If A = <6 ,8 ,-3 >, B = <7 ,1 ,-4 > and C=A-B, what is the angle between A and C?

Nov 8, 2016

The angle is 50.9º

Explanation:

Let's start by calculating $\vec{C} = \vec{A} - \vec{B}$
vecC=〈6,8,-3〉-〈7,1,-4〉=〈-1,7,1〉
The angle between $\vec{A}$ and $\vec{C}$ is given by the dot product definition
vecA.vecC=∥vecA∥*∥vecC∥costheta

where $\theta$ is the angle between $\vec{A}$ an $\vec{C}$

:. cos theta=vecA.vecC/(∥vecA∥*∥vecC∥)
The dot product is vecA.vecC=〈6,8,-3〉.〈-1,7,1〉=-6+56-3=47
The modulus of vecA=∥vecA∥=sqrt(36+64+9)=sqrt109
The modulus of vecC=∥vecC∥=sqrt(1+49+1)=sqrt51
So $\cos \theta = \frac{47}{\sqrt{109} \cdot \sqrt{51}} = 0.63$
theta=50.9º