A particle moves so that its position vector is given by `vec r = cos omega t hat x + sin omega t hat y`, where `omega` is a constant which of the following is true ?

Options

A. `vecv` and `veca` both are perpendiculart to `vecr`
B. `vecv ` and `veca` both are parallel to `vecr`
C. `vecv` is perpendicular to `vecr` and `veca` is directed aways form the origin.
D. velocity is perpendicular to`vecr` and `acceleration is directed towards origin

Answer

To solve the problem, we need to analyze the position vector of the particle given by:

\[ \vec{r} = \cos(\omega t) \hat{x} + \sin(\omega t) \hat{y} \]

### Step 1: Find the Velocity Vector
The velocity vector \(\vec{v}\) is the time derivative of the position vector \(\vec{r}\).

\[ \vec{v} = \frac{d\vec{r}}{dt} \]

Differentiating each component:

\[ \vec{v} = \frac{d}{dt}(\cos(\omega t)) \hat{x} + \frac{d}{dt}(\sin(\omega t)) \hat{y} \]

Using the chain rule:

\[ \vec{v} = -\sin(\omega t) \cdot \omega \hat{x} + \cos(\omega t) \cdot \omega \hat{y} \]

Thus, we can factor out \(\omega\):

\[ \vec{v} = \omega(-\sin(\omega t) \hat{x} + \cos(\omega t) \hat{y}) \]

### Step 2: Find the Acceleration Vector
The acceleration vector \(\vec{a}\) is the time derivative of the velocity vector \(\vec{v}\).

\[ \vec{a} = \frac{d\vec{v}}{dt} \]

Differentiating each component:

\[ \vec{a} = \frac{d}{dt}(-\sin(\omega t) \cdot \omega) \hat{x} + \frac{d}{dt}(\cos(\omega t) \cdot \omega) \hat{y} \]

Using the chain rule again:

\[ \vec{a} = -\omega \cos(\omega t) \cdot \omega \hat{x} - \sin(\omega t) \cdot \omega \cdot \omega \hat{y} \]

Thus, we can write:

\[ \vec{a} = -\omega^2(\cos(\omega t) \hat{x} + \sin(\omega t) \hat{y}) \]

Notice that this can be expressed in terms of the position vector \(\vec{r}\):

\[ \vec{a} = -\omega^2 \vec{r} \]

### Step 3: Check Perpendicularity
To check if \(\vec{v}\) and \(\vec{r}\) are perpendicular, we calculate the dot product \(\vec{r} \cdot \vec{v}\):

\[ \vec{r} \cdot \vec{v} = (\cos(\omega t) \hat{x} + \sin(\omega t) \hat{y}) \cdot \left(-\omega \sin(\omega t) \hat{x} + \omega \cos(\omega t) \hat{y}\right) \]

Calculating the dot product:

\[ = \cos(\omega t)(-\omega \sin(\omega t)) + \sin(\omega t)(\omega \cos(\omega t)) \]

This simplifies to:

\[ = -\omega \cos(\omega t) \sin(\omega t) + \omega \sin(\omega t) \cos(\omega t) = 0 \]

Since the dot product is zero, \(\vec{r}\) and \(\vec{v}\) are perpendicular.

### Step 4: Conclusion
From our calculations:
- The position vector \(\vec{r}\) is perpendicular to the velocity vector \(\vec{v}\).
- The acceleration vector \(\vec{a}\) is directed towards the origin and is anti-parallel to the position vector \(\vec{r}\).

Thus, the correct statement is that the velocity is perpendicular to the position vector, and the acceleration is directed towards the origin.

### Final Answer
The correct option is that the velocity vector is perpendicular to the position vector, and the acceleration vector is directed towards the origin.

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