What is the angle between #<1 ,3,-8># and #< 4,1,5>#?

1 Answer

#126,294^@#

Explanation:

There are 2 methods we can use to calculate this algebraically, either using the vector cross product or the vector inner product.
I shall use the latter method as it is quicker and also more general.

The angle between any 2 vectors #A and B# in any dimensional vector space may be given by the inverse cosine of the Euclidean inner product of the 2 vectors divided by the product of the norms of the 2 vectors.
ie. #costheta=(A*B)/(||A||*||B||)#

#therefore theta=cos^(-1) (([(1,3,-8) * (4,1,5)])/(||(1,3,-8)|| * ||(4,1,5)||))#

#=cos^(-1)((4+3-40)/(sqrt(74sqrt42)))#

#=126,294^@#.