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malsch 
Posted: December 10, 2011 12:03 pm

Newbie Group: Members+ Posts: 1 Member No.: 36,216 Joined: December 10, 2011 
Hi,
I have the following situation: n = px + qy + rz = qi  5qj  11pk where w, x, y and z are vectors and: x = 4i + 2j  3k y = 5i  3j + 8k z = 2i  j + 4k i need to find all the possible values of p, q and r. I substituted x, y and z in the first equation and compared coefficients of i, j and k and ended up with the following equations: 4p + 4q  2r = 0 2p + 2q  r = 0 8p + 8q +4r = 0 This implies that one solution is obviously 0 for all the 3 unknowns. I really do not know how to continue from here. i used gaussian elimination and ended up with the following matrix: (4 4 2  0) (0 0 0  0) (0 0 8  0) Any help would be greatly appreciated. 10q This post has been edited by malsch on December 11, 2011 08:50 am 
tashirosgt 
Posted: December 10, 2011 04:35 pm

Member Group: Trusted Members Posts: 246 Member No.: 21,543 Joined: December 17, 2008 
You could continue the elimination process to get
(4 4 0  0) (0 0 0  0) (0 0 1  0) and (1 1 0  0) (0 0 0  0) (0 0 1  0) These equations imply that r = 0 and p = q. So there is an infinite family of solutions. The general form of a solution is (p,q,r) = (a,a,0) where a is any number. 
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