Question:
Prove that $\vec{A} \cdot(\vec{A} \times \vec{B})=0$
Solution:
$(\mathbf{A} \times \mathbf{B})=\mathrm{AB} \sin \ominus \hat{\mathbf{u}}$, where รป is a unit vector perpendicular to both $\mathrm{A}$ and $\mathrm{B}$.
Now, $\mathbf{A} .(\mathbf{A} \times \mathbf{B})$ is basically a dot product between two vectors which are perpendicular to each other.
Then $\cos 90^{\circ}=0$, and thus
A. $(\mathbf{A} \times \mathbf{B})=0$