Perpendicular Distance from Point to Plane
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Show that the perpendicular distance D from the point P0(x0, y0, z0) to the plane ax + by + cz = d is:
D = | ax0 + by0 + cz0 − d | / sqrt(a^2 + b^2 + c^2)
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Solution Summary
The perpendicular distance from point to plane is determined.
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First, let us establish that the plane is perpendicular to the vector formed by the coefficients of its equation, n=(a,b,c).
(It may in fact be written somewhere in the books you have available, but in case you cannot find it i give the proof).
Any line in the plane can be defined by two distinct points on it. Let these points be
f = (f_x, f_y, f_z) and g = (g_x, g_y, g_z).
The vector connecting these two points is f-g = (f_x-g_x, f_y-g_y, f_z-g_z).
Now, since both ...
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