If $A = <7 ,-5 ,-2 >$ , $B = <4 ,-8 ,-3 >$ and $C=A-B$ , what is the angle between A and C?

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Answer 1 · 2017-01-10T06:33:29

The angle is $=84.04$ º

Let's start by calculating

$vecC=vecA-vecB$

$vecC=〈7,-5,-2〉-〈4,-8,-3〉=〈3,3,1〉$

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

$vecA.vecC=〈7,-5,-2〉.〈3,3,1〉=21-15-2=4$

The modulus of $vecA$ = $∥〈7,-5,-2〉∥=sqrt(49+25+4)=sqrt78$

The modulus of $vecC$ = $∥〈3,3,1〉∥=sqrt(9+9+1)=sqrt19$

So,

$costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=4/(sqrt78*sqrt19)=0.104$

$theta=84.04$ º

If A = <7 ,-5 ,-2 >A = <7 ,-5 ,-2 >, B = <4 ,-8 ,-3 >B = <4 ,-8 ,-3 > and C=A-BC=A-B, what is the angle between A and C?