Question

Find the equation of the plane which contains two parallel lines\[\frac{x - 4}{1} = \frac{y - 3}{- 4} = \frac{z - 2}{5}\text{  and }\frac{x - 3}{1} = \frac{y + 2}{- 4} = \frac{z}{5} .\]

Answer 1

\[ \text{ We know that the equation of the plane containing two given parallel lines } \]
\[\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} \text{ and  }\frac{x - x_2}{a} = \frac{y - y_2}{b} = \frac{z - z_2}{c} is\]
\[\begin{vmatrix}x - x_1 & y - y_1 & z - z_1 \\ x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a & b & c\end{vmatrix} = 0\]
\[\text{ Here } ,\]
\[ x_1 = 4; y_1 = 3; z_1 = 2; x_2 = 3; y_2 = - 2; z_2 = 0; l_1 = 1; m_1 = - 4; n_1 = 5; l_2 = 1; m_2 = - 4; n_2 = 5\]
\[\text{ Now } ,\]
\[\begin{vmatrix}x - 4 & y - 3 & z - 2 \\ 3 - 4 & - 2 - 3 & 0 - 2 \\ 1 & - 4 & 5\end{vmatrix} = 0\]
\[ \Rightarrow \begin{vmatrix}x - 4 & y - 3 & z - 2 \\ 3 - 4 & - 2 - 3 & 0 - 2 \\ 1 & - 4 & 5\end{vmatrix} = 0\]
\[ \Rightarrow - 33 \left( x - 4 \right) + 3 \left( y - 3 \right) + 9 \left( z - 2 \right) = 0\]
\[ \Rightarrow 11 \left( x - 4 \right) - \left( y - 3 \right) - 3 \left( z - 2 \right) = 0\]
\[ \Rightarrow 11x - y - 3z = 35\]