A) collinear
B) skew-lines
C) coplanar lines
D) parallel lines
Correct Answer: C
Solution :
Given equation of lines are \[r=(\hat{i}+\hat{j})+\lambda (\hat{i}+2\hat{j}-\hat{k})\](vector form) \[\Rightarrow \] \[\frac{x-1}{2}=\frac{y-1}{2}=\frac{z-0}{1}\] (Cartesian form) ?.(i) and \[r=(\hat{i}+\hat{j})+\mu (-\hat{i}+\hat{j}-2\hat{k})\] (vector form) \[\Rightarrow \] \[\frac{x-1}{-1}=\frac{y-1}{1}=\frac{z-0}{-2}\] (Cartesian form) ?.(ii) If two lines are coplanar, then \[\left| \begin{matrix} 1-1 & 1-1 & 0-0 \\ 1 & 2 & -1 \\ -1 & 1 & -2 \\ \end{matrix} \right|=\left| \begin{matrix} 0 & 0 & 0 \\ 1 & 2 & -1 \\ -1 & 1 & -2 \\ \end{matrix} \right|=0\] Hence, given lines are coplanar lines \[\left( \because \,\,\left| \begin{matrix} {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\ {{l}_{1}} & {{m}_{1}} & {{n}_{1}} \\ {{l}_{2}} & {{m}_{2}} & {{n}_{2}} \\ \end{matrix} \right|=0 \right)\]
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