Question

Suppose u= <1,-5,4> and, v= <-1,1,1>. Then: The projection of u along v is? The projection of u orthogonal to v is?

Answer 1

Answer:

$Proj_{v}U=\frac{u\cdot v}{\left\lVert v \right\rVert^2}v=\frac{-1-5+4}{\left(\sqrt{(-1)^2+(1)^2+(1)^2}\right)^2}(-1,1,1)=\left(\frac{2}{3},-\frac{2}{3},-\frac{2}{3}\right).$

Step-by-step explanation:

The orthogonal projection of U along V is a vector defined by $Proj_{v}u=\lambda v$ such that $v\cdot(u-\lambda v)=0.$. From here we obtain that $\lambda=\frac{u\cdot v}{\left\lVert v \right\rVert ^{2}}.$