Let `vecF` be the force acting on a particle having position vector `vecr and vecT` be the torque of this force about the origin. Then

Options

A. `vecr.vecGamma=0 and vecF.vecGamma=0`
B. `vecr.vecGamma=0 but vecF.vecGamma!=0`
C. `vecr.vecGamma !=0but vecF.vecGamma=0`
D. `vecr.vecGamma!=0 and vecF.vecGamma!=0`

Answer

To solve the problem, we need to analyze the torque \(\vec{T}\) of a force \(\vec{F}\) acting on a particle with position vector \(\vec{r}\) about the origin. We will check the validity of the statements given in the question regarding the dot products of these vectors.

### Step 1: Understanding Torque
The torque \(\vec{T}\) about the origin due to a force \(\vec{F}\) acting on a particle located at position \(\vec{r}\) is given by the cross product:
\[ \vec{T} = \vec{r} \times \vec{F} \]

### Step 2: Analyzing the Dot Product \( \vec{r} \cdot \vec{T} \)
We need to evaluate the dot product \( \vec{r} \cdot \vec{T} \):
\[ \vec{r} \cdot \vec{T} = \vec{r} \cdot (\vec{r} \times \vec{F}) \]
Using the property of the dot product, we know that the dot product of a vector with a vector that is perpendicular to it is zero. Since \(\vec{T}\) (torque) is perpendicular to both \(\vec{r}\) and \(\vec{F}\):
\[ \vec{r} \cdot \vec{T} = 0 \]

### Step 3: Analyzing the Dot Product \( \vec{F} \cdot \vec{T} \)
Next, we evaluate the dot product \( \vec{F} \cdot \vec{T} \):
\[ \vec{F} \cdot \vec{T} = \vec{F} \cdot (\vec{r} \times \vec{F}) \]
Again, using the property of the dot product, since \(\vec{T}\) is perpendicular to \(\vec{F}\):
\[ \vec{F} \cdot \vec{T} = 0 \]

### Conclusion
Both dot products \( \vec{r} \cdot \vec{T} = 0 \) and \( \vec{F} \cdot \vec{T} = 0 \) hold true. Therefore, the statements in the question are correct.

### Final Answer
The correct option is that both \( \vec{r} \cdot \vec{T} = 0 \) and \( \vec{F} \cdot \vec{T} = 0 \).

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