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Finding the normal to a plane
As promised, we return the the question of finding the equation for a plane from the location of three points, say
Q(x1,y1,z1),R(x2,y2,z2),S(x3,y3,z3)
The fact that the cross-product a×b is perpendicular to both a and b makes it very useful when dealing with normals to planes.
Let
b=⟨x1,y1,z1⟩, r=⟨x2,y2,z2⟩, s=⟨x3,y3,z3⟩.
The vectors
QR−→− = r−b,QS−→ = s−b,
then lie in the plane. The normal to the plane is given by the cross product n=(r−b)×(s−b). Once this normal has been calculated, we can then use the point-normal form to get the equation of the plane passing through Q,R, and S.

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