Is the cross product of two unit vectors itself a unit vector? Or, in general, what does the magnitude of the cross product mean? How would you prove or disprove this?
 A: $|\vec{a}\times\vec{b}|=|\vec{a}||\vec{b}||\sin(\theta)|$
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.
A: 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.

A: If you know that the unit vectors you start with are perpendicular to each other (the dot product $\vec{a}\cdot\vec{b}$ is zero), then the cross product $\vec{a}\times\vec{b}$ will be a unit vector (a length of one).
If you don't have the above constraint for the input unit vectors, then the output of the cross product $\vec{a}\times\vec{b}$ is not guaranteed to be a unit vector, since the length depends on the angle (the length is $sin(\theta)$ where $\theta$ is the angle between the input vectors).
This answer focuses on the special cases, for the generalizations see the other great answers.
