Let M and N be foots of the perpendiculars drawn from the point P (a, a, a) on the lines x- y =0, z =1 and x+ y =0, z =-1 respectively and if MPN =90^ then a^2=

Step-by-step solution

Given point : P (a, a, a) . For line L₁ with equations x - y = 0 , z = 1 , a general point is M (t, t, 1) . The direction vector is d₁ = (1, 1, 0) . The condition PM d₁ , where PM = (t - a, t - a, 1 - a) , gives: 2(t - a) = 0 t = a Thus, M = (a, a, 1) . For line L₂ with x + y = 0 , z = -1 , a general point is N (s, -s, -1) . The direction vector is d₂ = (1, -1, 0) . The condition PN d₂ , where PN = (s - a, -s - a, -1 - a) , gives: 2s = 0 s = 0 Thus, N = (0, 0, -1) . Given MPN = 90^ , the vectors PM = (0, 0, 1 - a)