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

Answer 1

Answer: (A)

$-2$

$x=1+s,y=-3-\lambda s,z=1+\lambda s$

$\Rightarrow L_1:\frac{x-1}{1}=\frac{y+3}{-\lambda }=\frac{z-1}{\lambda }$

$x=\frac{t}{2},y=1+t,y=2+t$

$L_2:\frac{x}{\frac{1}{2}}=\frac{y-1}{1}=\frac{z-2}{-1}$

$L_1$ & $L_2$ are coplanar

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

$\left|\begin{array}{ccc}-1& 4& 1\\ 1& -\lambda & \lambda \\ \frac{1}{2}& 1& -1\end{array}\right|=0-1(\lambda -\lambda )-y(-1-\frac{\lambda }{2})+1(1+\frac{\lambda }{2})=0$

$\Rightarrow y=(1+\frac{\lambda }{2})=0$

$\therefore 1+\frac{\lambda }{2}=0$

$\therefore \lambda =-2$