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MHT CET · Maths · Vector Algebra

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. A \(30 \hat{\mathrm{i}}-\hat{\mathrm{j}}-14 \hat{\mathrm{k}}\)
  2. B \(90 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}-42 \hat{\mathrm{k}}\)
  3. C \(90 \hat{\mathrm{i}}+\hat{\mathrm{j}}-7 \hat{\mathrm{k}}\)
  4. D \(30 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(90 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}-42 \hat{\mathrm{k}}\)

Step-by-step Solution

Detailed explanation

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.