Answer:
The angular momentum of the particle about the origin is [tex]\vec l = -19.305\,k\,\left[kg\cdot \frac{m}{s} \right][/tex].
Explanation:
Vectorially speaking, the angular momentum is given by the following cross product:
[tex]\vec l = \vec r \times m\vec v[/tex]
This cross product can be solved with the help of determinants and its properties, that is:
[tex]\vec l = \left|\begin{array}{ccc}i&j&k\\r_{x}&r_{y}&0\\m\cdot v_{x}&m\cdot v_{y}&0\end{array}\right|[/tex]
[tex]\vec l = m\left|\begin{array}{ccc}i&j&k\\r_{x}&r_{y}&0\\v_{x}& v_{y}&0\end{array}\right|[/tex]
The 3 x 3 determinant is solved by the Sarrus Law:
[tex]\vec l = m \cdot (r_{x}\cdot v_{y} - r_{y}\cdot v_{x})k[/tex]
If [tex]m = 1.30\,kg[/tex], [tex]\vec r = 1.50\,i + 2.20\,j\,[m][/tex] and [tex]\vec v = 4.50\,i-3.30\,j\,\left[\frac{m}{s} \right][/tex], the angular momentum of the particle about the origin is:
[tex]\vec l = (1.30\,kg)\cdot \left[\left(1.50\,m\right)\cdot\left(-3.30\,\frac{m}{s} \right)-\left(2.20\,m\right)\cdot\left(4.50\,\frac{m}{s} \right)\right]k[/tex]
[tex]\vec l = -19.305\,k\,\left[kg\cdot \frac{m}{s} \right][/tex]
The angular momentum of the particle about the origin is [tex]\vec l = -19.305\,k\,\left[kg\cdot \frac{m}{s} \right][/tex].
