March 18th, 2017, 05:00 PM  #11 
Math Team Joined: Jul 2011 From: Texas Posts: 2,659 Thanks: 1327 
$L = m r\,v \sin{\theta}$ $\theta = \arccos\left(\dfrac{4}{5}\right) \implies \sin{\theta} = \dfrac{3}{5}$ $L = 9 \cdot 4\sqrt{5} \cdot 3\sqrt{5} \cdot \dfrac{3}{5} = 324$ 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 xz plane ... Looking at the graph, the point of view is from the negative side of the yaxis. As $\vec{r}$ rotates CCW toward $\vec{v}$, the direction of $\vec{L}$ is out of the page in the negative direction of the yaxis, hence $\vec{L} = 324 \, kg\, m^2/s$ This is nice thing about using the 3x3 determinant to get the crossproduct ... it provides both magnitude & direction in one calculation. 

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angular, determine, momentum 
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