If $A = <-5 ,2 ,1 >$ , $B = <-8 ,7 ,2 >$ and $C=A-B$ , what is the angle between A and C?

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Answer 1 · 2017-05-12T12:12:13

The angle is $=143.36$ º

Let's start by calculating

$vecC=vecA-vecB$

$vecC=〈-5,2,1〉-〈-8,7,2〉=〈3,-5,-1〉$

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

$vecA.vecC=〈-5,2,1〉.〈3,-5,-1〉=-15-10-1=-26$

The modulus of $vecA$ = $∥〈-5,2,1〉∥=sqrt(25+4+1)=sqrt30$

The modulus of $vecC$ = $∥〈3,-5,1〉∥=sqrt(9+25+1)=sqrt35$

So,

$costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=-26/(sqrt30*sqrt35)=-0.80$

$theta=143.36$ º

If A = <-5 ,2 ,1 >A = <-5 ,2 ,1 >, B = <-8 ,7 ,2 >B = <-8 ,7 ,2 > and C=A-BC=A-B, what is the angle between A and C?