Question

If the straight lines $x=1+s,y=-3-\lambda s,z=1+\lambda s$ and $x=\frac{t}{2},y=1,z=2-t$, with parameters s and trespectively, are co-planar, then $\lambda$ equals
[AIEEE 2004]

• A. 0
• B. -1
• C. -1/2
• D. -2

Answer 1

The correct option is D

-2

We have, $\frac{x-1}{1}=\frac{y+3}{-\lambda }=\frac{z-1}{\lambda }=s$
and $\frac{x-0}{1/2}=\frac{y-1}{1}=\frac{z-2}{-1}=t$
Since, lines are coplanar then
$\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ l_1& m_1& n_1\\ l_2& m_2& n_2\end{array}\right|=0\Rightarrow \left|\begin{array}{ccc}-1& 4& 1\\ 1& -\lambda & \lambda \\ 1/2& 1& -1\end{array}\right|=0$
On Solving, $\lambda =-2.$