Question

Find the equation of a plane passing through the point (−1, −1, 2) and perpendicular to the planes 3x + 2y − 3z = 1 and 5x − 4y + z = 5.

 

Answer 1

\[\text{ The equation of any plane passing through  (-1, -1, 2) is } \]
\[a \left( x + 1 \right) + b \left( y + 1 \right) + c \left( z - 2 \right) = 0 . . . \left( 1 \right)\]
\[ \text{ It is given that (1) is perpendicular to each of the planes 3x + 2y - 3z = 1 and 5x - 4y + z = 5 . Then } ,\]
\[3a + 2b - 3c = 0 . . . \left( 2 \right)\]
\[5a - 4b + c = 0 . . . \left( 3 \right)\]
\[\text{ Solving (1), (2) and (3), we get } \]
\[\begin{vmatrix}x + 1 & y + 1 & z - 2 \\ 3 & 2 & - 3 \\ 5 & - 4 & 1\end{vmatrix} = 0\]
\[ \Rightarrow - 10 \left( x + 1 \right) - 18 \left( y + 1 \right) - 22 \left( z - 2 \right) = 0\]
\[ \Rightarrow 5 \left( x + 1 \right) + 9 \left( y + 1 \right) + 11 \left( z - 2 \right) = 0\]
\[ \Rightarrow 5x + 5 + 9y + 9 + 11z - 22 = 0\]
\[ \Rightarrow 5x + 9y + 11z - 8 = 0\]