Question

If $\overrightarrow{a}$ and $\overrightarrow{b}$ are perpendicular, then $\overrightarrow{a}\times (\overrightarrow{a}\times (\overrightarrow{a}\times (\overrightarrow{a}\times \overrightarrow{b})))$ is equal to :

• A. $\frac{1}{2}|\overrightarrow{a}{|}^4\overrightarrow{b}$
• B. $\overrightarrow{a}\times \overrightarrow{b}$
• C. $|\overrightarrow{a}{|}^4\overrightarrow{b}$
• D. $\overrightarrow{0}$

Answer 1

The correct option is C $|\overrightarrow{a}{|}^4\overrightarrow{b}$

$\overrightarrow{a}\times (\overrightarrow{a}\times (\overrightarrow{a}\times (\overrightarrow{a}\times \overrightarrow{b})))$$=\overrightarrow{a}\times (\overrightarrow{a}\times ((\overrightarrow{a}\cdot \overrightarrow{b})\overrightarrow{a}-|\overrightarrow{a}{|}^2\overrightarrow{b})$
$=\overrightarrow{a}\times (-|\overrightarrow{a}{|}^2(\overrightarrow{a}\times \overrightarrow{b}\left)\right)$
$=-|\overrightarrow{a}{|}^2\left(\right(\overrightarrow{a}\cdot \overrightarrow{b})\overrightarrow{a}-|\overrightarrow{a}{|}^2\overrightarrow{b}))$
$=-(\overrightarrow{a}\cdot \overrightarrow{b})\overrightarrow{a}|\overrightarrow{a}{|}^2+|\overrightarrow{a}{|}^4\overrightarrow{b}$
$=|\overrightarrow{a}{|}^4\overrightarrow{b}(\because \overrightarrow{a}\cdot \overrightarrow{b}=0)$