If A = <4 ,2 ,-5 >, B = <6 ,5 ,9 > and C=A-B, what is the angle between A and C?

1 Answer
May 22, 2017

The angle is =54.7º

Explanation:

Let's start by calculating

vecC=vecA-vecB

vecC=〈4,2,-5〉-〈6,5,9〉=〈-2,-3,-14〉

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=〈4,2,-5〉.〈-2,-3,-14〉=-8-6+70=56

The modulus of vecA= ∥〈4,2,-5〉∥=sqrt(16+4+25)=sqrt45

The modulus of vecC= ∥〈-2,-3,-14〉∥=sqrt(4+9+196)=sqrt209

So,

costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=56/(sqrt45*sqrt209)=0.58

theta=54.7º