If #A = <2 ,-5 ,-4 >#, #B = <-9 ,1 ,-6 ># and #C=A-B#, what is the angle between A and C?

1 Answer
Jan 26, 2017

The angle is #58.9#º

Explanation:

Let's start by calculating

#vecC=vecA-vecB#

#vecC=〈2,-5,-4〉-〈-9,1,-6〉=〈11,-6,2〉#

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=〈2,-5,-4〉.〈11,-6,2〉=22+30-8=44#

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

The modulus of #vecC#= #∥〈11,-6,2〉∥=sqrt(121+36+4)=sqrt161#

So,

#costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=44/(sqrt45*sqrt161)=0.52#

#theta=58.9#º