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Question
Find the equation of a plane passing through the points (0, 0, 0) and (3, −1, 2) and parallel to the line \[\frac{x - 4}{1} = \frac{y + 3}{- 4} = \frac{z + 1}{7} .\]
Solution
\[\text{ The equation of the plane through (0,0, 0) is } \]
\[a \left( x - 0 \right) + b \left( y - 0 \right) + c \left( z - 0 \right) = 0 \]
\[ax + by + cz = 0 . . . \left( 1 \right)\]
\[\text{ This plane passes through (3, -1, 2). So } ,\]
\[3a - b + 2c = 0 . . . \left( 2 \right)\]
\[ \text{ Again plane (1) is parallel to the given line } .\]
\[\text{ It means that the normal to plane (1) is perpendicular to the line.}\]
\[ \Rightarrow a \left( 1 \right) + b \left( - 4 \right) + c \left( 7 \right) = 0 . . . \left( 3 \right) (\text{ Because } a_1 a_2 + b_1 b_2 + c_1 c_2 = 0)\]
\[\text{ Solving (1), (2) and (3), we get } \]
\[\begin{vmatrix}x & y & z \\ 3 & - 1 & 2 \\ 1 & - 4 & 7\end{vmatrix} = 0\]
\[ \Rightarrow x - 19y - 11z = 0\]
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