Answer
We know that if $u$ and $v$ are orthogonal, their included angle is $\theta=90^\circ$.
Thus, $||u\times v||=||u||\ ||v||\sin{\theta}=||u||\ ||v||\sin{90^\circ}=||u||\ ||v||(1)=||u||\ ||v||.$
If $u$ and $v$ are unit vectors, then $||u||=||v||=1$.
Thus, $||u\times v||=||u||\ ||v||=(1)(1)=1$, so it must also be a unit vector, because its magnitude is $1$.
Work Step by Step
We know that if $u$ and $v$ are orthogonal, their included angle is $\theta=90^\circ$.
Thus, $||u\times v||=||u||\ ||v||\sin{\theta}=||u||\ ||v||\sin{90^\circ}=||u||\ ||v||(1)=||u||\ ||v||.$
If $u$ and $v$ are unit vectors, then $||u||=||v||=1$.
Thus, $||u\times v||=||u||\ ||v||=(1)(1)=1$, so it must also be a unit vector, because its magnitude is $1$.
