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

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Answer 1 · 2017-04-06T13:51:18

The angle is $=66.6$ º

Let's start by calculating

$vecC=vecA-vecB$

$vecC=〈4,2,5〉-〈-8,2,8〉=〈12,0,-3〉$

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

$vecA.vecC=〈4,2,5〉.〈12,0,-3〉=48+0-15=33$

The modulus of $vecA$ = $∥〈4,2,5〉∥=sqrt(16+4+25)=sqrt45$

The modulus of $vecC$ = $∥〈12,0,-3〉∥=sqrt(144+0+9)=sqrt153$

So,

$costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=33/(sqrt45*sqrt153)=0.4$

$theta=66.6$ º

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