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April 20th, 2017, 04:12 AM  #1 
Member Joined: Apr 2014 From: Greece Posts: 58 Thanks: 0  Correlation between cross and dot product
I have to prove that $\displaystyle \left  \vec{v} \times \vec{w}\right \^{2}=\begin{vmatrix} \left \langle \vec{v},\vec{v} \right \rangle & \langle\vec{v},\vec{w}\rangle\\ \langle \vec{w},\vec{v}\rangle& \langle\vec{w},\vec{w}\rangle \end{vmatrix}$ the only corellation of cross and dot product I've found is the Lagrange identity $\displaystyle \parallel\vec{v} \times \vec{w}\parallel^{2}=\parallel\vec{v}\parallel^{2} \cdot\parallel\vec{w}\parallel^{2}\langle \vec{v}\cdot\vec{w}\rangle^{2} $ I don't see how I could end with what I'm asked to prove with this. Does anybody have any ideas? 
April 20th, 2017, 04:20 AM  #2 
Senior Member Joined: Jun 2015 From: England Posts: 829 Thanks: 244 
The expression on the right of you first equation is a determinant. Wolfram is probably easier to follow than Wiki for this Look where they discuss 2 x 2 determinants a couple of paragraphs down. Does the formula look familiar or compare with anything you see in the question? Determinant  from Wolfram MathWorld 
April 20th, 2017, 05:28 AM  #3 
Member Joined: Apr 2014 From: Greece Posts: 58 Thanks: 0 
I tried to analyze the cross product in the form of a determinant but the problem was that it was a 3x3 determinant while on the right side of the equality there is a 2x2 determinant. I tried to solve the determinant further but with no luck. Here's what I got: I don't know how to proceed after that and I'm pretty sure it's futile to even try, as I think I'm on a completly wrong track 
April 21st, 2017, 10:51 AM  #4 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 
vXw$\displaystyle ^{2}$=v$\displaystyle ^{2}$w$\displaystyle ^{2}$sin$\displaystyle ^{2}$$\displaystyle \theta$=v$\displaystyle ^{2}$w$\displaystyle ^{2}$(1cos$\displaystyle ^{2}$$\displaystyle \theta$)= (v.v)(w.w)(v.w)(w.v) which is the determinant. Nothing to it if you remember you saw it. 
April 21st, 2017, 03:54 PM  #5 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 
OK. If you want to do it your way, which is the logical, nonshortcut, nontricky way: $\displaystyle \vec{v}x\vec{w}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ v_{1} &v_{2} & v_{3}\\ w_{1}&w_{2} & w_{3} \end{vmatrix}=(v_{2}w_{3}v_{3}w_{2})\vec{i}(v_{1}w_{3}v_{3}w_{1})\vec{j}+(v_{1}w_{2}v_{2}w_{1})\vec{k}\\ $ $\displaystyle =A\vec{i}+B\vec{j}+C\vec{k}\\ \vec{v}x\vec{w}^{2}= A^{2}+B^{2}+C^{2}=v^{2}w^{2}(\vec{v}\cdot \vec{w})^{2}$ You just have to wade through a lot of algebra. 
April 21st, 2017, 07:20 PM  #6 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 
There is a way around all the algebra of the previous thread by noting the result is independent of the coordinate system so that: $\displaystyle \vec{v}=(v_{1},0,0)\\ \vec{w}=(w_{1},w_{2},w_{3})\\ \vec{v}x\vec{w}=v_{1}w_{3}\vec{j}+v_{1}w_{2}\vec{k}\\ \vec{v}x\vec{w}^{2}=(v_{1}w_{3})^{2}+(v_{1}w_{2} )^{2}\\ =v^{2}_{1}(w^{2}_{1}+w^{2}_{2}+w^{2}_{3})v^{2}_{1}w^{2}_{1}\\ =v^{2}w^{2}(\vec{v}\cdot\vec{w})^{^{2}} $ ok, a little tricky, but educational. 
April 21st, 2017, 07:27 PM  #7 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,797 Thanks: 715 Math Focus: Wibbly wobbly timeywimey stuff.  
April 24th, 2017, 10:25 AM  #8 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 
You can also take the xaxis in direction $\displaystyle \vec{v}$ and then rotate coordinate system about xaxis till $\displaystyle \vec{w}$ is in the xy plane. Then $\displaystyle \vec{v}=(v_{1},0,0)=v_{1}\vec{i}\\ \vec{w}= (w_{1},w_{2},0)=w_{1}\vec{i}+w_{2}\vec{j}\\ \vec{v}\times\vec{w}=v_{1}w_{2}\vec{k}\\ \vec{v}\times\vec{w}^{2}=v_{1}^{2}w_{2}^{2}=v_{1 }^{2}(w_{1}^{2}+w_{2}^{2})v_{1}^{2}w_{1}^{2}\\ =v^{2}w^{2}(\vec{v}\cdot \vec{w})^{2} $ 

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