In ABC , if the midpoints of the sides AB , BC and CA are respectively (l, 0,0),(0, ~m , 0) and (0,0, n ) , then AB ^2+ BC ^2+ CA ^2 l^2+ m ^2+ n ^2 =

Step-by-step solution

(l, 0,0) is mid point of AB. x₁+x₂=2 l ; y₁+y₂=0 ; z₁+z₂=0 ...(i) (0, m, 0) is mid point of BC . ( x₂+x₃ 2 , y₂+y₃ 2 , z₂+z₃ 2 )=(0, ~m , 0) x₂+x₃=0, y₂+y₃=2 m, z₂+z₃=0 ...(ii) Since (0,0, n) is mid point of AC. ( x₁+x₃ 2 , y₁+y₃ 2 , z₁+z₃ 2 )=(0,0, n) x₁+x₃=0, y₁+y₃=0, z₁+z₃=2 n ...(iii) Solving eq ^ n (i), (ii) & (iii), we get: aligned & x₁=l, y₁=m, z₁=-n & x₂=l, y₂=-m, z₂=n & x₃=-l, y₃=m, z₃=n aligned So, the coordinate of A (l, m,-n), B (l,-m, n) and C (-l, m, n) aligned & AB ^2+ BC ^2+ CA ^2 l^2+m^2+n^2 = [ ar