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Home  >>  TN XII Math  >>  Vector Algebra
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Show that the following two lines are skew lines: $\overrightarrow{r}=(\overrightarrow{3i}+\overrightarrow{5j}+\overrightarrow{7k})+ t (\overrightarrow{i}-\overrightarrow{2j}+\overrightarrow{k})$ and $\overrightarrow{r}=(\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k}) + s (\overrightarrow{7i}-\overrightarrow{6j}+\overrightarrow{7k})$

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  • The lines $ \overrightarrow r=\overrightarrow a_1+t\overrightarrow u, \: \overrightarrow r=\overrightarrow a_2+s\overrightarrow u$ intersect if $ [(\overrightarrow a_2-\overrightarrow a_1)\overrightarrow u\: \overrightarrow v]=0$
$ \overrightarrow r=(3\overrightarrow i+5\overrightarrow j+7\overrightarrow k)+t(\overrightarrow i-2\overrightarrow j+\overrightarrow k) $ (i)
$ \overrightarrow r = (\overrightarrow i+\overrightarrow j+\overrightarrow k)+s(7\overrightarrow i+6\overrightarrow j+7\overrightarrow k)$ (ii)
(i) Passes through the point with pv $ \overrightarrow a_1 = 3\overrightarrow i+5\overrightarrow j+7\overrightarrow k$ and is parallel to $ \overrightarrow u=\overrightarrow i-2\overrightarrow j+\overrightarrow k$
(ii) Passes through the point with pv $ \overrightarrow a_2=\overrightarrow i+\overrightarrow j+\overrightarrow k$ and is parallel to $ \overrightarrow v = 7\overrightarrow i+6\overrightarrow j+7\overrightarrow k$
Since the vectors $ \overrightarrow u\: and \overrightarrow v$ are not parallel.
(i) and (ii) are either intersecting or skew lines.
Now $ \overrightarrow a_2-\overrightarrow a_1=-2\overrightarrow i-4\overrightarrow j-6\overrightarrow k$
Consider $[\overrightarrow a_2-\overrightarrow a_1\: \: \overrightarrow u\: \overrightarrow v]$
$ = \begin{vmatrix} -2 & -4 & -6 \\ 1 & -2 & 1 \\ 7 & 6 & 7 \end{vmatrix} = (-2)(-14-6)-(-4)(7-7)-6(6+14)$
$ = 40+0-120=-80 \neq 0$
Since $ [\overrightarrow a_2-\overrightarrow a_1\: \: \overrightarrow u\overrightarrow v] \neq 0$ the lines do not intersect.
$ \therefore $ they are skew lines.
answered Jun 12, 2013 by thanvigandhi_1
 

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