If the direction ratio of two lines are given by 3 l m-4 l n+m n=0 and l+2 m+3 n=0 , then the angle between the lines, is :

Step-by-step solution

We have, 3 l m-4 l n+m n=0 (i) and l+2 m+3 n=0 (ii) From Eq. (ii), l=-(2 m+3 n) Using in Eq. (i), we get -3(2 m+3 n) m+4(2 m+3 n) n+m n=0 -6 m^2-9 m n+8 m n+12 n^2+m n=0 -6 m^2+12 n^2=0 Now, m^2=2 n^2 m= 2 n Now, m= 2 n l=-(2 2 n+3 n)=-(2 2 +3) n l: m: n=-(3+2 2 ) n: 2 n: n=-(3+2 2 ): 2 : 1 Also, m=- 2 n l=-(-2 2 +3) nl: m: n=-(3-2 2 ) n:- 2 n: n=-(3-2 2 ):- 2 : 1 If is the angle between the lines, then = l₁ l₂+m₁ m₂+n₁ n₂ l₁^2+m₁^2+n₁^2 l₂^2+m₂^2+n₂^2 = (3+2 2 )(3-2 2 )+( 2 )(- 2 )+1.1 (3+2 2 )^2+( 2 )^2+1^2 (3-2