How do you find the angle between the vectors #u=2i-3j# and #v=i-2j#?
1 Answer
So the acute angle between the vectors is
Explanation:
The angle
# 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
#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
# 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