What is the angle between <5,0,-2 > <5,0,−2> and <-8,3,-6> <−8,3,−6>?

1 Answer
Ratnaker Mehta
Jun 22, 2016

Reqd. Angle =arccos(-28/sqrt3161),=arccos(−28√3161), or pi-arccos(28/sqrt3161).π−arccos(28√3161).

Explanation:

If thetaθ is the reqd. angle between the given vectors, say vecx→x=<5,0,-2><5,0,−2> and vecy→y=<-8,3,-6>,<−8,3,−6>, then, we have by defn. of Dot Product

vecx.vecy=||vecx||||vecy||costheta..........(1)

Now vecx.vecy=<5,0,-2>.<-8,3,-6>=5*(-8)+0*3+(-2)*(-6)=-40+0+12=-28.
||vecx||=sqrt{5^2+0^2+(-2)^2}=sqrt29.
||vecy|\=sqrt{(-8)^2+3^2+(-6)^2}=sqrt109.

Substituting these values in (1), we get, costheta=-28/(sqrt29*sqrt109)=-28/sqrt3161.

Hence, Reqd. Angle =arccos(-28/sqrt3161), or pi-arccos(28/sqrt3161).

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