If #A = <6 ,-7 ,5 >#, #B = <1 ,-3 ,-8 ># and #C=A-B#, what is the angle between A and C?

1 Answer
Jul 8, 2017

The angle is #=36º#

Explanation:

Let's start by calculating

#vecC=vecA-vecB#

#vecC=〈6,-7,5〉-〈1,-3,-8〉=〈5,-4,13〉#

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,-7,5〉.〈5,-4,13〉=30+28+65=123#

The modulus of #vecA#= #∥〈6,-7,5〉∥=sqrt(36+49+25)=sqrt110#

The modulus of #vecC#= #∥〈5,-4,13〉∥=sqrt(25+16+169)=sqrt210#

So,

#costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=123/(sqrt110*sqrt210)=0.81#

#theta=36#º