What is the angle between #<8,-1,2> # and #<-4,-1,3>#?

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Answer 1 · 2017-01-16T05:33:22

The angle is #=53.8#º

The angle is given by the dot product definition

#veca.vecb=∥veca∥*∥vecb∥costheta#

Where, #theta# is the angle between the 2 vectors

Here, we have

#veca=〈8,-1,2〉#

#vecb=〈-4,-1,3〉#

The dot product is #〈8,-1,2〉.〈-4,-1,3〉=-32+1+6=25#

The modulus of #veca# is #=∥〈8,-1,2〉∥=sqrt(64+1+4)=sqrt69#

The modulus of #vecb# is #=∥〈-4,-1,3〉∥=sqrt(16+1+9)=sqrt26#

Therefore,

#costheta=(veca.vecb)/(∥veca∥*∥vecb∥)=25/(sqrt69*sqrt26)=0.59#

#theta=53.8#º