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

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Answer 1 · 2016-11-11T06:23:15

The angle is $=77.6º$

Let'scalculate $vecC=vecA-vecB=〈1,6,-8〉-〈-9,4,-8〉=〈10,2,0〉$

The angle is given by the dot product definition
$vecA.vecC=∥vecA∥*∥vecC∥costheta$
where $theta$ is the angle between the two vector.

$vecA.vecC=〈1,6,-8〉.〈10,2,0〉=10+12-0=22$

The modulus of $vecA=∥〈1,6,-8〉∥=sqrt(1+36+64)=sqrt101$

The modulus of $vecC=∥〈10,2,0〉∥=sqrt(100+4)=sqrt104$

So, $costheta=22/(sqrt101*sqrt104)=0.214$

$theta=77.6º$

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