What is the angle between $<-4,2,-1 >$ and $<-8,0,0>$ ?

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Answer 1 · 2017-03-02T07:32:08

The angle is $=29.2$ 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

$vecA.vecB=〈-4,2,-1〉.〈-8,0,0 〉 = 32$

The modulus of $vecA$ = $∥〈-4,2,-1〉∥=sqrt(16+4+1)=sqrt21$

The modulus of $vecB$ = $∥〈-8,0,0〉∥=sqrt64=8$

So,

$costheta=(vecA.vecB)/(∥vecA∥*∥vecB∥)=32/(sqrt21*8)=0.87$

$theta=29.2$ º

What is the angle between <-4,2,-1 > <-4,2,-1 > and <-8,0,0> <-8,0,0> ?