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The Wikipedia article for scalar triple product says the following:

Geometrically, the scalar triple product $$\mathbf{a}\cdot(\mathbf{b}\times \mathbf{c})$$ is the (signed) volume of the parallelepiped defined by the three vectors given. Here, the parentheses may be omitted without causing ambiguity, since the dot product cannot be evaluated first. If it were, it would leave the cross product of a scalar and a vector, which is not defined.

$(\mathbf{b}\times \mathbf{c})$ is the vector normal to vectors $\mathbf{b}$ and $\mathbf{c}$. What is the geometric reasoning that leads us to understand that the dot product of $\mathbf{a}$ with this normal vector is equal to the volume of the parallelepiped defined by the three vectors? I would greatly appreciate it if people would please take the time to explain this.

The Pointer
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  • Does this answer your question? Scalar triple product - why equivalent to determinant? – Raskolnikov Jun 19 '20 at 05:38
  • @Raskolnikov But this doesn't give any geometric explanation. – The Pointer Jun 19 '20 at 05:41
  • How about this one : https://math.stackexchange.com/questions/3069505/compute-volume-of-parallelepiped-using-triple-vector/3069581#3069581 – Raskolnikov Jun 19 '20 at 05:56
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    The length of $b\times c$ is the area of the parallelogram spanned by $b$ and $c$. The scalar product with $a$ projects $a$ onto the line along $b\times c$ which is orthogonal to the parallelogram and, hence, gives the height of the parallelepiped. So, the triple product is nothing but $base \times height$ (up to a sign). – trancelocation Jun 19 '20 at 05:59
  • @Raskolnikov Ahh, that's what I was looking for. But we're taking the dot product of the height of the parallelepiped with the normal vector $(\mathbf{b}\times \mathbf{c})$, rather than the magnitude of $(\mathbf{b}\times \mathbf{c})$; so where is the area of the base coming from if we're not taking the magnitude? – The Pointer Jun 19 '20 at 06:01
  • @trancelocation Again, we're not taking the magnitude of $\mathbf{b}\times \mathbf{c}$, so how is the area of the base being incorporated into the calculation? – The Pointer Jun 19 '20 at 06:03
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    @ThePointer : The dot-product does incorporate the magnitude of both vectors involved: $a \cdot b = |a||b|\cos \angle (a,b)$. – trancelocation Jun 19 '20 at 06:10
  • @trancelocation Oh, of course! Yes, it all makes sense now. Thanks for the clarification! – The Pointer Jun 19 '20 at 06:11

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To close out this question, here's a link to another answer (of a related question), pointed out by @Raskolnikov, that just so happens to be a perfect answer to this question.

Lee Mosher
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