Question

If $\overline{u},\overline{v}$ and $\overline{w}$ are three non coplanar vectors, then the value of $(\overline{u}+\overline{v}-\overline{w})\cdot (\overline{u}-\overline{v})\times (\overline{v}-\overline{w})$

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

Answer 1

Answer: (A)

$\overline{u}\cdot \overline{v}\times \overline{w}$

we have $(\overline{u}+\overline{v}-\overline{w})\cdot (\overline{u}-\overline{v})\times (\overline{v}-\overline{w})$

$=$$(\overline{u}+\overline{v}-\overline{w})\cdot [\overline{u}\times \overline{v}-\overline{u}\times \overline{w}+\overline{v}\times \overline{w}]\because \overline{v}\times \overline{v}=0$

$=\overline{u}\cdot (\overline{u}\times \overline{v})-\overline{u}\cdot (\overline{u}\times \overline{w})+\overline{u}\cdot (\overline{v}\times \overline{w})+\overline{v}\cdot (\overline{u}\times \overline{v})-\overline{v}\cdot (\overline{u}\times \overline{w})+\overline{v}\cdot (\overline{v}\times \overline{w})-\overline{w}\cdot (\overline{u}\times \overline{v})-\overline{w}\cdot (\overline{u}\times \overline{w})-\overline{w}\cdot (\overline{v}\times \overline{w})$

$=\overline{u}\cdot (\overline{v}\times \overline{w})+\overline{v}\cdot (\overline{w}\times \overline{u})-\overline{w}\cdot (\overline{u}\times \overline{v})$

$=\overline{u}\cdot (\overline{v}\times \overline{w})+\overline{u}\cdot (\overline{v}\times \overline{w})-\overline{u}\cdot (\overline{u}\times \overline{w})$

$=\overline{u}\cdot (\overline{v}\times \overline{w})$ $(\because [abc]=[bca]=[cab\left]\right)$