If $\bar{a}, \bar{b}$ and $\bar{c}$ are any three non-zero vectors, then $(\bar{a}+2 \bar{b}+\bar{c})…

If $\bar{a}, \bar{b}$ and $\bar{c}$ are any three non-zero vectors, then $(\bar{a}+2 \bar{b}+\bar{c}) \cdot[(\bar{a}-\bar{b}) \times(\bar{a}-\bar{b}-\bar{c})]=$
  1. $\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]$
  2. $2\left[\begin{array}{lll}-\bar{a} & \bar{b} & \bar{c}\end{array}\right]$
  3. $3\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]$
  4. $4\left[\begin{array}{lll}\vec{a} & \bar{b} & \bar{c}\end{array}\right]$

Solution

$\begin{aligned} & (\bar{a}+2 \bar{b}+\bar{c}) \cdot[(\bar{a}-\bar{b}) \times(\bar{a}-\bar{b}-\bar{c})] \\ & =(\bar{a}+2 \bar{b}+\bar{c}) \cdot(\bar{a} \times \bar{a}-\bar{a} \times \bar{b}-\bar{a} \times \bar{c}-\bar{b} \times \bar{a}+\bar{b} \times \bar{b}+\bar{b} \times \bar{c}) \\ & =(\bar{a}+2 \bar{b}+\bar{c}) \cdot(\overline{0}-\bar{a} \times \bar{b}-\bar{a} \times \bar{c}+\bar{a} \times \bar{b}+\overline{0}+\bar{b} \times \bar{c}) \\ & =(\overline{\mathrm{a}}+2 \overline{\mathrm{b}}+\overline{\mathrm{c}}) \cdot(\overline{\mathrm{c}} \times \overline{\mathrm{a}}+\overline{\mathrm{b}} \times \overline{\mathrm{c}}) \\ & =\bar{a} \cdot(\bar{c} \times \bar{a})+\bar{a} \cdot(\bar{b} \times \bar{c})+2 \bar{b} \cdot(\bar{c} \times \bar{a})+2 \bar{b} \cdot(\bar{b} \times \bar{c}) \\ & +\overline{\mathrm{c}} \cdot(\overline{\mathrm{c}} \times \overline{\mathrm{a}})+\overline{\mathrm{c}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}}) \\ & =\overline{0}+\bar{a} \cdot(\bar{b} \times \bar{c})+2 \bar{b} \cdot(\bar{c} \times \bar{a})+2 \times 0-0-0 \\ & =\left[\begin{array}{lll}\overline{\mathrm{a}} & \overline{\mathrm{b}} & \overline{\mathrm{c}}\end{array}\right]+2\left[\begin{array}{lll}\overline{\mathrm{b}} & \overline{\mathrm{c}} & \overline{\mathrm{a}}\end{array}\right] \\ & =\left[\begin{array}{lll}\overline{\mathrm{a}} & \overline{\mathrm{b}} & \overline{\mathrm{c}}\end{array}\right]+2\left[\begin{array}{lll}\overline{\mathrm{a}} & \overline{\mathrm{b}} & \overline{\mathrm{c}}\end{array}\right] \\ & =3\left[\begin{array}{lll}\overline{\mathrm{a}} & \overline{\mathrm{b}} & \overline{\mathrm{c}}\end{array}\right] \\ & \end{aligned}$

Asked in: MHT CET 2023 (09 May Shift 2)

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