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

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