If $\overline{\mathrm{a}}=\mathrm{a}_1 \hat{\mathrm{i}}+\mathrm{a}_2 \hat{\mathrm{j}}+\mathrm{a}_3…

If $\overline{\mathrm{a}}=\mathrm{a}_1 \hat{\mathrm{i}}+\mathrm{a}_2 \hat{\mathrm{j}}+\mathrm{a}_3 \hat{\mathrm{k}}, \overline{\mathrm{b}}=\mathrm{b}_1 \hat{\mathrm{i}}+\mathrm{b}_2 \hat{\mathrm{j}}+\mathrm{b}_3 \hat{\mathrm{k}}$ and $\bar{c}=c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}$ are non-zero non-coplanar vectors and $m$ is non-zero scalar such that value of $m$ is equal to
  1. 2
  2. 3
  3. 4
  4. 7

Solution

$\begin{aligned} & {\left[\begin{array}{lll} \overline{\mathrm{m}} \overline{\mathrm{a}}+\overline{\mathrm{b}} & \mathrm{~m} \overline{\mathrm{~b}}+\overline{\mathrm{c}} & \mathrm{~m} \overline{\mathrm{c}}+\overline{\mathrm{a}} \end{array}\right]=28\left[\begin{array}{lll} \overline{\mathrm{a}} & \overline{\mathrm{~b}} & \overline{\mathrm{c}} \end{array}\right]} \\ & \Rightarrow(\mathrm{ma}+\overline{\mathrm{b}})[(\mathrm{m} \overline{\mathrm{~b}}+\overline{\mathrm{c}}) \times(\mathrm{m} \overline{\mathrm{c}}+\overline{\mathrm{a}})]=28\left[\begin{array}{lll} \overline{\mathrm{a}} & \overline{\mathrm{~b}} & \overline{\mathrm{c}} \end{array}\right] \\ & \Rightarrow(m \bar{a}+\bar{b}) \\ & {\left[\mathrm{m}^2(\overline{\mathrm{~b}} \times \overline{\mathrm{c}})+\mathrm{m}(\overline{\mathrm{~b}} \times \overline{\mathrm{a}})+\mathrm{m}(\overline{\mathrm{c}} \times \overline{\mathrm{c}})+(\overline{\mathrm{c}} \times \overline{\mathrm{a}})\right]} \\ & =28\left[\begin{array}{lll} \overline{\mathrm{a}} & \overline{\mathrm{~b}} & \overline{\mathrm{c}} \end{array}\right] \end{aligned}$ $\begin{aligned} & \Rightarrow(\mathrm{ma}+\overline{\mathrm{b}})\left[\mathrm{m}^2(\overline{\mathrm{~b}} \times \overline{\mathrm{c}})+\mathrm{m}(\overline{\mathrm{~b}} \times \overline{\mathrm{a}})+(\overline{\mathrm{c}} \times \overline{\mathrm{a}})\right] \\ & =28\left[\begin{array}{lll} \overline{\mathrm{a}} & \overline{\mathrm{~b}} & \overline{\mathrm{c}} \end{array}\right] \\ & \Rightarrow m \bar{a} \cdot m^2(\overline{\mathrm{~b}} \times \overline{\mathrm{c}})+\mathrm{m}^2 \overline{\mathrm{a}}(\overline{\mathrm{~b}} \times \overline{\mathrm{a}})+\mathrm{ma}(\overline{\mathrm{c}} \times \overline{\mathrm{a}}) \\ & +\overline{\mathrm{b}} \cdot \mathrm{~m}^2(\overline{\mathrm{~b}} \times \overline{\mathrm{c}})+\overline{\mathrm{b}} \cdot \mathrm{~m}(\overline{\mathrm{~b}} \times \overline{\mathrm{a}})+\overline{\mathrm{b}} \cdot(\overline{\mathrm{c}} \times \overline{\mathrm{a}}) \\ & =28\left[\begin{array}{lll} \overline{\mathrm{a}} & \overline{\mathrm{~b}} & \overline{\mathrm{c}} \end{array}\right] . \end{aligned}$ $\begin{aligned} & \Rightarrow \mathrm{m}^3\left[\begin{array}{lll}\overline{\mathrm{a}} & \overline{\mathrm{b}} & \overline{\mathrm{c}}\end{array}\right]+\left[\begin{array}{lll}\overline{\mathrm{b}} & \overline{\mathrm{a}} & \overline{\mathrm{c}}\end{array}\right]=28\left[\begin{array}{lll}\overline{\mathrm{a}} & \overline{\mathrm{b}} & \overline{\mathrm{c}}\end{array}\right] \\ & \Rightarrow \mathrm{m}^3+1=28 \\ & \Rightarrow \mathrm{~m}^3=27 \\ & \Rightarrow \mathrm{~m}=3\end{aligned}$

Asked in: MHT CET 2024 (03 May Shift 2)

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