Q62 / 150 · MHT CET 2024 (02 May Shift 1) · Mathematics

The scalar $\overline{\mathrm{a}} \cdot[(\overline{\mathrm{b}}+\overline{\mathrm{c}}) \times(\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}})]$ equals

  • Correct answer
    A
    0
  • B
    $[\bar{a} \overline{\mathrm{~b}} \overline{\mathrm{c}}]+[\overline{\mathrm{b}} \overline{\mathrm{c}} \overline{\mathrm{a}}]$
  • C
    $[\bar{a} \overline{\mathrm{~b}} \overline{\mathrm{c}}]$
  • D
    1

Solution

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

Hence, the correct answer is option: 'A'.

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