Question

Prove that $\overrightarrow{\mathrm{A}}.\left(\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}}\right)=0$.

Answer 1

To prove: $\overrightarrow{\mathrm{A}}.\left(\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}}\right)=0$

Proof: Vector product is given by $\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}}=\left|\overrightarrow{\mathrm{A}}\right|\left|\overrightarrow{\mathrm{B}}\right|\sin \hat{n}$.

$\left|\overrightarrow{\mathrm{A}}\right|\left|\overrightarrow{\mathrm{B}}\right|\sin \hat{n}$ is a vector which is perpendicular to the plane containing $\overrightarrow{\mathrm{A}}\mathrm{and}\overrightarrow{\mathrm{B}}$. This implies that it is also perpendicular to $\overrightarrow{\mathrm{A}}$. We know that the dot product of two perpendicular vectors is zero.
∴ $\overrightarrow{\mathrm{A}}.\left(\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}}\right)=0$

Hence, proved.