How do you find the angle between the vectors #v=3i-2j, w=2i+2j#?

2 Answers
Douglas K.
May 31, 2017

Compute the dot-product:

#vecv*vecw = 3(2)+(-2)(2)#

#vecv*vecw = 2#

Compute the magnitudes:

#|vecv| = sqrt(3^2+ (-2)^2)#

#|vecv| = sqrt13#

#|vecw| = sqrt(2^2+2^2)#

#|vecw| = sqrt8#

Use the formula, #vecv*vecw = |vecv||vecw|cos(theta)# to solve for the angle between the vectors, #theta#:

#2 = sqrt13sqrt8cos(theta)#

#2 = sqrt(104)cos(theta)#

#cos(theta) = 2/sqrt104#

#theta ~~ 78.69^@#

Jim G.
May 31, 2017

#78.69^@#

Explanation:

#"from the definition of the "color(blue)"scalar (dot) product"#

#• ulv.ulw=|ulv||ulw|costheta#

#rArrcostheta=(ulv.ulw)/(|ulv||ulw|)to(color(red)(1))#

#"where " theta" is the angle between " ulv" and " ulw#

#"also if " ulv=((x_1),(y_1))" and " ulw=((x_2),(y_2))#

#rArrulv.ulw=x_1x_2+y_1y_2#

#|ulv|=sqrt(x_1^2+y_1^2),|ulw|=sqrt(x_2^2+y_2^2)#

#"here" ulv=((3),(-2))" and " ulw=((2),(2))#

#rArrulv.ulw=(3xx2)+(-2xx2)=2#

#rArr|ulv|=sqrt(3^2+(-2)^2)=sqrt13#

#rArr|ulw|=sqrt(2^2+2^2)=sqrt8#

#"substitute these results into "(color(red)(1))#

#rArrcostheta=2/(sqrt13xxsqrt8)#

#rArrtheta~~78.69^@" to 2 dec. places"#

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