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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.
Solution
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
\[ l_1 = 4, m_1 = 4, n_1 = - 5, l_2 = 7, m_2 = 1, n_2 = 3\]
\[ = \begin{vmatrix}8 - 5 & 4 - 7 & 5 - \left( - 3 \right) \\ 4 & 4 & - 5 \\ 7 & 1 & 3\end{vmatrix}\]
\[ = \begin{vmatrix}3 & - 3 & 8 \\ 4 & 4 & - 5 \\ 7 & 1 & 3\end{vmatrix}\]
\[ = 3\left( 12 + 5 \right) + 3\left( 12 + 35 \right) + 8\left( 4 - 28 \right)\]
\[ = 51 + 141 - 192\]
\[ = 0\]
So, the given lines are coplanar.
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