Question

If the lines  \[x =\]  5 ,  \[\frac{y}{3 - \alpha} = \frac{z}{- 2}\] and   \[x = \alpha\] \[\frac{y}{- 1} = \frac{z}{2 - \alpha}\] are coplanar, find the values of  \[\alpha\].

 

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 

\[\frac{x - 5}{0} = \frac{y}{3 - \alpha} = \frac{z}{- 2}\] and 

\[\frac{x - \alpha}{0} = \frac{y}{- 1} = \frac{z}{2 - \alpha}\] are coplanar.

\[\therefore \begin{vmatrix}\alpha - 5 & 0 - 0 & 0 - 0 \\ 0 & 3 - \alpha & - 2 \\ 0 & - 1 & 2 - \alpha\end{vmatrix} = 0\]

\[ \Rightarrow \begin{vmatrix}\alpha - 5 & 0 & 0 \\ 0 & 3 - \alpha & - 2 \\ 0 & - 1 & 2 - \alpha\end{vmatrix} = 0\]

\[ \Rightarrow \left( \alpha - 5 \right)\left[ \left( 3 - \alpha \right) \times \left( 2 - \alpha \right) - 2 \right] - 0 + 0 = 0\]

\[ \Rightarrow \left( \alpha - 5 \right)\left( \alpha^2 - 5\alpha + 4 \right) = 0\]

\[ \Rightarrow \left( \alpha - 5 \right)\left( \alpha - 1 \right)\left( \alpha - 4 \right) = 0\]

\[\Rightarrow \alpha - 1 = 0 \text{ or }  \alpha - 4 = 0 \text{ or }  \alpha - 5 = 0\]
\[ \Rightarrow \alpha = 1 \text { or }  \alpha = 4 \text{ or }  \alpha = 5\]   Thus, the values of \[\alpha\]  are 1, 4 and 5.