Question

Find the equation of the plane that is perpendicular to the plane 5x + 3y + 6z + 8 = 0 and which contains the line of intersection of the planes x + 2y + 3z − 4 = 0, 2x + y − z + 5 = 0.

 

Answer 1

\[ \text{ The equation of the plane passing through the line of intersection of the given planes is } \]

\[x + 2y + 3z - 4 + \lambda \left( 2x + y - z + 5 \right) = 0 \]

\[\left( 1 + 2\lambda \right)x + \left( 2 + \lambda \right)y + \left( 3 - \lambda \right)z - 4 + 5\lambda = 0 . . . \left( 1 \right)\]

\[\text{ This plane is perpendicular to 5x + 3y + 6z + 8 = 0 . So } ,\]

\[5 \left( 1 + 2\lambda \right) + 3 \left( 2 + \lambda \right) + 6 \left( 3 - \lambda \right) = \text{ 0 } (\text{ Because }  a_1 a_2 + b_1 b_2 + c_1 c_2 = 0)\]

\[ \Rightarrow 5 + 10\lambda + 6 + 3\lambda + 18 - 6\lambda = 0\]

\[ \Rightarrow 7\lambda + 29 = 0\]

\[ \Rightarrow \lambda = \frac{- 29}{7}\]

\[ \text{ Substituting this in (1), we get } \]

\[\left( 1 + 2 \left( \frac{- 29}{7} \right) \right)x + \left( 2 - \frac{29}{7} \right)y + \left( 3 + \frac{29}{7} \right)z - 4 + 5 \left( \frac{- 29}{7} \right) = 0\]

\[ \Rightarrow - 51x - 15y + 50z - 173 = 0\]

\[ \Rightarrow 51x + 15y - 50z + 173 = 0\]