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प्रश्न
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.
उत्तर
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
For k = −2,
\[ \Rightarrow \left( x - 1 \right)\left( 4 - 4 \right) - \left( y + 1 \right)\left( - 4 - 4 \right) + z\left( 4 + 4 \right) = 0\]
\[ \Rightarrow 8\left( y + 1 \right) + 8z = 0\]
\[ \Rightarrow y + z + 1 = 0\]
Thus, the equation of the plane containing the given lines is y + z + 1 = 0.
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