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I know the vector identity $\vec{A} \times (\vec{B} \times \vec{C}) = (\vec{A} \cdot \vec{C}) \vec{B} - (\vec{A} \cdot \vec{B}) \vec{C}$

Now, is there a succinct way of obtaining $|\vec{A} \times (\vec{B} \times \vec{C})|^2$ using vector algebra? I know we can expand, multiply and group the terms back, but is there a neater way of obtaining the result?

Using Levi-Civita symbols, would this be easier?

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    $\begingroup$ You could dot it with itself - is there something wrong with the answer you get from doing that? $\endgroup$ – preferred_anon Apr 21 at 10:34
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    $\begingroup$ $|\vec{A} \times (\vec{B} \times \vec{C})|\cdot |\vec{A} \times (\vec{B} \times \vec{C})| = (\vec{A} \cdot \vec{C})^2 B^2 - 2 (\vec{A} \cdot \vec{B})(\vec{A} \cdot \vec{C}) \vec{B} \cdot \vec{C} + (\vec{A} \cdot \vec{B}) C^2$ is what I get. But, this is there a neater way of representing the terms? $\endgroup$ – Exp ikx Apr 21 at 10:39
  • $\begingroup$ Just in terms of dot products? Probably not - is there anything you're particularly expecting to see? You could say something like $|A||B||C|\sin \theta \sin \phi$ but those angles are not easy to find. $\endgroup$ – preferred_anon Apr 21 at 10:57
  • $\begingroup$ Yes, just in terms of dot products. The angles involved are not relevant for the question I am trying to solve which includes this squared term of triple product. $\endgroup$ – Exp ikx Apr 21 at 11:22
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    $\begingroup$ Is this for the purpose of memorising it? I think the dot product expression is quite clear, but it depends on what you want to do with the expression. i did look to see if there were any obvious simplifications you could make, but I don't think there are. If you express it with Levi Cevita (or deltas) then you might get something easier to remember, but it will be harder to read. $\endgroup$ – preferred_anon Apr 21 at 12:32
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Use triple product square equation or exterior algebra inner product to get

$$ (\vec{A} \cdot (\vec{B} \times \vec{C}))^2 = \begin{vmatrix} \vec{A}\cdot\vec{A} & \vec{A}\cdot\vec{B} & \vec{A}\cdot\vec{C} \\ \vec{B}\cdot\vec{A} & \vec{B}\cdot\vec{B} & \vec{B}\cdot\vec{C} \\ \vec{C}\cdot\vec{A} & \vec{C}\cdot\vec{B} & \vec{C}\cdot\vec{C} \end{vmatrix}.\tag{1} $$

Now we also have the equation

$$ |\vec{A} \times (\vec{B} \times \vec{C})|^2 = |\vec{A}|^2\ | \vec{B} \times \vec{C} |^2 - (\vec{A} \cdot (\vec{B} \times \vec{C}))^2 \tag{2} $$

and a little bit of inspection leads to the result

$$ |\vec{A} \times (\vec{B} \times \vec{C})|^2 = - \begin{vmatrix} 0 & \vec{A}\cdot\vec{B} & \vec{A}\cdot\vec{C} \\ \vec{B}\cdot\vec{A} & \vec{B}\cdot\vec{B} & \vec{B}\cdot\vec{C} \\ \vec{C}\cdot\vec{A} & \vec{C}\cdot\vec{B} & \vec{C}\cdot\vec{C} \end{vmatrix}.\tag{3} $$

The advantage of exterior algebra is that it works in $\ \mathbb{R}^n\ $ while the usual cross product is restricted to $\ \mathbb{R}^3.$

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  • $\begingroup$ Thanks, I did not know about exterior algebra. $\endgroup$ – Exp ikx Apr 21 at 16:55

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