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Home  >>  TN XII Math  >>  Vector Algebra
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Find the shortest distance between the parallel lines $\overrightarrow{r}=(\overrightarrow{2i}-\overrightarrow{j}-\overrightarrow{k}) + t (\overrightarrow{i}-\overrightarrow{2j}+\overrightarrow{3k})$ and $\overrightarrow{r}=(\overrightarrow{i}+\overrightarrow{2j}+\overrightarrow{k}) + s (\overrightarrow{i}-\overrightarrow{2j}+\overrightarrow{3k})$

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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|}$
$ \overrightarrow r = (2\overrightarrow i-\overrightarrow j-\overrightarrow k) + t (\overrightarrow i-2\overrightarrow j+3\overrightarrow k)$ and
$ \overrightarrow r = (\overrightarrow i-2\overrightarrow j+\overrightarrow k)+s(\overrightarrow i-2\overrightarrow j+3\overrightarrow k)$
Both lines are parallel to $ \overrightarrow u = \overrightarrow i-2\overrightarrow j+3\overrightarrow k$
$ \overrightarrow a_1 = 2\overrightarrow i-\overrightarrow j-\overrightarrow k\: and \: \overrightarrow a_2 = \overrightarrow i-2\overrightarrow j=\overrightarrow k$
The shortest distance between the line is $ d = \large\frac{\overrightarrow u \times (\overrightarrow a_2-\overrightarrow a_1)}{|\overrightarrow u|}$
$ |\overrightarrow u| = \sqrt{1+4+9} = \sqrt {14}$
$ \overrightarrow a_2-\overrightarrow a_1 = (1-2)\overrightarrow i+(-2+1)\overrightarrow j+(1+1)\overrightarrow k$
$ = -\overrightarrow i-\overrightarrow j+2\overrightarrow k$
$ \overrightarrow u \times (\overrightarrow a_2-\overrightarrow a_1) = \begin{vmatrix} \overrightarrow i & \overrightarrow j & \overrightarrow k \\ 1 & -2 & 3 \\ -1 & -1 & 2 \end{vmatrix}=(-4+3)\overrightarrow i-(2+-3)\overrightarrow j+(-1-2)\overrightarrow k$
$ = -\overrightarrow i-5\overrightarrow j-\overrightarrow k$
$ |\overrightarrow u \times (\overrightarrow a_2-\overrightarrow a_1)| = \sqrt{1+25+19} = \sqrt{35}$
$ d = \large\frac{\sqrt{35}}{\sqrt{14}} = \sqrt{\large\frac{5}{2}}$ units

 

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

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