My name is Lindsay. I am an OAC student (which is the Ontario equivalent to Grade 12 in most other states and provinces). I am in an Algebra and Geometry course and am currently studying a unit on equations of planes. Our teacher has given us this question that my friend and I have attempted several times, but we are still unable to solve it. My teacher has also suggested using the internet as a resource. The question is:

Prove that a necessary condition that the three planes

-x + ay + bz = 0
ax -  y + cz = 0
bx + cy -  z = 0
have a line in common is that
a^2 + b^2 + c^2 + 2abc = 1

Thank you very much.

Hi Lindsay

Here is are two approaches:

Take two of the planes and find the line of intersection. Now take the points on this line of intersection and substitute them into the third equation. If they fit the third equation - then the three planes share the line.
   Notice that your teacher has made this easier - bacause you already know ONE of the common points - the origin (0,0,0) fits all three planes. Since a line is determined by TWO distinct points - you only need ONE more point.
   When you solve the first two equations, that is find x and y in terms of z, and substitute into the third, you will find that the substitution gives a^2 + b^2 + c^2 + 2abc - 1 for the LHS and 0 for the RHS.
   As an alternative (if you know about determinants) would be:
Three homogeneous equations in 3 unknowns. These have extra common solutions (i.e. the second point in addition to (0,0,0) I mentioned above) if and only if the determinant = 0.
   Take the determinant and see what you get.

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