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.

Answer 1

\[\text { We know that the lines } \frac{x - x_1}{a_1} = \frac{y - y_1}{b_1} = \frac{z - z_1}{c_1} \text { and } \frac{x - x_2}{a_2} = \frac{y - y_2}{b_2} = \frac{z - z_2}{c_2} are coplanar if \]

\[\begin{vmatrix}x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2\end{vmatrix} = 0\]

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

Thus, the given lines are coplanar.