If $\overline{\mathrm{u}}, \overline{\mathrm{v}}$ and $\overline{\mathrm{w}}$ are three non-coplanar vectors…

If $\overline{\mathrm{u}}, \overline{\mathrm{v}}$ and $\overline{\mathrm{w}}$ are three non-coplanar vectors, then $(\bar{u}+\bar{v}-\bar{w}) \cdot[(\bar{u}-\bar{v}) \times(\bar{v}-\bar{w})]$ is equal to
  1. $\overline{\mathrm{u}} \cdot(\overline{\mathrm{v}} \times \overline{\mathrm{w}})$
  2. $\overline{\mathrm{u}} \cdot(\overline{\mathrm{w}} \times \overline{\mathrm{v}})$
  3. $3 \overline{\mathrm{u}} \cdot(\overline{\mathrm{v}} \times \overline{\mathrm{w}})$
  4. 0

Solution

$\begin{aligned} & \begin{aligned}( & \bar{u}+\bar{v}-\bar{w}) \cdot[(\bar{u}-\bar{v}) \times(\bar{v}-\bar{w})] \\ = & \bar{u} \cdot(\bar{u} \times \bar{v})-\bar{u} \cdot(\bar{u} \times \bar{w})+\bar{u} \cdot(\bar{v} \times \bar{w})+\bar{v} \cdot(\bar{u} \times \bar{v}) \\ & \quad-\bar{v} \cdot(\bar{u} \times \bar{w})+\bar{v} \cdot(\bar{v} \times \bar{w})-\bar{w} \cdot(\bar{u} \times \bar{v}) \\ & \quad+\bar{w} \cdot(\bar{u} \times \bar{w})-\bar{w} \cdot(\bar{v} \times \bar{w}) \\ = & {[\bar{u} \cdot \bar{v} \bar{w}]-[\bar{v} \bar{u} \bar{w}]-[\bar{w} \bar{u} \bar{v}] }\end{aligned} \\ & =[\bar{u} \cdot \bar{v} \bar{w}]+[\bar{u} \bar{v} \bar{w}]-[\bar{u} \bar{v} \bar{w}] \\ & = \\ & =\bar{u} \cdot\left(\bar{v} \times \overline{w_w}\right)\end{aligned}$

Asked in: MHT CET 2024 (15 May Shift 1)

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