Find the angle between two vectors `vec(A)=2i+j-k` and `vec(B)=i-k`.

Explanation

`cos theta =(vec(A)vec(B))/(|vec(A)||vec(B)|)=(2x1+1xx0+(-1)xx(-1))/(sqrt(2^(2)+1^(2)+(-1)^(2))sqrt(1^(2)+0^(2)+(-1)^(2)))=(sqrt(3))/(2)`
`30^(@)`