My Math Forum Correlation between cross and dot product

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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: $\parallel\vec{v}\times\vec{w}\parallel=\begin{vmatrix} \vec{e_{1}} & \vec{e_{2}} & \vec{e_{3}} \\\vec{v_{1}}& \vec{v_{2}} &\vec{v_{3}} \\\vec{w_{1}} &\vec{w_{2}} &\vec{w_{3}} \end{vmatrix}=\vec{e_{1}}\begin{vmatrix}\vec{v_{2}}&\vec{v_{3}} \\\vec{w_{2}}&\vec{w_{3}}\end{vmatrix}-\vec{e_{2}}\begin{vmatrix}\vec{v_{1}}&\vec{v_{2}} \\\vec{w_{1}}&\vec{w_{3}}\end{vmatrix}+\vec{e_{3}}\begin{vmatrix}\vec{v_{1}}&\vec{v_{2}} \\\vec{w_{1}}&\vec{w_{2}}\end{vmatrix}$ 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}$$\displaystyle \theta=v\displaystyle ^{2}w\displaystyle ^{2}(1-cos\displaystyle ^{2}$$\displaystyle \theta$)= (v.v)(w.w)-(v.w)(w.v) which is the determinant. Nothing to it if you remember you saw it. Thanks from topsquark and studiot
 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, non-short-cut, non-tricky 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. Thanks from topsquark
April 21st, 2017, 07:27 PM   #7
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Quote:
 Originally Posted by zylo $\displaystyle \vec{v}x\vec{w}$
Quick LaTeX tip: \times works better here. $\displaystyle \vec{v} \times \vec{w}$

-Dan

 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 x-axis in direction $\displaystyle \vec{v}$ and then rotate coordinate system about x-axis till $\displaystyle \vec{w}$ is in the x-y 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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