If $\overline{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ and $\bar{b}=\hat{i}…

If $\overline{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ and $\bar{b}=\hat{i} \times(\bar{a} \times \hat{i})+\hat{j} \times(\bar{a} \times \hat{j})+\hat{k} \times(\bar{a} \times \hat{k})$ then $|\vec{b}|$ is
  1. $\sqrt{12}$
  2. $2 \sqrt{12}$
  3. $3 \sqrt{14}$
  4. $2 \sqrt{14}$

Solution

$\begin{aligned} & \hat{\mathrm{i}} \times(\overline{\mathrm{a}} \times \hat{\mathrm{i}}) \\ & =(\hat{\mathrm{i}} \cdot \hat{\mathrm{i}}) \overline{\mathrm{a}}-(\hat{\mathrm{i}} \cdot \overline{\mathrm{a}}) \hat{\mathrm{i}} \\ & =1(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})-[\hat{\mathrm{i}} \cdot(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})] \hat{\mathrm{i}} \\ & =\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}-\hat{\mathrm{i}}=2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\end{aligned}$ $\begin{aligned} & \hat{\mathrm{j}} \times(\overline{\mathrm{a}} \times \hat{\mathrm{j}})=(\hat{\mathrm{j}} \cdot \hat{\mathrm{j}}) \overline{\mathrm{a}}-(\hat{\mathrm{j}} \cdot \overline{\mathrm{a}}) \hat{\mathrm{j}} \\ & =\hat{i}+3 \hat{k} \\ & \hat{\mathrm{k}} \times(\overline{\mathrm{a}} \times \hat{\mathrm{k}})=(\hat{\mathrm{k}} \cdot \hat{\mathrm{k}}) \overline{\mathrm{a}}-(\hat{\mathrm{k}} \cdot \overline{\mathrm{a}}) \hat{\mathrm{k}} \\ & =\hat{i}+2 \hat{j} \\ & \therefore \quad \bar{b}=2 \hat{j}+3 \hat{k}+\hat{i}+3 \hat{k}+\hat{i}+2 \hat{j} \\ & =2 \hat{i}+4 \hat{j}+6 \hat{k} \\ & \therefore \quad|\bar{b}|=\sqrt{4+16+36}=\sqrt{56}=2 \sqrt{14} \end{aligned}$

Asked in: MHT CET 2024 (02 May Shift 2)

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