The scalar $\overline{\mathrm{a}} \cdot[(\overline{\mathrm{b}}+\overline{\mathrm{c}})…
The scalar $\overline{\mathrm{a}} \cdot[(\overline{\mathrm{b}}+\overline{\mathrm{c}}) \times(\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}})]$ equals
- 0
- $[\bar{a} \overline{\mathrm{~b}} \overline{\mathrm{c}}]+[\overline{\mathrm{b}} \overline{\mathrm{c}} \overline{\mathrm{a}}]$
- $[\bar{a} \overline{\mathrm{~b}} \overline{\mathrm{c}}]$
- 1
Solution
$\begin{aligned} & \overline{\mathrm{a}}[(\overline{\mathrm{b}} \times \overline{\mathrm{c}}) \times(\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}})] \\ & =\overline{\mathrm{a}}[(\overline{\mathrm{b}} \times \overline{\mathrm{a}}+\overline{\mathrm{b}} \times \overline{\mathrm{b}}+\overline{\mathrm{b}} \times \overline{\mathrm{c}}+\overline{\mathrm{c}} \times \overline{\mathrm{a}}+\overline{\mathrm{c}} \times \overline{\mathrm{b}}+\overline{\mathrm{c}} \times \overline{\mathrm{c}})] \\ & =\overline{\mathrm{a}}(\overline{\mathrm{b}} \times \overline{\mathrm{a}})+\overline{\mathrm{a}}(\overline{\mathrm{b}} \times \overline{\mathrm{c}})+\overline{\mathrm{a}}(\overline{\mathrm{c}} \times \overline{\mathrm{a}})+\overline{\mathrm{a}}(\overline{\mathrm{c}} \times \overline{\mathrm{b}}) \\ & =\left[\begin{array}{lll}\overline{\mathrm{a}} & \overline{\mathrm{b}} & \overline{\mathrm{c}}\end{array}\right]+\left[\begin{array}{lll}\overline{\mathrm{a}} & \bar{c} & \overline{\mathrm{~b}}\end{array}\right] \\ & =\left[\begin{array}{lll}\overline{\mathrm{a}} & \overline{\mathrm{b}} & \overline{\mathrm{c}}\end{array}\right]-\left[\begin{array}{lll}\overline{\mathrm{a}} & \overline{\mathrm{b}} & \overline{\mathrm{c}}\end{array}\right] \\ & =0\end{aligned}$
Asked in: MHT CET 2024 (02 May Shift 1)
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