Question
The angular momentum $\vec{L}=\vec{r} \times \vec{p}$, where $\vec{r}$ is a position vector and $\vec{p}$ is linear momentum of a body.
If $\vec{r}=4 \vec{i} \times 6 \vec{j}-3 \hat{k}$ and $\vec{p}=2 \vec{i} \times 4 \vec{j}-5 \hat{k}$, find $\vec{L}$

Answer

$\vec{A} \times \vec{B}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z\end{array}\right|$
Using determinant to find cross-product
$
\overrightarrow{ L }=\overrightarrow{ r } \times \overrightarrow{ p }=\left|\begin{array}{ccc}
\hat{ i } & \hat{ j } & \hat{ k } \\
4 & 6 & -3 \\
2 & 4 & -5
\end{array}\right|  
$
$\begin{aligned} \therefore \quad \overrightarrow{ L } & =[(6 \times(-5)-(-3) \times 4] \hat{ i } \\ & +[(-3) \times 2-4 \times(-5)] \hat{ j }+[4 \times 4-6 \times 2] \hat{ k } \\ & =(-30+12) \hat{ i }+(-6+20) \hat{ j }+(16-12) \hat{ k } \\ & =- 1 8 \hat{ i }+ 1 4 \hat{ j }+ 4 \hat{ k }\end{aligned}$
$L$ is $- 1 8 \hat{ i }+ 1 4 \hat{ j }+ 4 \hat{ k }$

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