How do you find the angle between the vectors #u=6i+3j# and #v=-4i+4j#?

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:

# 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=6 ulhati+3ulhatj=((6),(3))# and #vec v=-4ulhati+4ulhatj=((-4),(4))#

The modulus is given by;

# |vec u| = |((6),(3))| \ \ \ = sqrt(6^2+3^2)=sqrt(36+9)=sqrt(45)=3sqrt(5) #
# |vec v| = |((-4),(4))| = sqrt(-4^2+4^2)=sqrt(16+16)=sqrt(32)=4sqrt(2) #

And the scaler product is:

# vec u * vec v = ((6),(3)) * ((-4),(4))#
# \ \ \ \ \ \ \ \ \ \ = (6)(-4) + (3)(4)#
# \ \ \ \ \ \ \ \ \ \ = -24+12#
# \ \ \ \ \ \ \ \ \ \ = -12#

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

# -12 = 3sqrt(5) * 4sqrt(2) * cos theta #
# :. cos theta = (-12)/(12sqrt(5) sqrt(2))#
# :. cos theta = -0.316227 ... #
# :. theta = 108.4349 °#
# :. theta = 108 °#(3sf)

So the acute angle between the vectors is #72°#(3sf)