A) - 3
B) 2
C) 1
D) - 2
Correct Answer: D
Solution :
The given lines are \[\frac{x-1}{1}\,=\frac{y+3}{-\lambda }\,=\frac{z-1}{\lambda }=(a)\] and \[\frac{x-0}{1/2}\,=\frac{y-1}{1}=\frac{z-2}{-1}=(b)\] \[\therefore \] for coplanarity, we must have \[\left| \begin{matrix} -1 & 4 & 1 \\ 1 & -\lambda & \lambda \\ 1/2 & 1 & -1 \\ \end{matrix} \right|=0\] \[\Rightarrow \,-1(\lambda -\lambda )-4\,\left( -1-\frac{\lambda }{2} \right)\,+\left( 1+\frac{\lambda }{2}\, \right)=0\] \[\Rightarrow \,4+2\lambda \,+1+\frac{\lambda }{2}=0\] \[\Rightarrow \,\,5+\frac{5\lambda }{2}\,=0\] \[\therefore \,\lambda =-2\]You need to login to perform this action.
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