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The following two lines are $ \large\frac{x-1}{2}=\frac{y-1}{-1}=\frac{z}{1}$ and $\large\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z-1}{2}$ is

\[\begin{array}{1 1}(1)parallel &(2)intersecting\\(3)skew&(4)perpendicular\end{array}\]

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Let $x_1=1,y_1=1,z_1=0,l_1=2,m_1=-1,n_1=1$
$x_2=2,y_2=1,z_2=1,l_2=3,m_2=-5,n_2=2$
$\large\frac{l_1}{l_2}=\frac{2}{3}$
$\large\frac{m_1}{m_2}=\frac{1}{5}$
$\large\frac{n_1}{n_2}=\frac{1}{2}$
$\large\frac{l_1}{l_2} \neq \large\frac{m_1}{m_2} \neq \large\frac{n_1}{n_2} $
$\therefore $ The lines are not parallel
$l_1l_2 +m_1m_2 +n_1n_2= 2 \times 3 +(-1) (-5) +1 \times 2 \neq 0$
$\therefore $ The lines are not perpedicular
$\begin{vmatrix} 2-1 & 1-1 & 1-0 \\ 2 & -1 & 1 \\ 3 & -5 & 2 \end{vmatrix}=\begin{vmatrix} 1 & 0 & 1 \\ 2 & -1 & 1 \\3 & -5 & 2 \end{vmatrix}$
$\qquad= 1(-2+5)-0+1(-10+3)$
$\qquad= 3-7 \neq 0$
The lines does not intersect
So the lines are skew lines.
Hence 3 is the correct answer.
answered May 13, 2014 by meena.p
 

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