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STD 12 - 11. three dimension geometry — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 11. three dimension geometry1 Mark
MCQ
If two lines $L_1$ and $L_2$ in space, are defined by ${L_1} = \{ x = \sqrt \lambda  y + \left( {\sqrt \lambda   - 1} \right),z = \left( {\sqrt \lambda   - 1} \right)y + \sqrt \lambda  \} $ and ${L_2} = \{ x = \sqrt \mu  y + \left( {1 - \sqrt \mu  } \right),z = \left( {1 - \sqrt \mu  } \right)y + \sqrt \mu  \} $ then $L_1$ is perpendicular to $L_2$, for all non-negative reals $\lambda $ and $ \mu $, such that
  • A
    $\sqrt \lambda   + \sqrt \mu   = 1$
  • B
    $\lambda  \ne \mu $
  • C
    $\lambda  + \mu  = 0$
  • ✓
    $\lambda  = \mu $
✓

Answer

Correct option: D.
$\lambda  = \mu $
d
For $L_{1}$

$x=\sqrt{\lambda} y+(\sqrt{\lambda}-1) \Rightarrow y=\frac{x-(\sqrt{\lambda}-1)}{\sqrt{\lambda}}$          .....$(i)$

$z=(\sqrt{\lambda}-1) y+\sqrt{\lambda} \Rightarrow y=\frac{z-\sqrt{\lambda}}{\sqrt{\lambda}-1}$         ......$(ii)$

From $(i)$ and $(ii)$

$\frac{x-(\sqrt{\lambda}-1)}{\sqrt{\lambda}}=\frac{y-0}{1}=\frac{z-\sqrt{\lambda}}{\sqrt{\lambda}-1}$         ....$(A)$

The equation $(\mathrm{A})$ is the equation of line $\mathrm{L}_{1}$

Similarly equation ofline $\mathrm{L}_{2}$ is

$x-\frac{(1-\sqrt{\mu})}{\sqrt{\mu}}=\frac{y-0}{1}=\frac{z-\sqrt{\mu}}{1-\sqrt{\mu}}$     ....$(B)$

Since $\mathrm{L}_{1} \perp \mathrm{L}_{2},$ therefore

$\sqrt{\lambda} \sqrt{\mu}+1 \times 1+(\sqrt{\lambda}-1)(1-\sqrt{\mu})=0$

$\Rightarrow \sqrt{\lambda}+\sqrt{\mu}=0 \Rightarrow \sqrt{\lambda}=-\sqrt{\mu}$

$\Rightarrow \lambda=\mu$

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