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Home  >>  TN XII Math  >>  Vector Algebra
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Find the shortest distance between the parallel lines $\large\frac{x-1}{-1}=\frac{y}{3}=\frac{z+3}{2} $ and $\large\frac{x-3}{-1}=\frac{y+1}{3}=\frac{z-1}{2} $

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  • Distance between two lines
  • (i) Shortest = 0 if they are intersecting.
  • (ii) Parallel lines $ \overrightarrow r = \overrightarrow a_1+t\overrightarrow u_1, \: \overrightarrow s=\overrightarrow a_2+s\overrightarrow u$
  • (iii) Skew line $ \overrightarrow r = \overrightarrow a_1+t\overrightarrow u, \: \overrightarrow r = \overrightarrow a_2+s\overrightarrow v$
  • $ d = \large\frac{[(\overrightarrow a_2-\overrightarrow a_1)\overrightarrow u\: \overrightarrow v]}{|\overrightarrow u \times \overrightarrow v|}$
$ \large\frac{x-1}{-1} = \large\frac{y}{3}=\large\frac{z+3}{2} ; \large\frac{x-3}{-1}=\large\frac{y+1}{3} = \large\frac{z-1}{2}$
The lines are parallel to the vector with D.r.s $ -1, 3, 2, $ i.e., $ \overrightarrow u = -\overrightarrow i+3\overrightarrow j+2\overrightarrow k$
The first line passes through the point $ A(1, 0, -3) $ i.e., $ \overrightarrow a_1 = \overrightarrow i-3\overrightarrow k$
The second line passes through the point (3, -1, 1) i.e., $ \overrightarrow a_2 = 3\overrightarrow i-\overrightarrow j+\overrightarrow k$
$ \overrightarrow a_2-\overrightarrow a_1 = (3-1)\overrightarrow i+(-1-0)\overrightarrow j+(1+3)\overrightarrow k = 2\overrightarrow i-\overrightarrow j+4\overrightarrow k$
The distance between them $ d = \large\frac{|\overrightarrow u \times (\overrightarrow a_2-\overrightarrow a_1)|}{|\overrightarrow u|}$
$|\overrightarrow u| = \sqrt{1+9+4} = \sqrt{14}$
$ \overrightarrow u \times (\overrightarrow a_2-\overrightarrow a_1) = \begin{vmatrix} \overrightarrow i & \overrightarrow j & \overrightarrow k \\ -1 & 3 & 2 \\ 2 & -1 & 4 \end{vmatrix} = 14\overrightarrow i+8\overrightarrow j-5\overrightarrow k$
$ |\overrightarrow u \times (\overrightarrow a_2 - \overrightarrow a_1) | =\sqrt{196+64+25} = \sqrt{\large\frac{285}{14}} $ units

 

answered Jun 12, 2013 by thanvigandhi_1
edited Jun 24, 2013 by thanvigandhi_1
 

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