How do you find the angle between the vectors #u=2i-3j# and #v=i-2j#?

1 Answer
Apr 14, 2017

So the acute angle between the vectors is #7.122°#(3sf)

Explanation:

The angle #theta# between two vectors #vec A# and #vec B# is related to the modulus (or magnitude) and scaler (or dot) product of #vec A# and #vec A# by the relationship:

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# vec A * vec B = |A| |B| cos theta #

By convention when we refer to the angle between vectors we choose the acute angle.

So for this problem, let the angle betwen #vecu# and #vecv# be #theta# then:

#vec u=2hati-3hatj# and #vec v=hati-2hatj#

The modulus is given by;

# ||vec u|| = sqrt((2)^2+(-3)^2)=sqrt(4+9)=sqrt(13) #
# ||vec v|| = sqrt((1)^2+ (-2)^2) =sqrt(1+4) =sqrt(5) #

And the scaler product is:

# vec u * vec v = (2)(1) + (-3)(-2)#
# \ \ \ \ \ \ \ \ \ \ = 2+6#
# \ \ \ \ \ \ \ \ \ \ = 8#

And so using # vec A * vec B = ||A|| * ||B|| * cos theta # we have:

# 8 = sqt(13)sqrt(5) cos theta #
# :. cos theta = 8/(sqrt(13)sqrt(5))#
# :. cos theta = 0.992277 ... #
# :. theta = 7.1250 ... °#
# :. theta = 7.13 °#(3sf)

So the acute angle between the vectors is #7.13°#(3sf)