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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