A particle of mass 1 kg is subjected to a force which depends on the position as $F = -k(x i + y j)$ kg ms$^{-2}$ with $k = 1$ kg $s^{-2}.$ At time $t = 0,$ the particle's position $r = (1/√2 i + √2 j) m$ and its velocity $v = (-√2 i + √2 j + 2/π k)$ ms$^{-1}.$ Let $v_{x}$ and $v_{y}$ denote the x and the y components of the particle's velocity, respectively. Ignore gravity. When $z = 0.5 m,$ the value of $(x v_{y} - y v_{x})$ is ______ $m^{2} s^{-1}.$

Step 1: Analyze the force and torque.

The force acting on the particle is directed towards the $z$-axis.

$$\vec{F} = -k(x\hat{i} + y\hat{j})$$

The torque $\vec{\tau}$ about the origin is the cross product of the position vector and the force vector.

$$\vec{\tau} = \vec{r} \times \vec{F}$$ $$\vec{\tau} = (x\hat{i} + y\hat{j} + z\hat{k}) \times (-k(x\hat{i} + y\hat{j}))$$

Step 2: Determine the conservation of angular momentum.

Calculate the $z$-component of the torque.

$$\tau_z = x F_y - y F_x$$ $$\tau_z = x(-ky) - y(-kx) = 0$$

Since $\tau_z = 0$, the $z$-component of the angular momentum $L_z$ is conserved.

$$L_z = m(xv_y - yv_x) = \text{constant}$$

Given $m = 1\,\text{kg}$, the quantity $(xv_y - yv_x)$ is constant throughout the motion.

$$xv_y - yv_x = \text{constant}$$

Step 3: Calculate the value using initial conditions.

Extract the initial position and velocity components at $t = 0$.

$$x_0 = \frac{1}{\sqrt{2}}\,\text{m}, \quad y_0 = \sqrt{2}\,\text{m}$$ $$v_{x0} = -\sqrt{2}\,\text{ms}^{-1}, \quad v_{y0} = \sqrt{2}\,\text{ms}^{-1}$$

Evaluate the constant value of the angular momentum.

$$xv_y - yv_x = x_0 v_{y0} - y_0 v_{x0}$$ $$xv_y - yv_x = \left(\frac{1}{\sqrt{2}}\right)(\sqrt{2}) - (\sqrt{2})(-\sqrt{2})$$ $$xv_y - yv_x = 1 + 2 = 3$$

Since this quantity is conserved, it remains $3\,\text{m}^2\text{s}^{-1}$ at $z = 0.5\,\text{m}$.

Answer: 3