Determine the acute angle between two intersecting lines

[justinvh]

Homework Statement

A basketball gymnasium is 37 meters high, 80 meters wide and 200 meters long. For a half time stunt, the cheerleaders want to run two strings, one from each of the two corners above one basket to the diagonally opposite corners of the gym floor. What is the cosine of the acute angle made by the strings as they cross? Round your answer to 3 decimal places.

Homework Equations

I believe that you just have to equate:
$\left\|\vec{v}\right\| \left\|\vec{w}\right\|\cos{\theta} = \vec{v} \cdot \vec{w}$

I thought that the sign of the dot product determines whether an angle is obtuse or acute.

The Attempt at a Solution

Vectors:
$\vec{v} = \left\{200\hat{i}, 80\hat{j}, -37\hat{k}\right\}$
$\vec{w} = \left\{200\hat{i}, -80\hat{j}, -37\hat{k}\right\}$

Magnitudes:
$\left\|\vec{v}\right\| = \sqrt{200^2 + 80^2 + (-37)^2} = \sqrt{47769}$
$\left\|\vec{w}\right\| = \sqrt{200^2 + (-80)^2 + (-37)^2} = \sqrt{47769}$
$\left\|\vec{v}\right\| \left|\vec{w}\right\| = 47769$

Dot Product:
$\vec{v} \cdot \vec{w} = 200*200 + (-80)*80 + (-37)*(-37) = 34969$

Equating the sides:
$\cos{\theta} = \displaystyle\frac{\vec{v} \cdot \vec{w}}{\left\|\vec{v}\right\| \left\|\vec{w}\right\|}$

$\cos{\theta} = \displaystyle\frac{34969}{47769}$

After solving for theta by taking the arccos, I have 0.749. With that all said and done, it is not the correct answer. I am not entirely sure where I went wrong. I checked my math and it seems to point to being the correct answer.

 

[tiny-tim]

welcome to pf!

hi justinvh! welcome to pf!

(have a theta: θ )

… What is the cosine of the acute angle made by the strings as they cross? Round your answer to 3 decimal places.
…
After solving for theta by taking the arccos, I have 0.749.

erm …

you're asked for cosθ, not θ ​

 

[justinvh]

you're asked for cosθ, not θ ​

Oh my dear lord. I can't believe I missed that. I've been driving myself crazy for the last day and a half. Thank you for pointing that out.