My Math Forum  

Go Back   My Math Forum > College Math Forum > Linear Algebra

Linear Algebra Linear Algebra Math Forum


Reply
 
LinkBack Thread Tools Display Modes
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=<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 ac?0 and bc?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.
Channeltsui is offline  
 
January 9th, 2013, 10:50 AM   #2
Math Team
 
Joined: Sep 2007

Posts: 2,409
Thanks: 5

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.
HallsofIvy is offline  
January 11th, 2013, 10:21 AM   #3
Newbie
 
Joined: Dec 2012

Posts: 16
Thanks: 0

Re: A Question about Cross Product

Quote:
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!
Channeltsui is offline  
Reply

  My Math Forum > College Math Forum > Linear Algebra

Tags
cross, product, question



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
A question about cross product Channeltsui Linear Algebra 1 January 26th, 2013 06:10 PM
Vectors Cross Product aaron-math Calculus 1 February 29th, 2012 02:57 PM
Cross product in cylindrical coordinates Agent007 Algebra 0 February 26th, 2012 10:52 PM
Cross Product aaron-math Calculus 1 February 6th, 2012 02:44 PM
cross product [a]x dimevit Linear Algebra 2 October 18th, 2007 03:45 AM





Copyright © 2017 My Math Forum. All rights reserved.