A) \[\hat{i}+\hat{j}-\hat{k}\]
B) \[2\hat{i}-\hat{j}+4\hat{k}\]
C) \[\hat{i}-\hat{j}+\hat{k}\]
D) \[\hat{i}-\hat{j}-\hat{k}\]
E) \[\hat{i}+\hat{j}+\hat{k}\]
Correct Answer: E
Solution :
Two given lines intersect, if \[7\hat{i}+10\hat{j}+13\hat{k}+s(2\hat{i}+3\hat{j}+4\hat{k})\] \[=3\hat{i}+5\hat{j}+7\hat{k}+t(\hat{i}+2\hat{j}+3\hat{k})\] \[\Rightarrow \]\[(7+2s)\hat{i}+(10+3s)\hat{j}+(13+4s)\hat{k}\] \[=(3+t)\hat{i}+(5+2t)\hat{j}+(7+3t)\hat{k}\] \[\Rightarrow \]\[7+2s=3+t\Rightarrow 2s-t=-4\] ....(i) \[10+3s=5+2t\Rightarrow 3s-2t=-5\] ...(ii) and \[13+4s=7+3t\Rightarrow 45-3t=-6\] ...(iii) On solving Eqs. (i) and (iii), we get \[\begin{align} & \underline{\begin{align} & 4s-2t=-8 \\ & 4s-3t=-6 \\ & -\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,+ \\ \end{align}} \\ & \,\,\,\,\,\,\,\,\,\,t=-2 \\ \end{align}\] \[\Rightarrow \] \[2s+2=-4\] \[\Rightarrow \] \[2s=-6\] \[\Rightarrow \] \[s=-3\] \[\therefore \]Required line is \[7\hat{i}+10\hat{j}+13\hat{k}+(-3)[2\hat{i}+3\hat{j}+4\hat{k}]\] \[\Rightarrow \]\[\hat{i}+\hat{j}+\hat{k}\]is the required line.
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