Q80 / 125 · BITSAT 2021 · Mathematics
If $\vec{a}$ is a vector of magnitude 50 , collinear with the vector $\vec{b}=6 \hat{i}-8 \hat{j}-\frac{15}{2} \hat{k}$ and makes an acute angle with the positive direction of $Z$ - axis, then $\vec{a}$ is equal to
- Correct answerA$-24 \hat{i}+32 \hat{j}+30 \hat{k}$
- B$24 \hat{i}-32 \hat{j}-30 \hat{k}$
- C$-12 \hat{i}+16 \hat{j}-15 \hat{k}$
- D$12 \hat{i}-16 \hat{j}-15 \hat{k}$
Solution
Since $\vec{a}=m \vec{b}$ for some scalar $m$ i.e.,
$\begin{array}{l}
\vec{a}=m\left(6 \hat{i}-8 \hat{j}-\frac{15}{2} \hat{k}\right) \\
\Rightarrow|a|=|m| \sqrt{36+64+\frac{225}{4}} \\
\Rightarrow 50=\frac{25}{2}|m| \\
\Rightarrow|m|=4 \\
\Rightarrow m=\pm 4
\end{array}$
Since, a makes an acute angle with the positive direction of $Z$ - axis, so its $z$ component must be positive and hence, $m$ must be $-4$.
$\therefore a=-4\left(6 \hat{i}+8 \hat{j}-\frac{15}{2} \hat{k}\right)=-24 \hat{i}+32 \hat{j}+30 \hat{k}$
Hence, the correct answer is option: 'A'.