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 coplanar then $\lambda$ equals

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

Answer 1

Answer: (C)

$-2$

From the given lines we have

$\frac{x-1}{1}=\frac{y+3}{-\lambda }=\frac{z-1}{\lambda }=s.....\left(A\right)$ and $\frac{x-0}{1}=\frac{y-1}{2}=\frac{z-2}{-2}......\left(B\right)$

As lines $\left(A\right)$ and $\left(B\right)$ are coplanar,

$\therefore \left|\begin{array}{ccc}1& -4& -1\\ 1& -\lambda & \lambda \\ 1& 2& -2\end{array}\right|=0$

$\Rightarrow (2\lambda -2\lambda )+4(-2-\lambda )-1(2+\lambda )=0$
$\Rightarrow 5\lambda =-10$

$\therefore \lambda =-2$