If #A = <6 ,6 ,-3 >#, #B = <-9 ,6 ,-2 ># and #C=A-B#, what is the angle between A and C?

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Answer 1 · 2016-11-26T11:08:58

The angle is #=46.6#º

Let's calculate #vecC#

#vecC=vecA-vecB#

#vec C=〈6,6,-3〉-〈-9,6,-2〉=〈15,0,-1〉#

To calculate the angle #theta# between 2 vectors, we use the dot product definition

#vecA.vecC=∥vecA∥*vecC∥costheta#

The dot product is #=〈6,6,-3〉.〈15,0,-1〉=90+0+3=93#

The modulus of #vecA=∥〈6,6,-3〉∥=sqrt(36+36+9)=sqrt81=9#

The modulus of #vecC=∥〈15,0,-1〉∥=sqrt(225+1)=sqrt226#

So, #costheta=93/(9sqrt226)=0.69#

And #theta=46.6#º