Question

Show that the lines  \[\frac{x + 3}{- 3} = \frac{y - 1}{1} = \frac{z - 5}{5}\] and  \[\frac{x + 1}{- 1} = \frac{y - 2}{2} = \frac{z - 5}{5}\]  are coplanar. Hence, find the equation of the plane containing these lines.

 

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 are \[\frac{x + 3}{- 3} = \frac{y - 1}{1} = \frac{z - 5}{5}\]  and \[\frac{x + 1}{- 1} = \frac{y - 2}{2} = \frac{z - 5}{5}\]

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

\[\begin{vmatrix}- 1 - \left( - 3 \right) & 2 - 1 & 5 - 5 \\ - 3 & 1 & 5 \\ - 1 & 2 & 5\end{vmatrix} = \begin{vmatrix}2 & 1 & 0 \\ - 3 & 1 & 5 \\ - 1 & 2 & 5\end{vmatrix} = 2\left( 5 - 10 \right) - 1\left( - 15 + 5 \right) + 0 = - 10 + 10 + 0 = 0\]

So, the given lines are coplanar.
The equation of the plane containing the given lines is

\[\begin{vmatrix}x - \left( - 3 \right) & y - 1 & z - 5 \\ - 3 & 1 & 5 \\ - 1 & 2 & 5\end{vmatrix} = 0\]
\[ \Rightarrow \begin{vmatrix}x + 3 & y - 1 & z - 5 \\ - 3 & 1 & 5 \\ - 1 & 2 & 5\end{vmatrix} = 0\]
\[ \Rightarrow \left( x + 3 \right)\left( 5 - 10 \right) - \left( y - 1 \right)\left( - 15 + 5 \right) + \left( z - 5 \right)\left( - 6 + 1 \right) = 0\]
\[ \Rightarrow - 5\left( x + 3 \right) + 10\left( y - 1 \right) - 5\left( z - 5 \right) = 0\]
\[ \Rightarrow x - 2y + z = 0\]

Thus, the equation of the plane containing the given lines is x − 2y + z = 0.