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I'm trying to prove the vector triple product expansion by magnitude and direction: $$ \vec{a} \times(\vec{b}\times \vec{c})=(\vec a\cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} $$ The equality of direction is easy: using the property of the cross product that $\vec{b} \times \vec{c} $ is orthogonal to both $ \vec{b} $ and $\vec{c} $. But I can't give a correct proof for the latter. Where is my error? My trying is following:

Let $ \theta $ be the angle between $ \vec{a} $ and the plane containing $ \vec{b} $ and $ \vec{c} $, $ \varphi $ be the angle between $ \vec{b} $ and $ \vec{c} $ in the plane containing them.

The left hand:

$$ \begin{align} \Vert \vec{a} \times(\vec{b}\times \vec{c}) \Vert &= \Vert \vec{a} \Vert \Vert \vec{b}\times \vec{c} \Vert \vert\sin \theta \vert \\&= \Vert \vec{a} \Vert \Vert \vec{b} \Vert \Vert \vec{c} \Vert \vert\sin \theta \vert \vert\sin\varphi \vert \end{align} $$

The right hand:

$$ \begin{align} \Vert(\vec a\cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}\Vert ^2 & = \Vert(\vec a\cdot \vec{c})\vec{b}\Vert^2 -2 (\vec a\cdot \vec{c})(\vec{a} \cdot \vec{b})(\vec{b}\cdot\vec{c})+ \Vert(\vec{a} \cdot \vec{b})\vec{c}\Vert ^2 \\&= 2\Vert \vec{a} \Vert ^2 \Vert \vec{b} \Vert ^2 \Vert \vec{c} \Vert ^2 \cos^2 (\frac{\pi}{2} - \theta) - 2\Vert \vec{a} \Vert ^2 \Vert \vec{b} \Vert ^2 \Vert \vec{c} \Vert ^2 \cos^2 (\frac{\pi}{2} - \theta)\cos\varphi \\ &= 2\Vert \vec{a} \Vert ^2 \Vert \vec{b} \Vert ^2 \Vert \vec{c} \Vert ^2 \vert\sin^2 \theta \vert (1-\cos\varphi ) \end{align} $$ figure

yunoa7
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  • No. The recommended question uses different approach. – yunoa7 May 07 '21 at 12:26
  • One answer uses "Comparing the magnitude of the vectors", which you want, and the answer by user10354138 uses $\sin(\theta)$ and $\cos(\theta)$ like you requested. – Dietrich Burde May 07 '21 at 12:29
  • I'd recognized "Comparing the magnitude of the vectors" in the recommended question, but what I want to know is why two magnitues in my question are unequal. I can't get the reason from the recommended. – yunoa7 May 07 '21 at 12:59
  • The right hand You seem to assume the angles between $\vec a, \vec c$ and $\vec a, \vec b$ are both $\pi/2 - \theta$. They are not. – dxiv May 08 '21 at 02:06
  • @dxiv I see. I just wrote graph and understood for your help. For example, moving only $ \vec{c} $ changes only the angle between $ \vec{a} $ and $ \vec{c} $. – yunoa7 May 08 '21 at 03:50
  • By the way how can I close this question? (I'll delete this comment after I close this question. ) – yunoa7 May 08 '21 at 03:51
  • @yunoa7 You can delete your own question as long as no one answered it, or you can post a self-answer. – dxiv May 08 '21 at 03:57
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    @dxiv Okay, I'll post a self-answer after I think a little more. – yunoa7 May 08 '21 at 04:10

1 Answers1

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[self answer]

Let point $A(\vec{a}), B(\vec{b}), C(\vec{c}), D(\vec{b}\times \vec{c})$,

then $\angle $ AOD is $\theta $ and $\angle $ BOC is $\varphi$. Using law of cosines,

  • $ \cos \angle AOB = \frac{\Vert\vec{a}\Vert^2 + \Vert\vec{b}\Vert^2 - \Vert\vec{a} - \vec{b} \Vert^2 }{2 \Vert\vec{a}\Vert \Vert\vec{b}\Vert} = \frac{ (\vec{a} \cdot \vec{b}) }{ \Vert\vec{a}\Vert \Vert\vec{b}\Vert} $
  • $\cos \angle AOC = \frac{ (\vec{a} \cdot \vec{c}) }{ \Vert\vec{a}\Vert \Vert\vec{c}\Vert} $

The left hand:

$$ \Vert \vec{a} \Vert ^2 \Vert \vec{b} \Vert ^2 \Vert \vec{c} \Vert ^2\vert \sin\theta \sin \varphi \vert =\Vert \vec{a} \Vert ^2 \Vert \vec{b} \Vert ^2 \Vert \vec{c} \Vert ^2 \vert \sin\angle AOD \vert^2 \vert\sin\angle BOC \vert^2 $$

The right hand:

$$ \begin{align}\Vert(\vec a\cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}\Vert ^2 &= \Vert(\vec a\cdot \vec{c})\vec{b}\Vert^2 -2 (\vec a\cdot \vec{c})(\vec{a} \cdot \vec{b})(\vec{b}\cdot\vec{c})+ \Vert(\vec{a} \cdot \vec{b})\vec{c}\Vert ^2 \\ &= \Vert \vec{a} \Vert ^2 \Vert \vec{b} \Vert ^2 \Vert \vec{c} \Vert ^2 \cos^2 \angle AOC - 2\Vert \vec{a} \Vert ^2 \Vert \vec{b} \Vert ^2 \Vert \vec{c} \Vert ^2 \cos \angle AOB \cos \angle AOC \\& + \Vert \vec{a} \Vert ^2 \Vert \vec{b} \Vert ^2 \Vert \vec{c} \Vert ^2 \cos^2 \angle AOB \\ &=\Vert \vec{a} \Vert ^2 \Vert \vec{b} \Vert ^2 \Vert \vec{c} \Vert ^2 \frac{ \vert(\vec{a} \cdot \vec{c})\vert^2}{ \Vert\vec{a}\Vert^2 \Vert\vec{c}\Vert ^2} - 2\Vert \vec{a} \Vert ^2 \Vert \vec{b} \Vert ^2 \Vert \vec{c} \Vert ^2 \vert \frac{ (\vec{a} \cdot \vec{b}) }{ \Vert\vec{a}\Vert \Vert\vec{b}\Vert} \frac{ (\vec{a} \cdot \vec{c}) }{ \Vert\vec{a}\Vert \Vert\vec{c}\Vert} + \Vert \vec{a} \Vert ^2 \Vert \vec{b} \Vert ^2 \Vert \vec{c} \Vert ^2 \vert \frac{ \vert(\vec{a} \cdot \vec{b}) \vert^2}{ \Vert\vec{a}\Vert ^2\Vert\vec{b}\Vert^2} \\ &=\Vert \vec{a} \Vert ^2 \Vert \vec{b} \Vert ^2 \Vert \vec{c} \Vert ^2 ( \cos^2 \angle AOC - 2\cos \angle AOB \cos \angle AOC + \cos^2 \angle AOB ) \\ &=\Vert \vec{a} \Vert ^2 \Vert \vec{b} \Vert ^2 \Vert \vec{c} \Vert ^2 ( \cos \angle AOC - \cos \angle AOB )^2 \end{align} $$

So, if the following is correct, the proof is copmleted: $$ \vert \sin\angle AOD \sin\angle BOC \vert=\vert \cos \angle AOC - \cos \angle AOB \vert $$ I'll ask another question about this. (I'll add its link in this article later. )

yunoa7
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