If A = <6 ,4 ,-7 >, B = <3 ,5 ,2 > and C=A-B, what is the angle between A and C?

1 Answer
Dec 2, 2016

The angle is $= 36.6$º

Explanation:

Let's start by calculating

$\vec{C} = \vec{A} - \vec{B}$

=〈6,4,-7〉-〈3,5,2〉=〈3,-1,-9〉

To calculate the angle $\theta$ between $\vec{A}$ and $\vec{C}$,

we use the dot product definition

vecA.vecC=∥vecA∥*∥vecC∥costheta

The dot product is

vecA.vecC=〈6,4,-7〉.〈3,-1,-9〉=(18-4+63)=77

The modulus of
vecA=∥〈6,4,-7〉∥=sqrt(36+16+49)=sqrt101

The modulus of
vecC=∥〈3,-1,-9〉∥=sqrt(9+1+81)=sqrt91

costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=77/(sqrt101*sqrt91)=0.8

$\theta = 36.6$º