What is the angle between #<-2,1,1 > # and #<1,3,8> #?

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Answer 1 · 2017-01-03T13:01:10

The angle is #=64.7#º

We use the dot product definition to calculate the angle #theta# between 2 vectors

#veca.vecb=∥veca∥*∥vecb∥*cos theta#

#cos theta=(veca.vecb)/(∥veca∥*∥vecb∥)#

We start by calculating the dot product,

#veca.vecb=〈-2,1,1〉*〈1,3,8〉=(-2+3+8)=9#

The modulus of #veca=∥〈-2,1,1〉∥=sqrt(4+1+1)=sqrt6#

The modulus of #vecb=∥〈1,3,8〉∥=sqrt(1+9+64)=sqrt74#

#cos theta=9/(sqrt6*sqrt74)=0.427#

#theta=64.7#º