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March 18th, 2017, 04:00 PM   #11
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$L = m |r|\,|v| \sin{\theta}$

$\theta = \arccos\left(-\dfrac{4}{5}\right) \implies \sin{\theta} = \dfrac{3}{5}$

$|L| = 9 \cdot 4\sqrt{5} \cdot 3\sqrt{5} \cdot \dfrac{3}{5} = 324$

The above calculation gives the magnitude of the angular momentum. To understand the direction for $\vec{L}$, reference the attached graph of the position and velocity vectors on the x-z plane ...

Looking at the graph, the point of view is from the negative side of the y-axis. As $\vec{r}$ rotates CCW toward $\vec{v}$, the direction of $\vec{L}$ is out of the page in the negative direction of the y-axis, hence $\vec{L} = -324 \, kg\, m^2/s$

This is nice thing about using the 3x3 determinant to get the cross-product ... it provides both magnitude & direction in one calculation.

Thanks from topsquark and The_Ys_Guy
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