Question

If a is a vector of magnitude $50$ , collinear with the vector $b =6 \hat{ i }-8 \hat{ j }-\frac{15}{2} \hat{ k }$ and makes an acute angle with the positive direction of Z-axis, then $a$ 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 $a = mb$ for some scalar $m$ i.e.
$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 1$
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 $-4$.
$\therefore a =-4\left(6 \hat{ i }+8 \hat{ j }-\frac{15}{2} \hat{ k }\right)=-24 \hat{ i }+32 \hat{ j }+30 \hat{ k }$