If #A = <6 ,-7 ,8 >#, #B = <9 ,-5 ,-2 ># and #C=A-B#, what is the angle between A and C?

1 Answer
Dec 5, 2016

The angle is #=54.1#º

Explanation:

Let's start by calculating

#vecC=vecA-vecB#

#vecC=〈6,-7,8〉-〈9,-5,-2〉=〈-3,-2,10〉#

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=〈6,-7,8〉.〈-3,-2,10〉=-18+14+80=76#

The modulus of #vecA#= #∥〈6,-7,8〉∥=sqrt(36+49+64)=sqrt149#

The modulus of #vecC#= #∥〈-3,-2,10〉∥=sqrt(9+4+100)=sqrt113#

So,

#costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=76/(sqrt149*sqrt113)=0.59#

#theta=54.1#º