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

Dec 25, 2015

$126 , {294}^{\circ}$

#### 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 $A \mathmr{and} B$ 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. $\cos \theta = \frac{A \cdot B}{| | A | | \cdot | | B | |}$

$\therefore \theta = {\cos}^{- 1} \left(\frac{\left[\left(1 , 3 , - 8\right) \cdot \left(4 , 1 , 5\right)\right]}{| | \left(1 , 3 , - 8\right) | | \cdot | | \left(4 , 1 , 5\right) | |}\right)$

$= {\cos}^{- 1} \left(\frac{4 + 3 - 40}{\sqrt{74 \sqrt{42}}}\right)$

$= 126 , {294}^{\circ}$.