A particle of mass 1 kg is subjected to a force which depends on the position as F=−k(xî+yĵ) kg m s⁻² with k=1 kg s⁻². At time t=0, the particle’s position r=(1/√2 î+√2 ĵ) m and its velocity v=(−√2 î+√2 ĵ+(2/π)k^) m s⁻¹. Let vₓ and vᵧ denote the x and y components of the particle’s velocity, respectively. Ignore gravity. When z=0.5 m, the value of (xvᵧ−yvₓ) is _____ m² s⁻¹.