Question

If $\overrightarrow{x}\times\overrightarrow{b} \,=\,\overrightarrow{c}\times\overrightarrow{b} $ and $ \overrightarrow{x}\,\bot\, \overrightarrow{a},$ then $\overrightarrow{x}$ is equal to

• A. $\frac{(\overrightarrow{b}\times\overrightarrow{c})\times \overrightarrow{a}}{\overrightarrow{b}.\overrightarrow{a}}$
• B. $\frac{ \overrightarrow{b}\times(\overrightarrow{a}\times\overrightarrow{c})}{\overrightarrow{b}.\overrightarrow{c}}$
• C. $\frac{(\overrightarrow{c}\times\overrightarrow{b})\times \overrightarrow{a}}{\overrightarrow{a}.\overrightarrow{b}}$
• D. none of these.

Answer 1

Answer: (A)

$\vec{x} \times \vec{b}=\vec{c} \times \vec{b}$
$\Rightarrow \vec{a} \times \left(\vec{x} \times \vec{b}\right)= \vec{a} \times \left(\vec{c} \times \vec{b}\right)$
$\Rightarrow \left(\vec{a}\cdot\vec{b}\right)\vec{x}-\left(\vec{a}\cdot \vec{x}\right) \vec{b}$
$=-\left(\vec{c} \times \vec{b}\right)\times\vec{a}$
$=\left(\vec{b} \times \vec{c}\right) \times \vec{a}$
$\Rightarrow \left(\vec{a}\cdot \vec{b}\right)\vec{x}=\left(\vec{b} \times \vec{c}\right) \times \vec{a}\quad[\because \vec{a} . \vec{x}$ as $=0\,\vec{x} \,\bot \,\vec{a}]$
$\Rightarrow \vec{x}=\frac{\left(\vec{b} \times \vec{c}\right)\times\vec{a}}{\vec{a} . \vec{b}}=\frac{\left(\vec{b} \times \vec{c}\right)\times \vec{a}}{\vec{b} . \vec{a}}$