If $A = <4 ,2 ,-5 >$ , $B = <6 ,5 ,9 >$ and $C=A-B$ , what is the angle between A and C?

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Answer 1 · 2017-05-22T13:12:10

The angle is $=54.7$ º

Let's start by calculating

$vecC=vecA-vecB$

$vecC=〈4,2,-5〉-〈6,5,9〉=〈-2,-3,-14〉$

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

$vecA.vecC=〈4,2,-5〉.〈-2,-3,-14〉=-8-6+70=56$

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

The modulus of $vecC$ = $∥〈-2,-3,-14〉∥=sqrt(4+9+196)=sqrt209$

So,

$costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=56/(sqrt45*sqrt209)=0.58$

$theta=54.7$ º

If A = <4 ,2 ,-5 >A = <4 ,2 ,-5 >, B = <6 ,5 ,9 >B = <6 ,5 ,9 > and C=A-BC=A-B, what is the angle between A and C?