Expert: Ahmed Salami - 11/7/2009

Question
I need help with this problem, please:

Suppose points A, B, and C have coordinates A (0, 0), B (1, –2), and C (4, 2).
Let u be the vector with initial point A and terminal point B.
Let v be the vector with initial point B and terminal point C.

Find the angle between u and v.  Round the result to the nearest tenth of a degree.  Thank  you!


Answer
Hi Cindy,
Vector AB, u = (1-0)i + (-2-0)j
            = i - 2j
where i and j are the unit vectors in the horizontal and vertical directions.
Vector BC, v = (4-1)i + (2- -2)j
            = 3i + 4j
The angle between the vectors can be found from the dot product formula
u.v = |u||v|cosθ
where θ is the angle between the vectors
|u| and |v| are the magnitude of the vectors
Now,
u.v = (i - 2j).(3i + 4j)
   = (1)(3) + (-2)(4)
   = 3 - 8
   = -5
|u| = √[(1)² + (-2)²] = √5
|v| = √[(3)² + (4)²] = 5
So,
cosθ = u.v/|u||v|
    = (-5)/5√5 = -√5/5
θ = 116.6°
OR
θ = 180 - 116.6
 = 63.4°

Regards