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 co-planar, then $\lambda$ equals:

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

Answer 1

The correct option is A $-2$

The given straight line is $x=1+s,y=-3-\lambda s,z=1+\lambda s$

$\frac{x-1}{1}=\frac{y+3}{-\lambda }=\frac{z-1}{\lambda }=s$

Also given equation of another straight line is

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

$\frac{x-0}{1}=\frac{y-1}{2}=\frac{z-2}{-2}=t$

These two lines are coplanar if

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

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

$\Rightarrow 1\left|\begin{array}{cc}-\lambda & \lambda \\ 2& -2\end{array}\right|+4\left|\begin{array}{cc}1& \lambda \\ 1& -2\end{array}\right|-1\left|\begin{array}{cc}1& -\lambda \\ 1& 2\end{array}\right|=0\Rightarrow (2\lambda -2\lambda )+4(-2-\lambda )-1(2=\lambda )=0\Rightarrow -8-4\lambda -2-\lambda =0\Rightarrow -10=5\lambda \Rightarrow \lambda =-2$