Question

If the straight lines $x=1+s,y=-3-\lambda s,z=1+\lambda s$ and $x=\frac{t}{2},y=1+t,z=2-t$, with parameter $s$ and $t$ respectively, are coplanar, then $\lambda$ equal

• A. $0$
• B. $-1$
• C. $-\frac{1}{2}$
• D. $-2$

Answer 1

The correct option is D $-2$

The given lines are $x-1=\frac{y+3}{-\lambda }=\frac{z-1}{\lambda }=s$ ...(1)

and $2x=y-1=\frac{z-2}{-1}=t$ ...(2)

The lines are coplanar if,

$\left|\begin{array}{ccc}0-(-1)& -1-3& -2-(-1)\\ 1& -\lambda & \lambda \\ \frac{1}{2}& 1& -1\end{array}\right|=0$

Applying $C_2\to C_2+C_3$

$\left|\begin{array}{ccc}1& -5& -1\\ 1& 0& \lambda \\ \frac{1}{2}& 0& -1\end{array}\right|=0$

$\Rightarrow 5(-1-\frac{\lambda }{2})=0$

$\Rightarrow \lambda =-2$