If $A = <3 ,-1 ,-8 >$ , $B = <5 ,6 ,-9 >$ and $C=A-B$ , what is the angle between A and C?

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Answer 1 · 2017-12-24T08:20:04

The angle is $=96.3^@$

Let's start by calculating

$vecC=vecA-vecB$

$vecC=〈3,-1,-8〉-〈5,6,-9〉=〈-2,-7,1〉$

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

$vecA.vecC=〈3,-1,-8〉.〈-2-7,1〉=-6+7-8=-7$

The modulus of $vecA$ = $∥〈3,-1,-8〉∥=sqrt(9+1+64)=sqrt74$

The modulus of $vecC$ = $∥〈-2,-7,1〉∥=sqrt(4+49+1)=sqrt54$

So,

$costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=-7/(sqrt74*sqrt54)=-0.11$

$theta=arccos (-0.11)=96.3^@$

If A = <3 ,-1 ,-8 >A = <3 ,-1 ,-8 >, B = <5 ,6 ,-9 >B = <5 ,6 ,-9 > and C=A-BC=A-B, what is the angle between A and C?