# What is the angle between <4,8,2 > and < 0,-7,1>?

Dec 11, 2016

The angle is $= 146.4$º

#### Explanation:

You have 2 vectors, $\vec{u}$ and $\vec{v}$

The angle is given by the dot product definition

vecu.vecv=∥vecu∥*∥vecv∥costheta

where $\theta =$ angle between the 2 vectors.

If vecu=〈u_1,u_2,u_3〉 and vecv=〈v_1,v_2,v_3〉

The dot product is

vecu.vecv=〈u_1,u_2,u_3〉.〈v_1,v_2,v_3〉

$= {u}_{1} {v}_{1} + {u}_{2} {v}_{2} + {u}_{3} {v}_{3}$

Here, the dot product is

〈4,8,2〉.〈0,-7,1〉=4*0-7*8+2*1=-54

The modulus of a vector is ∥vecu∥=sqrt(u_1^2+u_2^2+u_3^2)

The modulus of $\vec{u}$ is ∥vecu∥=∥〈4,8,2〉∥=sqrt(16+64+4)=sqrt84

The modulus of $\vec{v}$ is ∥vecu∥=∥〈0,-7,1〉∥=sqrt(0+49+1)=sqrt50

So,

costheta=(vecu.vecv)/(∥vecu∥*∥vecv∥)=-54/(sqrt84*sqrt50)=-0.83

$\theta = 146.4$º