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I have the angle between two 3D vectors $\beta$ and $\alpha$. Both their magnitude is 1. Can I calculate the vector $\beta$ and $\alpha$ from this angle and magnitude?

If anyone want to know how to calculate the angle between two vectors is simply $\theta = \arccos(\frac{\beta\cdot\alpha}{\left|\beta \right|\left|\alpha \right|})$

$\left|\beta \right|\left|\alpha \right| = 1$ for me, so the formula is simplified to $\theta = \arccos(\beta\cdot\alpha)$

Here is an example where an angle of 132 degrees is given. Example: NOT TO SCALE!!

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  • $\begingroup$ I am inclined to believe this is underdetermined. You can fix one vector and determine the other. It then turns into using the Law of Cosines. $\endgroup$ Dec 31, 2022 at 21:02
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    $\begingroup$ The way you have put it, this is impossible. $\alpha$ can be chosen arbitrarily, and then, even with a given angle there is plenty of freedom to choose $\beta$. I doubt that you were given the task exactly how you have put it: either there are additional constraints that you forgot to mention, or you are solving another problem and you think that solving this "bit" would help (while, instead, you should be pursuing a different approach). $\endgroup$
    – user700480
    Dec 31, 2022 at 21:02
  • $\begingroup$ So this is impossible to solve without using an arbitarily vector? $\endgroup$ Dec 31, 2022 at 21:06
  • $\begingroup$ Correct, and in fact it is impossible without use of two arbitrary vectors, in fact. $\alpha$ is totally arbitrary (with norm $1$) and $\beta$ still have one free variable to choose after choosing $\alpha$. (If $\alpha=\langle a,b,c\rangle$ and $\beta=\langle d,e,f\rangle$, then, given $\alpha$, to get $\beta$ you are still solving the system $ad+be+cf=\cos\theta, d^2+e^2+f^2=1$ which is a system of two equations in three unknowns, which generally has multiple solutions. In the coordinate system, it is an intersection of a plane and a sphere.) $\endgroup$
    – user700480
    Dec 31, 2022 at 21:35

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You can take $\alpha$ to be the vector

$ \alpha = ( \sin t \cos s , \sin t \sin s , \cos t ) $

Then vector $\beta$ makes an anlge of $\theta$ with $\alpha$, therefore, define

the two unit vectors

$u_1 = (\cos t \cos s, \cos t \sin s , - \sin t ) $

$u_2 = (-\sin s , \cos s , 0 ) $

Since $ \alpha \cdot \beta = \cos \theta $ , then

$ \beta = \cos \theta \ \alpha + \cos r \sin \theta \ u_1 + \sin r \sin \theta \ u_2 $

And thus $\alpha $ and $\beta$ are determined by the arbitrary choice of $t, s$ and $r$.

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