Exams Joint Entrance Examination Mathematics 3D Geometry Question 19 Question 19 of 521 Medium MCQ
Let the direction cosines of two lines satisfy the equations : 4 l + m − n = 0 4 l+m-n=0 4 l + m − n = 0 and 2 m n + 10 n l + 3 l m = 0 2 m n+10 n l+3 l m=0 2 mn + 10 n l + 3 l m = 0 .
Then the cosine of the acute angle between these lines is :
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n = 4 I + m n ( 2 m + 10 I ) + 3 I m = 0 ⇒ 2 ( 4 I + m ) ( m + 5 I ) + 3 I m = 0 ⇒ 8 I m + 2 m 2 + 10 m l + 40 l 2 + 3 I m = 0 \begin{gathered} n=4I+m \\ n(2m+10I)+3Im=0 \\ \Rightarrow 2(4I+m)(m+5I)+3Im=0 \\ \Rightarrow 8Im+2{m}^{2}+10ml+40{l}^{2}+3Im=0 \end{gathered} n = 4 I + m n ( 2 m + 10 I ) + 3 I m = 0 ⇒ 2 ( 4 I + m ) ( m + 5 I ) + 3 I m = 0 ⇒ 8 I m + 2 m 2 + 10 m l + 40 l 2 + 3 I m = 0
⇒ 40 l 2 + 21 m l + 2 m 2 = 0 ⇒ 40 l 2 + 16 m l + 5 m l + 2 m 2 = 0 ⇒ ( 8 l + m ) ( 5 l + 2 m ) = 0 \begin{gathered} \Rightarrow 40{l}^{2}+21ml+2{m}^{2}=0 \\ \Rightarrow 40{l}^{2}+16ml+5ml+2{m}^{2}=0 \\ \Rightarrow (8l+m)(5l+2m)=0 \end{gathered} ⇒ 40 l 2 + 21 m l + 2 m 2 = 0 ⇒ 40 l 2 + 16 m l + 5 m l + 2 m 2 = 0 ⇒ ( 8 l + m ) ( 5 l + 2 m ) = 0
m = − 8 I ; 2 m = − 5 I m=-8I;2m=-5I m = − 8 I ; 2 m = − 5 I
n = − 4 i ⇒ m = − 5 l 2 n=-4i\, \Rightarrow m=\frac{-5l}{2} n = − 4 i ⇒ m = 2 − 5 l
D ⋅ R 1 = ( l 1 , m 1 , n 1 ) = ( − 1 , − 8 , − 4 ) D R 2 = ( l 2 , m 2 , n 2 ) = ( l 2 , − 5 l 2 2 , 3 l 2 2 ) = ( 2 , − 5 , 3 ) cos θ = ∣ 2 + 40 − 12 9 ⋅ 38 ∣ cos θ = 10 3 38 \begin{gathered} D\cdot {R}_{1} =({l}_{1},{m}_{1},{n}_{1}) \\ =(-1,-8,-4) \\ D{R}_{2} =({l}_{2},{m}_{2},{n}_{2}) \\ =({l}_{2},\frac{-5{l}_{2}}{2},\frac{3{l}_{2}}{2}) \\ =(2,-5,3) \\ \cos \theta =|\frac{2+40-12}{9\cdot \sqrt{38}}| \\ \cos \theta =\frac{10}{3\sqrt{38}} \\ \end{gathered} D ⋅ R 1 = ( l 1 , m 1 , n 1 ) = ( − 1 , − 8 , − 4 ) D R 2 = ( l 2 , m 2 , n 2 ) = ( l 2 , 2 − 5 l 2 , 2 3 l 2 ) = ( 2 , − 5 , 3 ) cos θ = ∣ 9 ⋅ 38 2 + 40 − 12 ∣ cos θ = 3 38 10 Answer the question to unlock