## University Calculus: Early Transcendentals (3rd Edition)

$\lt \dfrac{8}{3}, \dfrac{4}{3},\dfrac{-4}{3}\gt$
The formula to calculate the vector projection of $u$ onto $v$ is given by: $proj_{v}u=(\dfrac{u \cdot v}{v \cdot v})v$ Now, $proj_{v}u=\dfrac{1(2)+(1)(1)+(-5)(-1)}{2^2+1^2+(-1)^2}\lt 2,1,-1 \gt$ $=\dfrac{8}{6}\lt 2,1,-1 \gt=\dfrac{4}{3}\lt 2,1,-1 \gt$ Thus, our final answer is: $\lt \dfrac{8}{3}, \dfrac{4}{3},\dfrac{-4}{3}\gt$