Answer
(a) $\vec v=(-2)\vec u$
(b) $\vec v=(-\frac{4}{3})\vec u$
(c) not parallel.
Work Step by Step
(a) Given vectors: $\vec u=\langle 3, -2, 4 \rangle$ and $\vec v=\langle -6, 4, -8 \rangle$
we can see that a factor of $-2$ can be used on $\vec u$ to get $\vec v$.
So the two vectors are parallel and we have $\vec v=(-2)\vec u=-2(\langle 3, -2, 4 \rangle)=\langle -6, 4, -8 \rangle$
(b) Given vectors: $\vec u=\langle -9, -6, 12 \rangle$ and $\vec v=\langle 12, 8, -16 \rangle$
we can see that a factor of $-\frac{4}{3}$ can be used on $\vec u$ to get $\vec v$.
So the two vectors are parallel and we have $\vec v=(-\frac{4}{3})\vec u=-\frac{4}{3}(\langle -9, -6, 12 \rangle)
=\langle 12, 8, -16 \rangle$
(c) Given vectors: $\vec u=\langle 1, 1, 1 \rangle$ and $\vec v=\langle 2, 2, -2 \rangle$
we can not find a factor that can be used on $\vec u$ to get $\vec v$.
So the two vectors are not parallel.
