Question

If a vector $\overrightarrow{a}$ of magnitude $50$ is collinear with vector $\overrightarrow{b}=6\hat{i}-8\hat{j}-\frac{15}{2}\hat{k}$ and makes an acute angle with positive $z$-axis then

• A. $\overrightarrow{a}=4\overrightarrow{b}$
• B. $\overrightarrow{a}=-4\overrightarrow{b}$
• C. $\overrightarrow{b}=4\overrightarrow{a}$
• D. $\overrightarrow{a}=-2\overrightarrow{b}$

Answer 1

Answer: (B)

$\overrightarrow{a}=-4\overrightarrow{b}$

$\overrightarrow{a}=k\overrightarrow{b}$
$|a|=|k||\overrightarrow{b}|$
So$=|k|\frac{25}{2}$
$|k|=4$
$\overrightarrow{a}.\hat{k}$=positive (acute angle with $z$ axis)
$k<o$
$k=-4$