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:

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

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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