A particle travels on a given straight line with a constant speed v...

Let the straight line path of the particle be represented by the equation $\vec{r} = \vec{r}_0 + \vec{v}t$, where $\vec{r}_0$ is the position vector of a fixed point on the line with respect to the origin $O$, and $\vec{v}$ is the constant velocity vector of the particle.

The position vector of the particle at any time $t$ is $\vec{OP} = \vec{r}$.

The given quantity is $\vec{OP} \times \vec{v}$. Substituting the expression for $\vec{OP}$, we get:

$\vec{OP} \times \vec{v} = (\vec{r}_0 + \vec{v}t) \times \vec{v}$

$\vec{OP} \times \vec{v} = \vec{r}_0 \times \vec{v} + t(\vec{v} \times \vec{v})$

Since the cross product of any vector with itself is a null vector ($\vec{v} \times \vec{v} = \vec{0}$), the equation simplifies to:

$\vec{OP} \times \vec{v} = \vec{r}_0 \times \vec{v}$

Since $\vec{r}_0$ is a fixed position vector and $\vec{v}$ is a constant velocity vector, their cross product $\vec{r}_0 \times \vec{v}$ is a constant vector.

Alternatively, the magnitude of this cross product is $|\vec{OP}| |\vec{v}| \sin \theta = v (OP \sin \theta) = v d$, where $d$ is the constant perpendicular distance from $O$ to the line of motion. The direction is also constant, being perpendicular to the plane containing $O$ and the line of motion.

Thus, the quantity $\vec{OP} \times \vec{v}$ is independent of the position $P$.