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 #=54.7#º

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=54.7#º