A) \[\frac{11}{2}\]
B) \[-\frac{11}{2}\]
C) \[\frac{9}{2}\]
D) \[-\frac{9}{2}\]
Correct Answer: A
Solution :
Since, the lines \[\frac{x+1}{2}=\frac{y-1}{1}=\frac{z+1}{3}\] and \[\frac{x+2}{2}=\frac{y-k}{3}=\frac{z}{4}\] are coplanar. \[\therefore \] \[\left| \begin{matrix} {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\ {{l}_{1}} & {{m}_{1}} & {{n}_{1}} \\ {{l}_{2}} & {{m}_{2}} & {{n}_{2}} \\ \end{matrix} \right|=\,\left| \begin{matrix} -2+1 & k-1 & 0+1 \\ 2 & 1 & 3 \\ 2 & 3 & 4 \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[\left| \begin{matrix} -1 & k-1 & 1 \\ 2 & 1 & 3 \\ 2 & 3 & 4 \\ \end{matrix} \right|=-(4-9)-(k-1)(8-6)+1(6-2)=0\] \[\Rightarrow \] \[5-2k+2+4=0\] \[\Rightarrow \] \[2k=11\,\,\,\,\Rightarrow \,\,\,\,k=\frac{11}{2}\]
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