does anyone have a good idea of how to start question 42 on HW1? Jroark

Heres what I did to start: You know the two planes are going to have the equations ax + by + cz + d = 0 and a'x + b'y + c'z +d' = 0 You also know that the points (1,1,1) and (3,5,0) are going to be on both planes. Plugging these points into either plane yields the equations a + b + c + d = 0 3a + 5b +0c + d = 0 a' + b' + c' + d' = 0 3a' + 5b' +0c' + d' = 0 Solve for a, b, c, and d (it will have 2 degrees of freedom obviously so you will only be able to solve for these in terms of two variables) then for a', b', c', and d' (same here, though to save you some work, notice that the coefficients in front of a', b', c' and d' are the same as for a, b, c, and d so the solutions will be the same). I solved for a, b, c, and d in terms of p and q and a', b', c', and d' in terms of r and s. Pick random values for p, q, r, and s such that you do not get non-trivial solutions (e.g. p=q=0) and such that p does not equal r and q does not equal s and you should be good. Hope that helps Rmcclur 22:30, 19 January 2010 (UTC)

thanks! that really helped a lot! Jroark

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