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 the value of $(3{\lambda }^2+2\lambda +1)=$

Answer 1

Given :
$L_1:x=1+s,y=-3-\lambda s,z=1+\lambda s\Rightarrow L_1:\frac{x-1}{1}=\frac{y+3}{-\lambda }=\frac{z-1}{\lambda }$
and $L_2:x=\frac{t}{2},y=1+t,z=2-t$
$\Rightarrow L_2:\frac{x}{1/2}=\frac{y-1}{1}=\frac{z-2}{-1}$
$D.r^{\prime }s$ of $L_1=(a_1,b_1,c_1)=(1,-\lambda ,\lambda )$
$D.r^{\prime }s$ of $L_2=(a_2,b_2,c_2)=(\frac{1}{2},1,-1)$
For lines to be coplanar :
$\left|\begin{array}{ccc}{x}_1-x_2& y_1-y_2& z_1-z_2\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|=0\Rightarrow \left|\begin{array}{ccc}-1& 4& 1\\ 1& -\lambda & \lambda \\ \frac{1}{2}& 1& -1\end{array}\right|=0\Rightarrow 0-4(-1-\frac{\lambda }{2})+1+\frac{\lambda }{2}=0\Rightarrow \lambda =-2$