What is the angle between $< 1, 5, 9 >$ $<1 , 5 , 9 >$ and $< - 2, 3, 2 >$ $< -2 , 3 , 2 >$ ?

Question

1641 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-23T21:38:44

$θ \approx 0.76 r a d i a n s$ $theta ~~ 0.76 radians$

Let $¯ ¯ ¯ A = < 1, 5, 9 >$ $barA = <1, 5, 9>$
Let $¯ ¯ ¯ B = < - 2, 3, 2 >$ $barB = <-2,3,2>$

The $¯ ¯ ¯ A \cdot ¯ ¯ ¯ B$ $barA*barB$ is:

$¯ ¯ ¯ A \cdot ¯ ¯ ¯ B = (1) (- 2) + (5) (3) + (9) (2) = 31$ $barA*barB = (1)(-2) + (5)(3) + (9)(2) = 31$

$∣ ∣ ¯ ¯ ¯ A ∣ ∣ = \sqrt{1^{2} + 5^{2} + 9^{2}} = \sqrt{107}$ $|barA| = sqrt(1^2 + 5^2 + 9^2) = sqrt(107)$

$∣ ∣ ¯ ¯ ¯ B ∣ ∣ = \sqrt{{(- 2)}^{2} + 3^{2} + 2^{2}} = \sqrt{17}$ $|barB| = sqrt((-2)^2 + 3^2 + 2^2) = sqrt(17)$

Use $¯ ¯ ¯ A \cdot ¯ ¯ ¯ B = ∣ ∣ ¯ ¯ ¯ A ∣ ∣ ∣ ∣ ¯ ¯ ¯ B ∣ ∣ cos (θ)$ $barA*barB = |barA||barB|cos(theta)$

Solve for $θ$ $theta$

$θ = {cos}^{- 1} (\frac{¯ ¯ ¯ A \cdot ¯ ¯ ¯ B}{∣ ∣ ¯ ¯ ¯ A ∣ ∣ ∣ ∣ ¯ ¯ ¯ B ∣ ∣})$ $theta = cos^-1((barA*barB)/(|barA||barB|))$

$θ = {cos}^{- 1} (\frac{31}{\sqrt{107} \sqrt{17}})$ $theta = cos^-1((31)/(sqrt(107)sqrt(17)))$

$θ \approx 0.76 r a d i a n s$ $theta ~~ 0.76 radians$

What is the angle between <1 , 5 , 9 > <1,5,9><1 , 5 , 9 > and < -2 , 3 , 2 > <−2,3,2> < -2 , 3 , 2 > ?