Question

Find the equation of the plane passing through the intersection of the planes 2x + 3y − z+ 1 = 0 and x + y − 2z + 3 = 0 and perpendicular to the plane 3x − y − 2z − 4 = 0.

 

Answer 1

\[\text{ The equation of the plane passing through the line of intersection of the given planes is } \]
\[2x + 3y - z + 1 + \lambda \left( x + y - 2z + 3 \right) = 0 \]
\[\left( 2 + \lambda \right)x + \left( 3 + \lambda \right)y + \left( - 1 - 2\lambda \right)z + 1 + 3\lambda = 0 . . . \left( 1 \right)\]
\[\text{This plane is perpendicular to 3x - y - 2z - 4 = 0 . So } ,\]
\[3 \left( 2 + \lambda \right) - \left( 3 + \lambda \right) - 2 \left( - 1 - 2\lambda \right) = \text{ 0 }(\text{ Because } a_1 a_2 + b_1 b_2 + c_1 c_2 = 0)\]
\[ \Rightarrow 6 + 3\lambda - 3 - \lambda + 2 + 4\lambda = 0\]
\[ \Rightarrow 6\lambda + 5 = 0\]
\[ \Rightarrow \lambda = \frac{- 5}{6}\]
\[\text{ Substituting this in (1), we get } \]
\[( 2 - \frac{5}{6} ) x + ( 3 - \frac{5}{6} )y + ( - 1 - 2 ( \frac {- 5}{6} ) )z + 1 + 3 ( \frac{- 5}{6}) = 0\]
\[ \Rightarrow 7X + 13Y + 4z - 9 = 0\]