If $\quad \overline{\mathrm{a}}=\hat{\mathrm{i}}+4 \hat{\mathrm{j}}+2 \hat{\mathrm{k}},…

If $\quad \overline{\mathrm{a}}=\hat{\mathrm{i}}+4 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}, \overline{\mathrm{b}}=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}, \overline{\mathrm{c}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}}$, then a vector $\bar{d}$ which is parallel to vector $\bar{a} \times \bar{b}$ and which $\overline{\mathrm{c}} \cdot \overline{\mathrm{d}}=15$, is
  1. $30 \hat{\mathrm{i}}-\hat{\mathrm{j}}-14 \hat{\mathrm{k}}$
  2. $90 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}-42 \hat{\mathrm{k}}$
  3. $90 \hat{\mathrm{i}}+\hat{\mathrm{j}}-7 \hat{\mathrm{k}}$
  4. $30 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}$

Solution

Here, $\bar{c}=2 \hat{i}-\hat{j}+4 \hat{k}$ And given that $\vec{c} \cdot \vec{d}=15$ We verify given options one by one to satisfy the above condition. Consider option (B), For $\overline{\mathrm{d}}=90 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}-42 \hat{\mathrm{k}}$ $\begin{aligned} \overline{\mathrm{c}} \cdot \overline{\mathrm{d}} & =(2)(90)+(-1)(-3)+(4)(-42) \\ & =180+3-168=15 \end{aligned}$ $\therefore \quad$ Option (B) is correct.

Asked in: MHT CET 2023 (11 May Shift 2)

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