If #A = <8 ,0 ,4 >#, #B = <6 ,7 ,4 ># and #C=A-B#, what is the angle between A and C?

1 Answer
Jun 12, 2017

The angle is #=75.8º#

Explanation:

Let's start by calculating

#vecC=vecA-vecB#

#vecC=〈8,0,4〉-〈6,7,4〉=〈2,-7,0〉#

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=〈8,0,4〉.〈2,-7,0〉=16+0+0=16#

The modulus of #vecA#= #∥〈8,0,4〉∥=sqrt(64+0+16)=sqrt80#

The modulus of #vecC#= #∥〈2,-7,0〉∥=sqrt(4+49+0)=sqrt53#

So,

#costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=16/(sqrt80*sqrt53)=0.25#

#theta=75.8#º