If is the angle between the lines whose direction cosines are given by 6 m n-2 n l+5 l m=0 and 3 l+m+5 n=0 , then =

Step-by-step solution

A line's direction cosines (l, m, n) satisfy both 6mn - 2nl + 5lm = 0 and 3l + m + 5n = 0 . Eliminating m = -3l - 5n via substitution yields the quadratic l^2 + 3ln + 2n^2 = 0 , which factors as (l + n)(l + 2n) = 0 . Case 1: l = -n gives m = -2n , producing normalized direction cosines ( -1 6 , -2 6 , 1 6 ) . Case 2: l = -2n gives m = n , producing normalized direction cosines ( -2 6 , 1 6 , 1 6 ) . The cosine of angle between these lines is = l₁l₂ + m₁m₂ + n₁n₂ = 2 6 - 2 6 + 1 6 = 1 6 . Using ^2 = 1 - ^2 , we find