Chapter 5 - Inner Product Spaces - Review Exercises - Page 286: 74

See below.

We know that $||u\times v||=||u||\cdot||v||\cdot \sin{\alpha}$ if $\alpha$ is the enclosed angle between $u$ and $v$. Hence, if $||u\times v||=||u||\cdot||v||$, then $\sin{\alpha}=1$, thus $\alpha=90^\circ$. Thus, we proved what we had to.