Answer
$ \lt \dfrac{8}{3}, \dfrac{4}{3},\dfrac{-4}{3}\gt$
Work Step by Step
The vector projections of $u$ onto $v$ is defined as: $proj_{v}u=(\dfrac{u \cdot v}{v \cdot v})v$
Thus,
$proj_{v}u=(\dfrac{u \cdot v}{v \cdot v})v=\dfrac{1(2)+(1)(1)+(-5)(-1)}{(2)^2+(1)^2+(-1)^2}\lt 2,1,-1 \gt=\dfrac{8}{6}\lt 2,1,-1 \gt$
