If $A = <4 ,1 ,8 >$ , $B = <5 ,7 ,3 >$ and $C=A-B$ , what is the angle between A and C?

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Answer 1 · 2018-03-31T13:05:10

The angle is $=65.0^@$

Start by calculating

$vecC=vecA-vecB$

$vecC=〈4,1,8〉-〈5,7,3〉=〈-1,-6,5〉$

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

$vecA.vecC=〈4,1,8〉.〈-1,-6,5〉=-4-6+40=30$

The modulus of $vecA$ = $∥〈4,1,8〉∥=sqrt(16+1+64)=sqrt81=9$

The modulus of $vecC$ = $∥〈-1,-6,5〉∥=sqrt(1+36+25)=sqrt62$

So,

$costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=30/(9*sqrt62)=0.42$

$theta=arccos(0.42)=65.0^@$

If A = <4 ,1 ,8 >A = <4 ,1 ,8 >, B = <5 ,7 ,3 >B = <5 ,7 ,3 > and C=A-BC=A-B, what is the angle between A and C?