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

1 Answer
Ratnaker Mehta
Jun 22, 2016

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

Explanation:

If #theta# is the reqd. angle between the given vectors, say #vecx#=#<5,0,-2># and #vecy#=#<-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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