SOLUTION: if x�b+c-a=y�c+a-b=z�a+b-c, then show that x(b-c)+y(c-a)+(a-b)=0

Question 1026678: if x�b+c-a=y�c+a-b=z�a+b-c, then show that x(b-c)+y(c-a)+(a-b)=0
Answer by robertb(5830) (Show Source): You can put this solution on YOUR website!
is the symmetric equation of a straight line in with direction vector {b+c-a, c+a-b, a+b-c}.
One point on the line is (b+c-a, c+a-b, a+b-c). Another point is the origin (0,0,0). (Why?) Incidentally, the point (b+c-a, c+a-b, a+b-c) is also contained in the plane
x(b-c)+y(c-a)+z(a-b) = 0,
since (b+c-a)(b-c) + (c+a - b)(c-a) + (a+b-c)(a-b)=0. (VERIFY!!)
Hence the line and the plane x(b-c)+y(c-a)+z(a-b) = 0 share two common points, namely (0,0,0) and (b+c-a,c+a-b,a+b-c), and thus the line is contained completely in the plane x(b-c)+y(c-a)+z(a-b) = 0.
Therefore if (x,y,z) satisfy , then (x,y,z) should also satisfy x(b-c)+y(c-a)+z(a-b) = 0, and the statement is proved.

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