During school today (Yr 12 Maths C), we covered finding the Dot Product (Scalar Product) of 2D Vectors of the form (Magnitude, Theta)
using the equation:
$$A.B=|A||B|\times cos \theta$$
Where $\theta$ is the angle between the two vectors. We then moved onto 3D Vectors of the form (Magnitude, Azimuth, Theta)
. The process described in our textbook said to converter from (Magnitude, Azimuth, Theta)
to (i, j, k)
components. However, I thought I would try just multiplying the above equation by $cos \phi$, just to see if it would work:
$$A.B=|A||B|\times cos \theta\times cos \phi$$
Surprisingly, this worked for the vectors $(15, 45 \circ, -33\circ)$ and $(8, 45 \circ, 57\circ)$, as well as $(9, 280 \circ, 16\circ)$ and $(4, 280 \circ, -28\circ)$.
I am a bit stumped as to why this method works. For me it did seem like the next logical step, however now I am having second thoughts. I was unable to test this on any vectors that did not share at least 1 identical value, which might be why.
Can you explain why, how this works, and if it does so for all cases.
Best Answer
If I understand your azimuthal and polar angles correctly as below
then for any given vector $\mathbf a = (r, \varphi, \theta) = (a_1, a_2, a_3)$ it follows $$ a_1 = r \cos \varphi \sin \theta \\ a_2 = r \sin \varphi \sin \theta \\ a_3 = r \cos \theta $$ Same for $\mathbf b = (\tilde r, \tilde \varphi, \tilde \theta) = (b_1, b_2, b_3)$ $$ b_1 = \tilde r \cos \tilde \varphi \sin \tilde \theta \\ b_2 = \tilde r \sin \tilde \varphi \sin \tilde \theta \\ b_3 = \tilde r \cos \tilde \theta $$ so $$ \mathbf a \cdot \mathbf b = r \tilde r \left [ \cos \varphi \sin \theta \cos \tilde \varphi \sin \tilde \theta + \sin \varphi \sin \theta \sin \tilde \varphi \sin \tilde \theta + \cos \theta \cos \tilde \theta \right ] = \\ r \tilde r \left [ \left ( \cos \varphi \cos \tilde \varphi + \sin \varphi \sin \tilde \varphi \right )\sin \theta \sin \tilde \theta + \cos \theta \cos \tilde \theta \right ] = \\ r \tilde r \left [\cos \left ( \varphi - \tilde \varphi\right )\sin \theta \sin \tilde \theta + \cos \theta \cos \tilde \theta \right ] $$ I don't think it can be simplified further. What you can get from here, it's that if $\varphi = \tilde \varphi$ as in your test pairs, expressions above becomes $$ \mathbf a \cdot \mathbf b = r \tilde r \left [ \sin \theta \sin \tilde \theta + \cos \theta \cos \tilde \theta\right ] = r \tilde r \cos \left ( \theta-\tilde \theta\right ) $$ so it's indeed $|\mathbf a||\mathbf b| \cos \left ( \varphi-\tilde\varphi\right)\cos \left ( \theta - \tilde \theta\right )$, since first $\cos$ is simply one. For other cases (when $\cos \left ( \varphi-\tilde\varphi\right) \ne 1$) your claim is not true.