The angle between two vectors is found using the formula
#cos(theta) = (vecu * vecv)/(||vecu|| ||vecv||)#, where #vecu * vecv# is known as the dot product, and #||vecu||# for instance is known as the magnitude of #vecu#.
The dot product is calculated the following way:
#vecu * vecv = u_1 v_1 + u_2 v_2 + ... u_n v_n#
The magnitude of the vector is found by squaring each term of vector #vecu = ‹ u_1, u_2, ... u_n ›#, add them all up, and take the square root of the result.
In our case we have
#vecu = 2hati-3hatj = ‹2, -3› -> ||vecu|| = sqrt((2)^2+(-3)^(2))=sqrt(13)#
#vecv = 4hati +3hatj = ‹4, 3› -> ||vecv|| = sqrt((4)^2+(3)^(2))=sqrt(25) = 5 #
#vecu * vecv = 2(4) + -3(3) = 8 - 9 = -1 #
Using the formula
#cos(theta) = (vecu * vecv)/(||vecu|| ||vecv||) = (-1)/(5*sqrt(13))#
So #theta = arccos((-1)/(5*sqrt(13))) ≈ 93.18°#
