What is the angle between $<8,7,6>$ and $<-1,6,1>$ ?

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Answer 1 · 2017-01-18T09:33:28

The angle is $57.9$ º

The angle between $vecA$ and $vecB$ is given by the dot product definition.

$vecA.vecB=∥vecA∥*∥vecB∥costheta$

Where $theta$ is the angle between $vecA$ and $vecB$

The dot product is

$vecA.vecB=〈8,7,6〉.〈-1,6,1〉=-8+42+6=40$

The modulus of $vecA$ = $∥〈8,7,6〉∥=sqrt(64+49+36)=sqrt149$

The modulus of $vecB$ = $∥〈-1,6,1〉∥=sqrt(1+36+1)=sqrt38$

So,

$costheta=(vecA.vecB)/(∥vecA∥*∥vecB∥)=40/(sqrt38*sqrt149)=0.53$

$theta=57.9$ º

What is the angle between <8,7,6> <8,7,6> and <-1,6,1><-1,6,1>?