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Question: Answered & Verified by Expert
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
MathematicsVector AlgebraMHT CETMHT CET 2023 (11 May Shift 2)
Options:
  • A $30 \hat{\mathrm{i}}-\hat{\mathrm{j}}-14 \hat{\mathrm{k}}$
  • B $90 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}-42 \hat{\mathrm{k}}$
  • C $90 \hat{\mathrm{i}}+\hat{\mathrm{j}}-7 \hat{\mathrm{k}}$
  • D $30 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}$
Solution:
1676 Upvotes Verified Answer
The correct answer is: $90 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}-42 \hat{\mathrm{k}}$
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.

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