If $A = <4 ,7 ,2 >$ , $B = <1 ,1 ,-4 >$ and $C=A-B$ , what is the angle between A and C?

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Answer 1 · 2017-03-08T18:09:45

The angle is $=28º$

Let's start by calculating

$vecC=vecA-vecB$

$vecC=〈4,7,2〉-〈1,1,-4〉=〈3,6,6〉$

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

$vecA.vecC=〈4,7,2〉.〈3,6,6〉=12+42+12=66$

The modulus of $vecA$ = $∥〈4,7,2〉∥=sqrt(16+49+4)=sqrt69$

The modulus of $vecC$ = $∥〈3,6,6〉∥=sqrt(9+36+36)=sqrt81=9$

So,

$costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=66/(sqrt69*9)=0.88$

$theta=28$ º

If A = <4 ,7 ,2 >A = <4 ,7 ,2 >, B = <1 ,1 ,-4 >B = <1 ,1 ,-4 > and C=A-BC=A-B, what is the angle between A and C?