What is the angle between $<-5,-3,6 >$ and $<3,9,-9>$ ?

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Answer 1 · 2018-03-19T09:07:53

The angle is $=151.3^@$

The angle between $2$ vectors, $vecA$ 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

$vecA.vecB=〈-5,-3,6〉.〈3,9,-9〉=-15-27-54=-96$

The modulus of $vecA$ = $∥〈-5,-3,6〉∥=sqrt(25+9+36)=sqrt70$

The modulus of $vecB$ = $∥〈3,9,-9〉∥=sqrt(9+81+81)=sqrt171$

So,

$costheta=(vecA.vecB)/(∥vecA∥*∥vecB∥)=-96/(sqrt70*sqrt171)=-0.877$

$theta=arccos(-0.877)=151.3^@$

What is the angle between <-5,-3,6 > <-5,-3,6 > and <3,9,-9> <3,9,-9> ?