If $A = <2 ,4 ,-1 >$ , $B = <-1 ,8 ,-5 >$ and $C=A-B$ , what is the angle between A and C?

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Answer 1 · 2017-03-03T17:50:44

The angle is $=118.5$ º

Let's start by calculating

$vecC=vecA-vecB$

$vecC=〈2,4,-1〉-〈-1,8,-5〉=〈3,-4,4〉$

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

$vecA.vecC=〈2,4,-1〉.〈3,-4,4〉=6-16-4=-14$

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

The modulus of $vecC$ = $∥〈3,-4,4〉∥=sqrt(9+16+16)=sqrt41$

So,

$costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=-14/(sqrt21*sqrt41)=-0.48$

$theta=118.5$ º

If A = <2 ,4 ,-1 >A = <2 ,4 ,-1 >, B = <-1 ,8 ,-5 >B = <-1 ,8 ,-5 > and C=A-BC=A-B, what is the angle between A and C?