Scalar Triple Product

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Scalar Triple Product

Postby Guest » Thu Jan 10, 2019 8:24 am

I am not currently in school, but need to relearn some vector algebra for an electrodynamics class. I have a textbook that I am using to teach myself the basics.

Here's the problem, in one section on the scalar triple product (A [tex]\cdot[/tex] (B X C)), it states that geometrically the magnitude of this (|A [tex]\cdot[/tex] (B X C)|)represents the volume of the parallelpiped generated by the 3 vectors A, B and C since |B X C| is the area of the base and |A cos ϴ|
is the height.

I get what they are saying, but mathematically it does not work out that way. Assume \vec{A} = (1x + 0y + 1z) and \vec{B} = (0x + 1y + 1z). The length (magnitude) of both A and B is \sqrt{2}. So the area of the base formed by these vectors would be 2; however, A x B = \sqrt{3}.

Can someone explain where I am going wrong? Or is the book wrong?
Guest
 

Re: Scalar Triple Product

Postby Baltuilhe » Thu Jan 10, 2019 2:25 pm

Good afternoon!

Well, the area should be 2 if both vectors were orthogonal between themselves, right?
The angle between these two vectors:
[tex]\cos\;\theta=\dfrac{\vec{A}\cdot\vec{B}}{\|\vec{A}\|\;\|\vec{B}\|}\\\\\cos\;\theta=\dfrac{(1,0,1)\cdot(0,1,1)}{\sqrt{1^2+0^2+1^2}\;\sqrt{0^2+1^2+1^2}}\\\\\cos\;\theta=\dfrac{1.0+0.1+1.1}{\sqrt{2}\;\sqrt{2}}=\dfrac{1}{2}\\\\\theta=60^{\circ}[/tex]

So, as these vectors have an angle of [tex]60^{\circ}[/tex] between them, lets calc the area as a paralelogram:
[tex]A=a\cdot b\cdot \sin\;\theta=\sqrt{2}\cdot\sqrt{2}\cdot\sin\;60^{\circ}=2\cdot\dfrac{\sqrt{3}}{2}=\sqrt{3}[/tex]

Right?

I hope I have helped! :)

Baltuilhe
 
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