Question

If $\overrightarrow{u},\overrightarrow{v} $ and $\overrightarrow{w} $ are three non coplanar vectors, then $(\overrightarrow{u}+\overrightarrow{v} -\overrightarrow{w}).[(\overrightarrow{u}-\overrightarrow{v})\times (\overrightarrow{v}-\overrightarrow{w} )]= $

• A. 0
• B. $\overrightarrow{u}.\overrightarrow{v} \times \overrightarrow{w}$
• C. $\overrightarrow{u}.\overrightarrow{w} \times \overrightarrow{v}$
• D. $3\,\overrightarrow{u}.\overrightarrow{v} \times \overrightarrow{w}$

Answer 1

Answer: (B)

$\left(\vec{u}+\vec{v}-\vec{w}\right)\cdot\left(\left(\vec{u}-\vec{v}\right)\times\left(\vec{v}-\vec{w}\right)\right)$
$=\left(\vec{u}+\vec{v}-\vec{w}\right)\cdot\left(\vec{u}\times\vec{v}-\vec{u}\times\vec{w}+\vec{v}\times\vec{w}\right)$
$\left[\because \vec{v}\times\vec{v}=\vec{0}\right]$
$=\vec{u}\cdot\vec{v}\times\vec{w}-\vec{v}\cdot\vec{u}\times\vec{w}-\vec{w}\cdot\vec{u}\times\vec{v}$
[All other variables]
$=\left[\vec{u}\,\vec{v}\,\vec{w}\right]+\left[\vec{u}\,\vec{v}\,\vec{w}\right]-\left[\vec{u}\,\vec{v}\,\vec{w}\right]$
$=\left[\vec{u}\,\vec{v}\,\vec{w}\right]$