# If the lines $\large\frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}$ and $\large\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}$ are co planar then the plane containing these two lines is?,

$\begin{array}{1 1} (a)\:\:y+2z+1=0\:\:\qquad\:(b)\:y+z+1=0\:\:\qquad\:(c)\:y-z+1=0\:\:\qquad\:(d)\:y-2z+1=0 \end{array}$

Toolbox:
• Equation of the plane containing the lines $\large\frac{x-x_1}{l_1}=\frac{y-y_1}{m_1}=\frac{z-z_1}{n_1}$ and $\large\frac{x-x_2}{l_2}=\frac{y-y_2}{m_2}=\frac{z-z_2}{n_2}$ is given by $\left|\begin {array}{ccc} x-x_1& y-y_1 &z-z_1 \\l_1 &m_1 & n_1\\ l_2 &m_2 & n_2\end {array}\right|=0$
• If the given two lines are coplanar, then $\left|\begin {array}{ccc} x_1-x_2& y_1-y_2 &z_1-z_2 \\l_1 &m_1 & n_1\\ l_2 &m_2 & n_2\end {array}\right|=0$
Given lines $\large\frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}$....(i) and $\large\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}$...(ii) are coplanar.
$\therefore \left|\begin {array}{ccc} 1+1& 0 &0 \\2 & k & 2\\ 5 & 2 & k\end {array}\right|=0$
$\Rightarrow\:2(k^2-4)=0\:\:\:or\:\:k=\pm2$
Equation of the plane containing them is given by
$\left|\begin {array}{ccc} x-1& y+1 &z \\2 & k & 2\\ 5 & 2 & k\end {array}\right|=0$
$\Rightarrow\: (x-1)(k^2-4)-(y+1)(2k-10)+z(4-5k)=0$
$\Rightarrow\:(k^2-4)x+(10-2k)y+z(4-5k)+(4-k^2-2k+10)=0$
Substituting the value of $k$ we get the equation of the planes to be
when $k=2$,
$6y-6z+6=0$ or $y-z+1=0$
when $k=-2$,
$14y+14z+14=0$ or $y+z+1=0$
answered Dec 25, 2013