Question

A particle has an angular momentum $\overrightarrow{L}$ and a linear momentum $\overrightarrow{P}$ about a reference point $O$. If $\overrightarrow{r}$ is the position vector of particle with respect to reference point, then identify correct option.

• A. $\overrightarrow{L}=\overrightarrow{P}\times \overrightarrow{r}$
• B. $|\overrightarrow{L}\times \overrightarrow{P}|=LP$
• C. $\overrightarrow{L}\times \overrightarrow{P}=0$
• D. None of these

Answer 1

The correct option is B $|\overrightarrow{L}\times \overrightarrow{P}|=LP$

We know,
$\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{P}$
$\therefore \overrightarrow{L}$ being the cross -product vector, It will be $\perp$ to the plane containing both vectors $\overrightarrow{P}$ and $\overrightarrow{r}$
$\Rightarrow$ Angle between vectors $\overrightarrow{L}$ and $\overrightarrow{P}$ will be ${90}^{\circ }$.
$\Rightarrow |\overrightarrow{L}\times \overrightarrow{P}|=LP\sin {90}^{\circ }=LP$
However, $\overrightarrow{L}.\overrightarrow{P}=LP\cos {90}^{\circ }=0$
$\therefore$ Option (b) is correct.