and normal vector n = (a, b, c). For any point

Let P be a plane with equation ax + by + cz = d and normal vector n = (a, b, c). For any point Q, there is a unique point P on P that is closest to Q, and is such that PQ is orthogonal to P(Figure 12).

Z 0 Q n P P X

Show that the point P on P closest to Q is determined by the equation

OP = 00 + d - 00.n n.n

Z 0 Q n P P X

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Since PO is orthogonal to the plane P it is parallel to the vector n a b c which is norm... View full answer

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