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Question: Answered & Verified by Expert
If $\mathbf{a}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}, \mathbf{b}=m \hat{\mathbf{i}}+n \hat{\mathbf{j}}+12 \hat{\mathbf{k}}$ and $\mathbf{a} \times \mathbf{b}=\mathbf{0}$, then $(m, n)$ is equal to
MathematicsVector AlgebraAP EAMCETAP EAMCET 2016
Options:
  • A $\left(\frac{-24}{5}, \frac{-36}{5}\right)$
  • B $\left(\frac{-24}{5}, \frac{36}{5}\right)$
  • C $\left(\frac{24}{5}, \frac{-36}{5}\right)$
  • D $\left(\frac{24}{5}, \frac{36}{5}\right)$
Solution:
1439 Upvotes Verified Answer
The correct answer is: $\left(\frac{-24}{5}, \frac{-36}{5}\right)$
We have, $a=2 \hat{i}+3 \hat{j}-5 \hat{k}$
$b=m \hat{i}+n \hat{j}+12 \hat{k}$
Since, $a \times b=0 \quad \Rightarrow a \| b$
$\begin{aligned}
& \therefore \quad \frac{2}{m}=\frac{3}{n}=\frac{-5}{12} \\
& \Rightarrow m=-\frac{24}{5} \text { and } n=-\frac{36}{5}
\end{aligned}
$Thus, $(m, n)=\left(-\frac{24}{5},-\frac{36}{5}\right)$

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