Question

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

• A. $-24 \hat{ i }+32 \hat{ j }+30 \hat{ k }$
• B. $24 \hat{ i }-32 \hat{ j }-30 \hat{ k }$
• C. $-12 \hat{ i }+16 \hat{ j }-15 \hat{ k }$
• D. $12 \hat{ i }-16 \hat{ j }-15 \hat{ k }$

Answer 1

Answer: (A)

Since $\vec{a}=m \vec{b}$ for some scalar $m$ i.e.,
$\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| $
$\Rightarrow |m|=4$
$\Rightarrow m=\pm 4$
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 $-4$.
$\therefore a=-4\left(6 \hat{i}+8 \hat{j}-\frac{15}{2} \hat{k}\right)$
$=-24 \hat{i}+32 \hat{j}+30 \hat{k}$