In three dimensional coordinate system (OXYZ), a concave mirror of radius of curvature 40cm is placed at x=80cm with the reflecting surface facing the origin. An object at origin is given a velocity v=(9i+6j+3k)cm/s. Find out the magnitude of velocity of its image.

A/c to mirror equation, 1/u + 1/v = 1/f ......(i)Here, u= -80 & f =R/2= -20On Solving, v= -80/3.Now Differentiation eq.(i) wrt t(time),-(1/u^2)du/dt - (1/v^2)dv/dt =0.Now du/dt = 9...And on solving, dv/dt = 1 cm/s.Now, m= -v/u = h(i)/h(o)....(ii)m= -1/3..Now, Differentiating eq ..(ii) wrt..t-(du/dt×h(i) + u× dh(i)/dt) = (dv/dt×h(o) + v×dh(o)/dt)Now putting du/dt = 9 ; dv/dt = 1 ;h(i) =0; h(o) =0 ; u = 80 & v= -80/3dh(i)/dt = 2cm/sSimilarly ...for Z axis solving eq (ii) and (iii) we get v(z) = 1cm/sHence req vel..= 1i+2j+1k...|v| = (6) ^1/2..ANS