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Let point $A(\vec{a}), B(\vec{b}), C(\vec{c}), D(\vec{b}\times \vec{c})$, then by using Vector Triple Product Expansion, I got the following equality:

$$ \vert \sin\angle AOD \sin\angle BOC \vert=\vert \cos \angle AOC - \cos \angle AOB \vert $$

Pyramid figure

How to get this from Vector Triple Product

My questions are the following:

  • Does this "Pyramid Theorem" have a specific name?
  • And how can I prove?
yunoa7
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1 Answers1

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Here's what to do to prove it (bold font represents vector): We know, $$|\mathbf{a}.(\mathbf{b}×\mathbf{c})|=|\mathbf{(a.c)b}-\mathbf{(a.b)c}|$$ Now all you have to do is, divide both sides by abc (this font represents magnitude). On the left hand side you get the product of sines, and on the right hand side you get two cosine angles. However, you will find that your proposition is erroneous, and there should be unit vectors along $\mathbf b$ and $\mathbf c$ multiplied to the cosine angles, respectively.

Ritam_Dasgupta
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  • "However, you will find that your proposition is erroneous, and there should be unit vectors along $\mathbf b$ and $\mathbf c$ multiplied to the cosine angles, respectively. " What you want to say would be like the curl (or rotation, a concept of vector analysis), I think. Is my understanding right? – yunoa7 May 12 '21 at 14:30
  • Sorry, I'm a high school student, I only have a rather rudimentary understanding of vector algebra. I got the expression through known methods, as stated in the answer, but I am not familiar with "curl" yet. – Ritam_Dasgupta May 12 '21 at 14:35
  • I see. $$|\mathbf{a}.(\mathbf{b}×\mathbf{c})|=|\mathbf{(a.c)b}-\mathbf{(a.b)c}|$$ and the following your words bring the solution. And I shoud take the direction of vectors into account as you say. I just noticed for your help that "curl"(otherwise known as "rotation") also may relate on this question. – yunoa7 May 12 '21 at 21:55
  • Yeah that's it. – Ritam_Dasgupta May 13 '21 at 06:50