Exams Joint Entrance Examination Mathematics 3D Geometry Question 133 Question 133 of 521 Medium NUMERICAL
Let a line passing through the point ( − 1 , 2 , 3 ) (-1,2,3) ( − 1 , 2 , 3 ) intersect the lines \[L_1: \frac{x-1}{3}=\frac{y-2}{2}=\frac{z+1}{-2}\] at \[M(\alpha, \beta, \gamma)\] and \[L_2: \frac{x+2}{-3}=\frac{y-2}{-2}=\frac{z-1}{4}\] at N ( a , b , c ) N(a, b, c) N ( a , b , c ) . Then, the ...
Answer: 196
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M ( 3 λ + 1 , 2 λ + 2 , − 2 λ − 1 ) ∴ α + β + γ = 3 λ + 2 N ( − 3 μ − 2 , − 2 μ + 2 , 4 μ + 1 ) ∴ a + b + c = − μ + 1 \begin{gathered} M(3\lambda +1,2\lambda +2,-2\lambda -1)\, \therefore \alpha +\beta +\gamma =3\lambda +2 \\ N(-3\mu -2,-2\mu +2,4\mu +1)\, \therefore a+b+c=-\mu +1 \end{gathered} M ( 3 λ + 1 , 2 λ + 2 , − 2 λ − 1 ) ∴ α + β + γ = 3 λ + 2 N ( − 3 μ − 2 , − 2 μ + 2 , 4 μ + 1 ) ∴ a + b + c = − μ + 1
3 λ + 2 − 3 μ − 1 = 2 λ − 2 μ = − 2 λ − 4 4 μ − 2 3 λ μ + 2 μ = 3 λ μ + λ 2 μ = λ 2 λ μ − λ = λ μ + 2 μ λ μ = λ + 2 μ ⇒ λ μ = 2 λ \begin{gathered} \frac{3\lambda +2}{-3\mu -1}=\frac{2\lambda }{-2\mu }=\frac{-2\lambda -4}{4\mu -2} \\ 3\lambda \mu +2\mu =3\lambda \mu +\lambda \\ 2\mu =\lambda \\ 2\lambda \mu -\lambda =\lambda \mu +2\mu \\ \lambda \mu =\lambda +2\mu \\ \Rightarrow \lambda \mu =2\lambda \end{gathered} − 3 μ − 1 3 λ + 2 = − 2 μ 2 λ = 4 μ − 2 − 2 λ − 4 3 λ μ + 2 μ = 3 λ μ + λ 2 μ = λ 2 λ μ − λ = λ μ + 2 μ λ μ = λ + 2 μ ⇒ λ μ = 2 λ
⇒ μ = 2 ( λ ≠ 0 ) ∴ λ = 4 α + β + γ = 14 a + b + c = − 1 ( α + β + γ ) 2 ( a + b + c ) 2 = 196 \begin{gathered} \Rightarrow \, \mu =2\, (\lambda \ne 0) \\ \therefore \, \lambda =4 \\ \alpha +\beta +\gamma =14 \\ a+b+c=-1 \\ \frac{(\alpha +\beta +\gamma {)}^{2}}{(a+b+c{)}^{2}}=196 \end{gathered} ⇒ μ = 2 ( λ = 0 ) ∴ λ = 4 α + β + γ = 14 a + b + c = − 1 ( a + b + c ) 2 ( α + β + γ ) 2 = 196 Answer the question to unlock