A) \[-2\]
B) \[-1\]
C) \[-\frac{1}{2}\]
D) \[0\]
Correct Answer: A
Solution :
The given straight line is \[x=1+s,y=-3-\lambda s,z=1+\lambda s\] Or \[\frac{x-1}{1}=\frac{y+3}{-\lambda }=\frac{z-1}{\lambda }=s\] Also given equation of another straight line is \[x=\frac{t}{2},y=1+t,z=2-t\] Or \[\frac{x-0}{1}=\frac{y-1}{2}=\frac{z-2}{-2}=t\] These two lines are coplanar, if \[\left| \begin{matrix} 1-0 & -3-1 & 1-2 \\ 1 & -\lambda & \lambda \\ 1 & 2 & -2 \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[\left| \begin{matrix} 1 & -4 & -1 \\ 1 & -\lambda & \lambda \\ 1 & 2 & -2 \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[1\left| \begin{matrix} -\lambda & \lambda \\ 2 & -2 \\ \end{matrix} \right|+4\left| \begin{matrix} 1 & \lambda \\ 1 & -2 \\ \end{matrix} \right|-1\left| \begin{matrix} 1 & -\lambda \\ 1 & 2 \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[(2\lambda -2\lambda )+4(-2-\lambda )-1(2+\lambda )=0\] \[\Rightarrow \] \[-8-4\lambda -2-\lambda =0\] \[\Rightarrow \] \[-10=5\lambda \Rightarrow \lambda =-2\]
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