If a moving particle have linear momentum `bar P` and position vector `bar r` then choose the correct relation between `bar r,bar p` and angular momentum `bar L ` of particle about the origin.

Options

A. ` bar r . bar L =0`
B. ` abs(bar r . bar L) =0`
C. ` bar L. bar P != 0`
D. ` abs(bar L . bar P) =0`

Answer

To solve the problem, we need to establish the relationship between the position vector \(\bar{r}\), linear momentum \(\bar{p}\), and angular momentum \(\bar{L}\) of a particle about the origin.

### Step-by-Step Solution:

1. **Understanding Angular Momentum**:
The angular momentum \(\bar{L}\) of a particle about a point (in this case, the origin) is defined as:
\[ \bar{L} = \bar{r} \times \bar{p} \]
where \(\bar{r}\) is the position vector of the particle and \(\bar{p}\) is the linear momentum of the particle.

2. **Properties of Cross Product**:
The cross product \(\bar{r} \times \bar{p}\) has a special property: it is always perpendicular to both \(\bar{r}\) and \(\bar{p}\). This means that:
\[ \bar{L} \cdot \bar{r} = 0 \]
This indicates that the angular momentum vector is orthogonal to the position vector.

3. **Dot Product with Angular Momentum**:
Since \(\bar{L} = \bar{r} \times \bar{p}\), we can also express the relationship in terms of the dot product:
\[ \bar{r} \cdot \bar{L} = \bar{r} \cdot (\bar{r} \times \bar{p}) = 0 \]
This confirms that the dot product of the position vector \(\bar{r}\) with the angular momentum \(\bar{L}\) is zero.

4. **Conclusion**:
From the above steps, we can conclude that the correct relation between the position vector \(\bar{r}\), linear momentum \(\bar{p}\), and angular momentum \(\bar{L}\) is:
\[ \bar{r} \cdot \bar{L} = 0 \]
Therefore, the correct answer from the options provided is:
\[ \text{Option A: } \bar{r} \cdot \bar{L} = 0 \]