# If A = <6 ,8 ,2 >, B = <7 ,1 ,-4 > and C=A-B, what is the angle between A and C?

Nov 22, 2016

THe angle is $49$º

#### Explanation:

Let's start by calculating $\vec{C}$

vecC=vecA-vecB=〈6,8,2〉-〈7,1,-4〉=〈-1,7,6〉

The angle between 2 vectors is given by the dot product.

vecA.vecC=∥vecA∥*∥vecC∥*costheta

where, $\theta$ is the angle between the two vectors.

The dot product is
=〈-1,7,6〉.〈6,8,2〉=-6+56+12=62

The modulus of $\vec{A}$ is

∥vecA∥=∥〈6,8,2〉∥=sqrt(36+64+4)=sqrt104

The modulus of $\vec{C}$ is

∥vecC∥=∥〈-1,7,6〉∥=sqrt(1+49+36)=sqrt86

Therefore,
costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=62/(sqrt104*sqrt86)=0.66

$\theta = 49$º