If A = <8 ,1 ,-5 >, B = <6 ,-2 ,4 >, and C=A-B, what is the angle between A and C?

1 Answer
Jan 5, 2017

The answer is =45.9º

Explanation:

Let's start by calculating

vecC=vecA-vecB

vecC=〈8,1,-5〉-〈6,-2,4〉=〈2,3,-9〉

The angle between vecA and vecC is given by the dot product definition.

vecA.vecC=∥vecA∥*∥vecC∥costheta

Where theta is the angle between vecA and vecC

The dot product is

vecA.vecC=〈8,1,-5〉.〈2,3,-9〉=16+3+45=64

The modulus of vecA= ∥〈8,1,-5〉∥=sqrt(64+1+25)=sqrt90

The modulus of vecC= ∥〈2,3,-9〉∥=sqrt(4+9+81)=sqrt94

So,

costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=64/(sqrt90*sqrt94)=0.696

theta=45.9º