# 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