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  • The line passes through the point (2,6,2) and is perpendicular to the plane <img alt="2x + y - 2z = 10" src="http

The line l_{1} passes through the point (2,6,2) and is perpendicular to the plane 2x + y - 2z = 10. Then the shortest distance between the linel_{1}  and the line \frac{x+1}{2}=\frac{y+4}{-3}=\frac{z}{2} is :

Option: 1

\frac{13}{3}


Option: 2

\frac{19}{3}


Option: 3

7


Option: 4

9


Answers (1)

best_answer

equation of l_{1} is \frac{x-2}{2}=\frac{y-6}{1}=\frac{z-2}{-2}

               Let l_2 is \frac{x+1}{2}=\frac{y+4}{-3}=\frac{z}{2}

Point on l_{1} is  \mathrm{a}=(2,6,2)$, direction $\overrightarrow{\mathrm{p}}=\langle 2,1,-2\rangle

Point on l_{2} is \mathrm{b}=(-1,-4,0)$ direction $\overrightarrow{\mathrm{q}}=\langle 2,-3,2\rangle

Shortest distance between l_{1} and l_{2} l_2=\left|\frac{(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}) \cdot(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}})}{|\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}|}\right|

\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \mathrm{k} \\ 2 & 1 & -2 \\ 2 & -3 & 2 \end{array}\right|=\hat{\mathrm{i}}(-4)-\hat{\mathrm{j}}(8)+\mathrm{k}(-8)

                   \begin{aligned} & =\left|\frac{\langle 3,10,2\rangle \cdot\langle-4,-8,-8\rangle}{\sqrt{16+64+64}}\right| \\ & =\left|\frac{-12-80-16}{\sqrt{144}}\right| \\ & =\frac{108}{12} \\ & =9 \end{aligned}

Shortest distance between the lines is 9.

 

 

Posted by

rishi.raj

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