Answer
See the proof below.
Work Step by Step
Our aim is to prove that $||u\times v||=||u||\ ||v||\sin{\theta}$
We know that $||u\times v||^2=||u||^2 \ ||v||^2 -(u \cdot v)^2$ and $u \cdot v =||u|| ||v|| \cos \theta $
Therefore, $||u \times v|| =\sqrt {||u \times v||^2}\\=\sqrt {||u|^2 ||v||^2 -(u \cdot v)^2}\\=\sqrt {||u||^2 \ ||v||^2 -||u||^2 ||v||^2 \cos^2 \theta}\\=\sqrt {||u||^2 ||v||^2 (1-\cos^2 \theta)}\\=\sqrt {||u||^2 ||v||^2 \sin^2 \theta)}\\=||u|| \ ||v|| \sin \theta$
Thus, the result has been proved.
