Find angle theta between u and v to the nearest tenth of a degree?

#vecu=5hati-4hatj#
#vecv=2hati+hatj#
Please show work/explain!

1 Answer
Apr 4, 2018

If two vectors are in the form:

#vecu = u_xhati+u_yhatj# and #vecv = v_xhati+v_yhatj#

then one computes the dot product as follows:

#vecu*vecv = (u_x)(v_x) + (u_y)(v_y)" [1]"#

If two vectors are in the form:

#vecu = |u|anglealpha# and #vecv = |v|anglebeta#

then one computes the dot product as follows:

#vecu*vecv= |vecu||vecv|cos(beta-alpha)#

where #beta-alpha# is the angle between the two vectors; you will often see the angle between the two vectors referred to as #theta#.

#vecu*vecv= |vecu||vecv|cos(theta)" [2]"#

We are given two vectors in the first form:

#vecu=5hati-4hatj# and vecv=2hati+hatj

therefore, we use form [1] to compute the dot product:

#vecu*vecv = (5)(2) + (-4)(1)#

#vecu*vecv = 6" [3]"#

We know how to compute the magnitudes of the two vectors:

#|vecu| = sqrt(u_x^2+u_y^2)#

#|vecv| = sqrt(v_x^2+v_y^2)#

#|vecu| = sqrt(5^2+(-4)^2)#

#|vecv| = sqrt(2^2+1^2)#

#|vecu| = sqrt(41)" [4]"#

#|vecv| = sqrt(5)" [5]"#

Substitute equation [3], [4], and [5] into equation [2]:

#6= sqrt41sqrt5cos(theta)#

Divide both sides by #sqrt41sqrt5#:

#cos(theta) = 6/(sqrt41sqrt5)#

Use the inverse cosine on both sides:

#theta = cos^-1(6/(sqrt41sqrt5))#

You can use a calculator on the above: