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JEE Main201815 Apr 2018Evening ShiftMathematicsVector AlgebraActual

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

Options

  1. A⁻¹ ( 1 8 )
  2. B⁻¹ ( 1 6 )
  3. C⁻¹ ( 1 3 )
  4. D⁻¹ ( 1 4 )

Correct answer

B. ⁻¹ ( 1 6 )

Step-by-step solution

Given aligned &l+3 m+5 n=0 & and 5 l m-2 m n+6 n l=0 aligned From eq. (1) we have l=-3 m-5 n Put the value of l in eq. (2), we get; aligned &5(-3 m-5 n) m-2 m n+6 n(-3 m-5 n)=0 & 15 m^2+45 m n+30 n^2=0 & m^2+3 m n+2 n^2=0 & m^2+2 m n+m n+2 n^2=0 & (m+n)(m+2 n)=0 & m=-n or m=-2 n & For m=-n, l=-2 n & And for m=-2 n, l=n aligned aligned & (l, m, n)=(-2 n,-n, n) Or (l, m, n) &=(n,-2 n, n) & (l, m, n)=(-2,-1,1) Or (l, m, n) &=(1,-2,1) aligned Therefore, angle between the lines is given as: aligned & ( )= (-2)(1)+(-1) (

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