Answer
(a) $\frac{1}{m}\vec v$.
(b) $\frac{1}{3}\langle 1, -2, 2 \rangle=\langle \frac{1}{3}, -\frac{2}{3}, \frac{2}{3} \rangle$
$\frac{\sqrt 2}{20}\langle -6, 8, 10 \rangle=\langle -\frac{3\sqrt 2}{10}, \frac{2\sqrt 2}{5}, \frac{\sqrt 2}{2} \rangle$
$\frac{\sqrt {142}}{142}\langle 6,5,9 \rangle=\langle \frac{6\sqrt {142}}{142},\frac{5\sqrt {142}}{142},\frac{9\sqrt {142}}{142} \rangle$
Work Step by Step
(a) If we multiply $\vec v$ with $\frac{1}{m}$, its magnitude will be 1 and $\vec v$ becomes a unit vector. So the answer is $\frac{1}{m}\vec v$.
(b) For $\langle 1, -2, 2 \rangle$, its amplitude is $m=\sqrt {1^2+(-2)^2+2^2}=3$, so $\frac{1}{3}\langle 1, -2, 2 \rangle=\langle \frac{1}{3}, -\frac{2}{3}, \frac{2}{3} \rangle$ is a unit vector.
For $\langle -6, 8, 10 \rangle$, its amplitude is $m=\sqrt {(-6)^2+(8)^2+10^2}=10\sqrt 2$, so
$\frac{\sqrt 2}{20}\langle -6, 8, 10 \rangle=\langle -\frac{3\sqrt 2}{10}, \frac{2\sqrt 2}{5}, \frac{\sqrt 2}{2} \rangle$ is a unit vector.
For $\langle 6, 5, 9 \rangle$, its amplitude is $m=\sqrt {6^2+(5)^2+9^2}=\sqrt {142}$, so
$\frac{\sqrt {142}}{142}\langle 6,5,9 \rangle=\langle \frac{6\sqrt {142}}{142},\frac{5\sqrt {142}}{142},\frac{9\sqrt {142}}{142} \rangle$ is a unit vector.
