Bobby Mitchell

2022-08-07

Show that the plane that passes through the three points
A=(a1,a2,a3)
B=(b1,b2,b3)
C=(c1,c2,c3)
consists of the points P=(x,y,z) given by
$|\begin{array}{ccc}{a}_{1}-x& {a}_{2}-y& {a}_{3}-z\\ {b}_{1}-x& {b}_{2}-y& {b}_{3}-z\\ {c}_{1}-x& {c}_{2}-y& {c}_{3}-z\end{array}|$

Irene Simon

Example:
(a1-x)[(b2-y)(c3-z)-(c2-y)(b3-z)]=(a1-x)[b2c3-(b2z+c3y)+yz-(c2b3-c2z-b3y+yz)]=
(a1-x)[b2c3-c2b3-(b2-c2)z-(c3-b3)y]. The higher order productsin xyz, xy, yz, xz cancel out and only linear terms remain. The determinate is in the normal form of the plane equation, which is the shape of a point. E+Fx+Gy+Hz=0.
D is the distance between A and the normal form of the cross product of B and C, where C is the plane's normal. A (dot) (BxC).
F=a2b1-a1b2-a2c1+b2c1+a1c2-b1c2.G=-a3b1+a1b3+a3c1-b3c1-a1c3+c1c3.H=a3b2-a2b3-a3c2+b3c2+a2c3-b2c3.
(F,G,H) is the normal of the plane containing the three pointsA,B,C.
The intermediate interpretation of the determinate is the parallelepiped product: (A-x) (dot) (B-x)X(C-x). So A is in theplane and BXC is the normal made two vectors in the plane. With xand A,B,C all vectors in the plane ful fill the condition. With x inthe plane the difference vectors are also in the plane, and thecross product forms a normal to the plane.

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