If $A = < 3, - 1, 8 >$ $A = <3 ,-1 ,8 >$ , $B = < 4, - 3, - 6 >$ $B = <4 ,-3 ,-6 >$ and $C = A - B$ $C=A-B$ , what is the angle between A and C?

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Answer 1 · 2017-06-02T07:31:18

The angle is $=28.7º$

Let's start by calculating

$vecC=vecA-vecB$

$vecC=〈3,-1,8〉-〈4,-3,-6〉=〈-1,2,14〉$

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

$vecA.vecC=〈3,-1,8〉.〈-1,2,14〉=-3-2+112=107$

The modulus of $vecA$ = $∥〈3,-1,8〉∥=sqrt(9+1+64)=sqrt74$

The modulus of $vecC$ = $∥〈-1,2,14〉∥=sqrt(1+4+196)=sqrt201$

So,

$costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=107/(sqrt74*sqrt201)=0.88$

$theta=28.7$ º

If A = <3 ,-1 ,8 >A=<3,−1,8>A = <3 ,-1 ,8 >, B = <4 ,-3 ,-6 >B=<4,−3,−6>B = <4 ,-3 ,-6 > and C=A-BC=A−BC=A-B, what is the angle between A and C?