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 parameters s and t respectively, are co-planar, then $\lambda$ equals

• A. -1
• B. $\frac{-1}{2}$
• C. -2

Answer 1

The correct option is C -2

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