My Math Forum A Question about Cross Product

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 January 8th, 2013, 04:23 PM #1 Newbie   Joined: Dec 2012 Posts: 16 Thanks: 0 A Question about Cross Product Given two nonzero vectors a= and b=, there's a nonzero vector c= 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*b3-a3*b1)*c1+(a2*b3-a3*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*b3-a3*b2 and c2=a3*b1-a1*b3 Subsitituting these values into (1) and (2), we then get c3=a1*b2-a2*b1 My question is how does c3 come out? Can somebody show me the algebraic process step by step? Thank you.
 January 9th, 2013, 10:50 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*b3-a3*b2 and c2=a3*b1-a1*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, 10:21 AM   #3
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Re: A Question about Cross Product

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 Originally Posted by HallsofIvy That's pretty much basic algebra, isn't it? You have, you say, c1=a2*b3-a3*b2 and c2=a3*b1-a1*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.
Ah, yes. Thank you!

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