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Let $\mathrm{l_{1}}$ be the line in xy-plane with x and y intercepts $\mathrm{\frac{1}{8}}$ and $\mathrm{\frac{1}{4\sqrt{2}}}$ respectively, and $\mathrm{l_{2}}$ be the line in zx-plane with x and z intercepts $\mathrm{-\frac{1}{8}}$ and $\mathrm{-\frac{1}{6\sqrt{3}}}$ respectively. If $\mathrm{d}$ is the shortest distance between the line $\mathrm{l_{1}\: and \: l_{2}}$ , then $\mathrm{d^{-2}}$ is equal to ___________.
Option: 1
51

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

$\mathrm{8 x+4 \sqrt{2} y=1, z=0}\\$

$\Rightarrow\mathrm{\frac{x-\frac{1}{8}}{1}=\frac{y-0}{-\sqrt{2}}=\frac{z-0}{0}=\lambda}\\$

$\mathrm{-8 x-6 \sqrt{3} z=1, \quad y=0}\\$

$\mathrm{\Rightarrow \frac{x+\frac{1}{8}}{3 \sqrt{3}}=\frac{y-0}{0}=\frac{z-0}{-4}}\\$

$\mathrm{\left|\begin{array}{ccc} \frac{1}{4} & 0 & 0 \\ 1 & -\sqrt{2} & 0 \\ 3 \sqrt{3} & 0 & -4 \end{array}\right|}\\$

$\mathrm{\Rightarrow\frac{1}{4}\left ( 4\sqrt{2}-0 \right )=\sqrt{2} }\\$

$\mathrm{d=\frac{1}{\sqrt{51}}}$

$\mathrm{\frac{1}{d^{2}}=51}$

Hence the correct answer is 51

Posted by

manish painkra

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