Question

Find the values of  \[\lambda\] for which the lines

\[\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z + 3}{\lambda^2}\]and  \[\frac{x - 3}{1} = \frac{y - 2}{\lambda^2} = \frac{z - 1}{2}\]  are coplanar . 

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 

 

\[\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z + 3}{\lambda^2}\] and

\[\frac{x - 3}{1} = \frac{y - 2}{\lambda^2} = \frac{z - 1}{2}\]

 are coplanar.

\[\therefore \begin{vmatrix}x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2\end{vmatrix} = 0\]

\[\Rightarrow \begin{vmatrix}3 - 1 & 2 - 2 & 1 - \left( - 3 \right) \\ 1 & 2 & \lambda^2 \\ 1 & \lambda^2 & 2\end{vmatrix} = 0\]
\[ \Rightarrow \begin{vmatrix}2 & 0 & 4 \\ 1 & 2 & \lambda^2 \\ 1 & \lambda^2 & 2\end{vmatrix} = 0\]
\[ \Rightarrow 2\left( 4 - \lambda^4 \right) - 0 + 4\left( \lambda^2 - 2 \right) = 0\]
\[ \Rightarrow - 2 \lambda^4 + 4 \lambda^2 = 0\]

\[\Rightarrow \lambda^2 \left( \lambda^2 - 2 \right) = 0\]

\[ \Rightarrow \lambda^2 = 0 \text{ or } \lambda^2 - 2 = 0\]

\[ \Rightarrow \lambda = 0 \text{ or }  \lambda = \pm \sqrt{2}\]

Thus, the values of \[\lambda\]  are 0, \[- \sqrt{2}\] and \[\sqrt{2}\] .