How do you find the angel between u=27i+14j, v=i-7j?
1 Answer
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=27 ulhati+14 ulhatj=((27),(14))# and#vec v=1 ulhati -7 ulhatj=((1),(-7))#
The modulus is given by;
# |vec u| = |((27),(14))| \ \ = sqrt(27^2+14^2)=sqrt(729+196)=sqrt(925) #
# |vec v| = |((1),(-7))| = sqrt(1^2+ -7^2) =sqrt(1+49) =sqrt(50) #
And the scaler product is:
# vec u * vec v = ((27),(14)) * ((1),(-7))#
# \ \ \ \ \ \ \ \ \ \ = (27)(1) + (14)(-7)#
# \ \ \ \ \ \ \ \ \ \ = 27-98#
# \ \ \ \ \ \ \ \ \ \ = -71#
And so using
# -71 = sqrt(925) * sqrt(50) * cos theta #
# :. cos theta = (-71)/(sqrt(46250))#
# :. cos theta = -0.33014 ... #
# :. theta = 109.277 °#
# :. theta = 109 °# (3sf)
So the acute angle between the vectors is