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I do not know what the sine of the angle between two vectors is. I think it may be the vector created by connecting the tips of the two vectors but I am not sure.

How do you find the sine of the angle between two vectors?

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In general for $\mathbf{a},\mathbf{b}\in\mathbb{R}^3$, we have $$\|\mathbf{a}\times \mathbf{b}\|=\|\mathbf{a}\|\|\mathbf{b}\|\sin\theta,$$ where $\theta$ is the angle between $\mathbf{a}$ and $\mathbf{b}$.

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Let $\theta$ be the angle between the two vectors. Now you can use, $$\cos\theta={\vec v\cdot\vec w\over |\vec v||\vec w|}$$ and $$\sin\theta=\sqrt{1-\cos^2\theta}.$$

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  • $\begingroup$ You don't need the $\pm$ because the angle between two vectors is always in $[0,\pi]$. $\endgroup$ – Thomas Andrews Mar 16 '15 at 21:58
  • $\begingroup$ Ah, yes, of course. $\endgroup$ – Tim Raczkowski Mar 16 '15 at 21:59
  • $\begingroup$ Provided $\sin \theta $ is positive. $\endgroup$ – loxaxs Dec 10 '17 at 1:32
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Translate your two vectors so that their tails are at the origin. Then draw a line through each of those two vectors. The smaller of the two angles is the called the "angle between the two vectors".

Hint on how to find it: The angle $\theta$ between two vectors $\vec u$ and $\vec v$ is given by the formula $$\theta = \arccos\left(\frac {\vec u\cdot\vec v}{|\vec u||\vec v|}\right)$$

Or -- here's a slightly harder way (because it involves calculating a cross product). You know that $|\vec u\times \vec v| = |\vec u||\vec v|\sin(\theta)$, so find the cross product and all of the norms and then plug in.

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    $\begingroup$ No, it's $\cos^{-1}{\vec u\cdot\vec v\over |\vec u||\vec v|}$. $\endgroup$ – Tim Raczkowski Mar 16 '15 at 21:58
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    $\begingroup$ Yep, sorry. I was editting in another method and didn't notice I forgot the $\arccos$. Whoops. $\endgroup$ – user224138 Mar 16 '15 at 22:00

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