Question

$\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}$ and $\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}$ are co-planar then the value of $k$ is

Answer 1

If two lines are co-planar, then

$\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ l_1& m_1& n_1\\ l_2& m_2& n_2\end{array}\right|=0$

where $x_1=2,y_1=3,z_1=4,l_1=1,m_1=1,n_1=-k$

and $x_2=1,y_2=4,z_2=5,l_2=k,m_2=2,n_2=1$

$=\left|\begin{array}{ccc}1-2& 4-3& 5-4\\ 1& 1& -k\\ k& 2& 1\end{array}\right|$

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

$=-1(1+2k)-1(1+k^2)+1(2-k)$

$=-2k-1-1-k^2+2-k$

$\Rightarrow -k^2-3k=0$

$\Rightarrow -k(k+3)=0$

$\therefore k=0,-3$