$\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k}$. A vector $\vec{b}$ is collinear with $\vec{a}$, has magnitude $3\sqrt{14}$ and makes an acute angle with the $z$-axis. What is $\vec{b}$?

Explanation

1. 01

   $\lvert \vec{a} \rvert = \sqrt{4 + 9 + 1} = \sqrt{14}$, so a collinear vector of magnitude $3\sqrt{14}$ is $\pm 3\vec{a}$.

2. 02

   An acute angle with the $z$-axis needs a positive $\hat{k}$ component; $3\vec{a}$ has $-3$, so $\vec{b} = -3\vec{a} = -6\hat{i} - 9\hat{j} + 3\hat{k}$.