Question

Show that the lines  \[\frac{5 - x}{- 4} = \frac{y - 7}{4} = \frac{z + 3}{- 5}\] and  \[\frac{x - 8}{7} = \frac{2y - 8}{2} = \frac{z - 5}{3}\] are coplanar.

 

Answer 1

The equations of the given lines can be re-written as\[\frac{x - 5}{4} = \frac{y - 7}{4} = \frac{z + 3}{- 5}\] and \[\frac{x - 8}{7} = \frac{y - 4}{1} = \frac{z - 5}{3}\]  We know that the lines  \[\frac{x - x_1}{l_1} = \frac{y - y_1}{m_1} = \frac{z - z_1}{n_1}\]  and 

\[\frac{x - x_2}{l_2} = \frac{y - y_2}{m_2} = \frac{z - z_2}{n_2}\]  are coplanar if

\[\begin{vmatrix}x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2\end{vmatrix} = 0\]  Here, 

\[x_1 = 5, y_1 = 7, z_1 = - 3, x_2 = 8, y_2 = 4, z_2 = 5\]
\[ l_1 = 4, m_1 = 4, n_1 = - 5, l_2 = 7, m_2 = 1, n_2 = 3\]

\[\therefore \begin{vmatrix}x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2\end{vmatrix}\]
\[ = \begin{vmatrix}8 - 5 & 4 - 7 & 5 - \left( - 3 \right) \\ 4 & 4 & - 5 \\ 7 & 1 & 3\end{vmatrix}\]
\[ = \begin{vmatrix}3 & - 3 & 8 \\ 4 & 4 & - 5 \\ 7 & 1 & 3\end{vmatrix}\]
\[ = 3\left( 12 + 5 \right) + 3\left( 12 + 35 \right) + 8\left( 4 - 28 \right)\]
\[ = 51 + 141 - 192\]
\[ = 0\]

So, the given lines are coplanar.