Mathematics

If two line $$L_1$$ and $$L_2$$ in space,are defined by
$$\begin{array}{l}{L_1} = \left\{ {x = \sqrt \lambda  y + \left( {\sqrt \lambda   - 1} \right),z = \left( {\sqrt \lambda   - 1} \right)y + \sqrt \lambda  } \right\}and\\{L_2} = \left\{ {x = \sqrt \mu  y + \left( {1 - \sqrt \mu  } \right),z = \left( {1 - \sqrt \mu  } \right)y + \sqrt \mu  } \right\}\end{array}$$,
then $$L_1$$ is perpendicular to $$L_2$$ for all non-negative reals $$\lambda $$ and $$\mu $$, such that:


ANSWER

$$\sqrt \lambda + \sqrt \mu = 0$$


SOLUTION
$$\begin{array}{l} { L_{ 1 } }=\{ x=\sqrt { \lambda  } y+(\sqrt { \lambda  } -1),z=(\sqrt { \lambda  } -1)y+\sqrt { \lambda  } \}  \\ \dfrac { { x-(\sqrt { \lambda  } -1) } }{ { \sqrt { \lambda  }  } } =y \\ \dfrac { { z-\sqrt { \lambda  }  } }{ { (\sqrt { \lambda  } -1) } } =y \\ \dfrac { { x-(\sqrt { \lambda  } -1) } }{ { \sqrt { \lambda  }  } } =y=\dfrac { { z-\sqrt { \lambda  }  } }{ { (\sqrt { \lambda  } -1) } }  \\ { L_{ 1 } }=\left( { \sqrt { \lambda  } -1,0,\sqrt { \lambda  }  } \right) +{ K_{ 1 } }\left( { \sqrt { \lambda  } ,1,\sqrt { \lambda  } -1 } \right)  \\ Now, \\ { L_{ 2 } }=\{ x=\sqrt { \mu  } y+(1-\sqrt { \mu  } ),z=(1-\sqrt { \mu  } )y+\sqrt { \mu  } \}  \\ \dfrac { { x-(1-\sqrt { \mu  } ) } }{ { \sqrt { \mu  }  } } =y \\ \dfrac { { z-\sqrt { \mu  }  } }{ { 1-\sqrt { \mu  }  } } =y \\ { L_{ 2 } }=\left( { 1-\sqrt { \mu  } ,0,\sqrt { \mu  }  } \right) +{ K_{ 2 } }\left( { \sqrt { \mu  } ,1,1-\sqrt { \mu  }  } \right)  \\ { L_{ 1 } }\bot { L_{ 2 } } \\ \left( { \sqrt { \lambda  } ,1,\sqrt { \lambda  } -1 } \right) .\left( { \sqrt { \mu  } ,1,1-\sqrt { \mu  }  } \right) =0 \\ \sqrt { \lambda \mu  } +1+\left( { \sqrt { \lambda  } -1 } \right) \left( { 1-\sqrt { \mu  }  } \right) =0 \\ \sqrt { \lambda \mu  } +1+\sqrt { \lambda  } -1-\sqrt { \lambda \mu  } +\sqrt { \mu  } =0 \\ \sqrt { \lambda  } +\sqrt { \mu  } =0 \\  \  \end{array}$$
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Single Correct Medium Published on 09th 09, 2020
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