Question

If the straight lines  \[\frac{x - 1}{2} = \frac{y + 1}{k} = \frac{z}{2}\] and \[\frac{x + 1}{2} = \frac{y + 1}{2} = \frac{z}{k}\] are coplanar, find the equations of the planes containing them.

 

Answer 1

The lines \[\frac{x - x_1}{a_1} = \frac{y - y_1}{b_1} = \frac{z - z_1}{c_1}\] and  \[\frac{x - x_2}{a_2} = \frac{y - y_2}{b_2} = \frac{z - z_2}{c_2}\] are coplanar if  \[\begin{vmatrix}x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2\end{vmatrix} = 0\]

The given lines  \[\frac{x - 1}{2} = \frac{y + 1}{k} = \frac{z}{2}\]and  \[\frac{x + 1}{2} = \frac{y + 1}{2} = \frac{z}{k}\] are coplanar.

\[\therefore \begin{vmatrix}- 1 - 1 & - 1 - \left( - 1 \right) & 0 - 0 \\ 2 & k & 2 \\ 2 & 2 & k\end{vmatrix} = 0\]
\[ \Rightarrow \begin{vmatrix}- 2 & 0 & 0 \\ 2 & k & 2 \\ 2 & 2 & k\end{vmatrix} = 0\]
\[ \Rightarrow - 2\left( k^2 - 4 \right) - 0 + 0 = 0\]
\[ \Rightarrow k^2 - 4 = 0\]
\[ \Rightarrow k = \pm 2\]

The equation of the plane containing the given lines is 

\[\begin{vmatrix}x - 1 & y + 1 & z \\ 2 & k & 2 \\ 2 & 2 & k\end{vmatrix} = 0\]

For k = 2,  \[\begin{vmatrix}x - 1 & y + 1 & z \\ 2 & k & 2 \\ 2 & 2 & k\end{vmatrix} =\]\[\begin{vmatrix}x - 1 & y + 1 & z \\ 2 & 2 & 2 \\ 2 & 2 & 2\end{vmatrix} = 0\] 

So, no plane exists for k = 2.

For k = −2,

\[\begin{vmatrix}x - 1 & y + 1 & z \\ 2 & k & 2 \\ 2 & 2 & k\end{vmatrix} = 0\]

\[\Rightarrow \begin{vmatrix}x - 1 & y + 1 & z \\ 2 & - 2 & 2 \\ 2 & 2 & - 2\end{vmatrix} = 0\]
\[ \Rightarrow \left( x - 1 \right)\left( 4 - 4 \right) - \left( y + 1 \right)\left( - 4 - 4 \right) + z\left( 4 + 4 \right) = 0\]
\[ \Rightarrow 8\left( y + 1 \right) + 8z = 0\]
\[ \Rightarrow y + z + 1 = 0\]

Thus, the equation of the plane containing the given lines is y + z + 1 = 0.