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April 20th, 2017, 04:12 AM   #1
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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?
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April 20th, 2017, 04:20 AM   #2
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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
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April 20th, 2017, 05:28 AM   #3
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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
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April 21st, 2017, 10:51 AM   #4
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|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.
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April 21st, 2017, 03:54 PM   #5
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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.
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April 21st, 2017, 07:20 PM   #6
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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.
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April 21st, 2017, 07:27 PM   #7
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Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
Originally Posted by zylo View Post
$\displaystyle \vec{v}x\vec{w}$
Quick LaTeX tip: \times works better here. $\displaystyle \vec{v} \times \vec{w}$

-Dan
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April 24th, 2017, 10:25 AM   #8
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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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