If $A = <3 ,-1 ,5 >$ , $B = <5 ,2 ,8 >$ and $C=A-B$ , what is the angle between A and C?

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Answer 1 · 2017-01-24T10:22:35

The angle is $=130.4$ º

Let's start by calculating

$vecC=vecA-vecB$

$vecC=〈3,-1,5〉-〈5,2,8〉=〈-2,-3,-3〉$

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

$vecA.vecC=〈3,-1,5〉.〈-2,-3,-3〉=-6+3-15=-18$

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

The modulus of $vecC$ = $∥〈-2,-3,-3〉∥=sqrt(4+9+9)=sqrt22$

So,

$costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=-18/(sqrt35*sqrt22)=-0.65$

$theta=130.4$ º

If A = <3 ,-1 ,5 >A = <3 ,-1 ,5 >, B = <5 ,2 ,8 >B = <5 ,2 ,8 > and C=A-BC=A-B, what is the angle between A and C?