# Prove the following

Question:

If $\vec{a} \& \vec{b}$ are perpendicular vactors, then $\overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})))$ is equal to

1. $\frac{1}{2}|\vec{a}|^{4} \vec{b}$

2. $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$

3. $|\vec{a}|^{4} \vec{b}$

4. \overrightarrow{0}

Correct Option: , 3

Solution:

$\bar{a} \times\left(\bar{a} \times\left((\bar{a} \cdot \bar{b})^{-}-\mid \vec{a}^{2} \bar{b}\right)\right)$

$\bar{a} \times\left(-|\vec{a}|^{2}(\vec{a} \times \vec{b})\right)=-|\vec{a}|^{2}\left((\vec{a} \cdot \vec{b}) \vec{a}-|\vec{a}|^{2} \vec{b}\right)$

$=-(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}) \overrightarrow{\mathrm{a}}|\overrightarrow{\mathrm{a}}|^{2}+|\overline{\mathrm{a}}|^{4} \overrightarrow{\mathrm{b}}$

$=|\overrightarrow{\mathrm{a}}|^{4} \overrightarrow{\mathrm{b}}(\because \overline{\mathrm{a}} \cdot \overline{\mathrm{b}}=0)$