If $A = <1 ,6 ,9 >$ , $B = <-9 ,4 ,-8 >$ and $C=A-B$ , what is the angle between A and C?

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Answer 1 · 2016-11-15T16:38:30

The angle is $=35.6$ º

Let's start by calculating $vecC$

$vecC=vecA-vecB=〈1,6,9〉-〈-9,4,-8〉=〈10,2,17〉$

The angle $theta$ is given by the dot product definition

$vecA.vecC=∥vecA∥*∥vecC∥*costheta$

$vecA.vecC=〈10,2,17〉.〈1,6,9〉=10+12+153=175$

The modulus of $vecA=∥vecA∥=∥〈1,6,9〉∥=sqrt(1+36+81)=sqrt118$

The modulus of $vecC=∥vecC∥=∥〈10,2,17〉∥=sqrt(100+4+289)=sqrt393$

Therefore, $costheta=175/(sqrt118*sqrt393)=0.81$

$theta=35.6$ º

If A = <1 ,6 ,9 >A = <1 ,6 ,9 >, B = <-9 ,4 ,-8 >B = <-9 ,4 ,-8 > and C=A-BC=A-B, what is the angle between A and C?