How do you find the angle between the vectors #u=2i-j# and #v=i-2j#?

2 Answers
Nov 2, 2016

Please see the explanation for a description of the process.

Explanation:

Compute the dot-product by multiplying the #hati# coefficients and then adding the product of the #hatj# coefficients:

#baru*barv = (2)(1) + (-2)(-1) = 4#

A second way to compute the dot-product uses the magnitude of the two vectors and the cosine of the angle between the two vectors:

#baru*barv = |baru||barv|cos(theta)# [1]

We know that the dot-product is 4, therefore, we need to compute the magnitude of the two vectors and then solve for the angle between, #(theta)#;

#|baru| = sqrt(2^2 + (-1)^2) = sqrt(5)#

#|barv| = sqrt(1^2 + (-2)^2) = sqrt(5)#

Substitute into equation [1] and then solve for #theta#:

#4 = sqrt(5)sqrt(5)cos(theta)#

#4 = 5cos(theta)#

#cos(theta) = 4/5#

#theta = cos^-1(4/5)#

#theta ~~ 37°#

Nov 2, 2016

The angle is #=36.9º#

Explanation:

The angle between two vectors is given by the dot product definition.
#vecu.vecv=∥vecu∥*∥vecv∥costheta#
where #theta# is the angle between the 2 vectors

so, #costheta= (vecu.vecv) / (∥vecu∥*∥vecv∥)#
The dot product is #=〈2,-1〉〈1,-2〉=2+2=4#
The modulus of #vecu=∥vecu∥=sqrt(4+1)=sqrt5#
The modulus of #vecv=∥vecv∥=sqrt(1+4)=sqrt5#

So, #costheta=4/(sqrt5*sqrt5)=4/5#
#theta=arccostheta=36.9º#