If #A = <6 ,4 ,9 >#, #B = <7 ,-1 ,5 ># and #C=A-B#, what is the angle between A and C?

1 Answer
May 17, 2018

The angle is #=42.9^@#

Explanation:

Start by calculating

#vecC=vecA-vecB#

#vecC=〈6,4,9〉-〈7,-1,5〉=〈-1,5,4〉#

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=〈6,4,9〉.〈-1,5,4〉=-6+20+36=50#

The modulus of #vecA#= #∥〈6,4,9〉∥=sqrt(36+16+81)=sqrt133#

The modulus of #vecC#= #∥〈-1,5,4〉∥=sqrt(1+25+9)=sqrt35#

So,

#costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=50/(sqrt133*sqrt35)=0.7328#

#theta=arccos(0.7328)=42.9^@#