Find the angle θθ between the vectors v=2i+kv=2i+k, w=j−3kw=j−3k.

Hello : 
v . w = | | v | | × | | w | |  cos(θ) (1)
v(2,0,1)   w(0,1,-3)

v . w = (2)(0)+(0)(1)+(1)(-3) = - 3
| | v | |  = √((2)²+(0)²+(1)²) = √5
| | w |  =  √((0)²+(1)²+(-3)²) = √10
by (1) : 
cos(θ) = (v . w ) /  | | v | | × | | w | |

cos(θ) = = - 3/√50
θ =   calculate by 2ind function ( calculator)

Iam not sure but i think the answer is b