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January 8th, 2013, 05:23 PM  #1 
Newbie Joined: Dec 2012 Posts: 16 Thanks: 0  A Question about Cross Product
Given two nonzero vectors a=<a1,a2,a3> and b=<b1,b2,b3>, there's a nonzero vector c=<c1,c2,c3> perpendicular to both a and b so we can get a·c?0 and b·c?0 such that a1*c1+a2*c2+a3*c3=0 (1) b1*c1+b2*c2+a3*c3=0 (2) to eliminate c3 we multiply (1) by b3 and (2) by a3 and subtract: (a1*b3a3*b1)*c1+(a2*b3a3*b2)c2=0 (3) Equation 3 has the form p*c1+q*c2=0, for which and obvious solution is c1=q and c2=p. So a solution of (3) is c1=a2*b3a3*b2 and c2=a3*b1a1*b3 Subsitituting these values into (1) and (2), we then get c3=a1*b2a2*b1 My question is how does c3 come out? Can somebody show me the algebraic process step by step? Thank you. 
January 9th, 2013, 11:50 AM  #2 
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: A Question about Cross Product
That's pretty much basic algebra, isn't it? You have, you say, c1=a2*b3a3*b2 and c2=a3*b1a1*b3. Putting those into a1*c1+a2*c2+a3*c3=0 we have a1(a2*b3 a3*b2)+ a2(a3*b1 a1*b3)+ a3*c3= 0 That is, a1a2b3 a1a3b2+ a2a3b1 a2a1b3+ a3c3= 0. Notice that the first term is a1a2b3 and the fourth is a2a1b3. Those cancel leaving a2a3b1 a1a3b2+ a3c3= 0. a3c3= a1a3b2 a2a3b1= a3(a1b2 a2b1). If a3 is not 0, we can divide both sides by it to get c3= a1b2 a2b1. 
January 11th, 2013, 11:21 AM  #3  
Newbie Joined: Dec 2012 Posts: 16 Thanks: 0  Re: A Question about Cross Product Quote:
 

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