If the direction ratios of two lines are given by 3lm − 4ln + mn = 0 and l + 2m + 3n = 0, then the angle between the lines is

Step-by-step solution

Given, 3lm − 4ln + mn = 0 … (i) l + 2m + 3n = 0 … (ii) l=-2 m-3 n From Eq. (i), we get 3(− 2m − 3n)m − 4(− 2m − 3n) n + mn = 0 -6 m^2-9 n m+8 m n+12 n^2+m n=0 gathered 6 m^2=12 n^2 m^2=2 n^2 m= 2 n gathered (l₁, m₁, n₁ ) and (l₂, m₂, n₂ ) are direction ratios where, m₁= 2 n₁ and m₂= 2 n₂ aligned l₁ & =-(2 2 +3) n₁ and l₂=-(-2 2 +3) n₂ & = l₁ l₂+m₁ m₂+n₁ n₂ l₁^2+m₁^2+n₁^2 l₂^2+m₂^2+n₂^2 & = n₁ n₂[-2+(9-8)+1] l₁^2+m₁^2+n₁^2 l₂^2+m₂^2+n₂^2 =0 & = / 2 = / 2 aligned