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 $-4$.
$\therefore a=-4\left(6\hat{i}+8\hat{j}-\frac{15}{2}\hat{k}\right)$
$=-24\hat{i}+32\hat{j}+30\hat{k}$