If $A = <2 ,-5 ,-4 >$ , $B = <-9 ,1 ,7 >$ and $C=A-B$ , what is the angle between A and C?

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Answer 1 · 2017-03-09T13:39:09

The angle is $=30.9$ º

Let's start by calculating

$vecC=vecA-vecB$

$vecC=〈2,-5,-4〉-〈-9,1,7〉=〈11,-6,-11〉$

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

$vecA.vecC=〈2,-5,-4〉.〈11,-6,-11〉=22+30+44=96$

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

The modulus of $vecC$ = $∥〈11,-6,-11〉∥=sqrt(121+36+121)=sqrt278$

So,

$costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=96/(sqrt45*sqrt278)=0.86$

$theta=30.9$ º

If A = <2 ,-5 ,-4 >A = <2 ,-5 ,-4 >, B = <-9 ,1 ,7 >B = <-9 ,1 ,7 > and C=A-BC=A-B, what is the angle between A and C?