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Or, in general, what does the magnitude of the cross product mean? How would you prove or disprove this?

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Every time you come up with a question like this, the very very first thing you have to do is to consider examples. It is very easy to construct unit vectors, and when you have two of them, it is also very easy to compute the magnitude of their cross product. If you do this a few times, you will find the answer to your question at once. – Mariano Suárez-Alvarez Feb 22 '11 at 19:32
The problem with "considering examples" is it doesn't give you the general rule. Fringe cases won't become evident. Looking it up in reference sources or asking experts is a more thorough approach. – Greg Dec 5 '15 at 21:20
up vote 17 down vote accepted

No - for example, the cross product of any unit vector with itself is 0. In general, the magnitude of the cross product of vectors $\vec{a}$ and $\vec{b}$ is $$|\vec{a}\times\vec{b}|=|\vec{a}||\vec{b}|\sin(\theta)$$ where $\theta$ is the angle between the vectors $\vec{a}$ and $\vec{b}$. Thus, the cross product of two unit vectors $\vec{u}$ and $\vec{v}$ is itself a unit vector if and only if $\vec{u}$ and $\vec{v}$ are orthogonal, i.e. meet at right angles (this makes $\sin(\theta)=\sin(\frac{\pi}{2})=1$).

As to the general interpretation of the magnitude of the cross product, see Wikipedia:

The magnitude of the cross product can be interpreted as the positive area of the parallelogram having $a$ and $b$ as sides.

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You mean $\pi/2$, don't you, instead of $\pi/4$? – Mike Spivey Feb 22 '11 at 19:21
Whoops, you're right - edited. – Zev Chonoles Feb 22 '11 at 19:24


Let $a,b$ unit vectors, so we have $|a| = |b| = 1$

$|\vec{a}\times\vec{b}|=|\sin(\theta)| \le 1$ (equality is when $|\sin(\theta)| = 1$ i.e. when a and b are perpendicular)

Therefore in general the result won't be a unit vector.

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