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Question
Show that the lines \[\frac{5 - x}{- 4} = \frac{y - 7}{4} = \frac{z + 3}{- 5} \text { and } \frac{x - 8}{7} = \frac{2y - 8}{2} = \frac{z - 5}{3}\] are coplanar.
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Solution
\[\text { We know that the lines } \frac{x - x_1}{a_1} = \frac{y - y_1}{b_1} = \frac{z - z_1}{c_1} \text { and } \frac{x - x_2}{a_2} = \frac{y - y_2}{b_2} = \frac{z - z_2}{c_2} are coplanar if \]
\[\begin{vmatrix}x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2\end{vmatrix} = 0\]
The given lines, when converted to their standard forms, are written as:
\[\frac{x - 5}{4} = \frac{y - 7}{4} = \frac{z + 3}{- 5} \text { and } \frac{x - 8}{7} = \frac{y - 4}{1} = \frac{z - 5}{3}\]
Here, x1 = 5; y1 =7; z1 = −3; x2 = 8; y2 = 4; z2 = 5; a1 = 4; b1 = 4; c1 = − 5; a2 = 7; b2 = 1 and c2 = 3
Now,
\[\begin{vmatrix}8 - 5 & 4 - 7 & 5 + 3 \\ 4 & 4 & - 5 \\ 7 & 1 & 3\end{vmatrix}\]
\[= \begin{vmatrix}3 & - 3 & 8 \\ 4 & 4 & - 5 \\ 7 & 1 & 3\end{vmatrix}\]
= 3(12 + 5) + 3(12 + 35) + 8(4 \[-\] 28)
= 51 + 141\[-\]192
= 0
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