How do you find the angle between u=<-1,9> and v=<3,12>?

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How do you find the angle between u=<-1,9> and v=<3,12>?

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Answer 1 · 2016-11-18T18:12:49

The angle between, $θ = {cos}^{- 1} (\frac{¯ u \cdot ¯ v}{| ¯ u | | ¯ v |}) \approx 1.5 radians$ $theta = cos^-1((baru*barv)/(|baru||barv|)) ~~ 1.5 " radians"$

Explanation:

Given: $¯ u = < - 1, 9 > and ¯ v = < 3, 12 >$ $baru = <-1, 9> and barv = <3, 12>$

Compute the dot-product:

$¯ u \cdot ¯ v = (- 1) (3) + (9) (12) = 105$ $baru*barv = (-1)(3) + (9)(12) = 105$

Compute the magnitude of $¯ u$ $baru$ :

$| ¯ u | = \sqrt{{(- 1)}^{2} + 9^{2}} = \sqrt{82}$ $|baru| = sqrt((-1)^2 + 9^2) = sqrt(82)$

Compute the magnitude of $¯ v$ $barv$ :

$| ¯ v | = \sqrt{3^{2} + 12^{2}} = \sqrt{153}$ $|barv| = sqrt(3^2 + 12^2) = sqrt(153)$

Another equation for the dot-product is:

$¯ u \cdot ¯ v = | ¯ u | | ¯ v | cos (θ)$ $baru*barv = |baru||barv|cos(theta)$

Solve for $θ$ $theta$ :

$θ = {cos}^{- 1} (\frac{¯ u \cdot ¯ v}{| ¯ u | | ¯ v |})$ $theta = cos^-1((baru*barv)/(|baru||barv|))$

Substitute in the computed values:

$θ = {cos}^{- 1} (\frac{105}{\sqrt{82} \sqrt{153}}) \approx 1.5 radians$ $theta = cos^-1((105)/(sqrt(82)sqrt(153))) ~~ 1.5 " radians"$

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How do you find the angle between u=<-1,9> and v=<3,12>?

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