Question

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

Answer 1

\[\text{ The equation of the plane passing through the line of intersection of the given planes is } \]
\[x + 2y + 3z - 4 + \lambda \left( 2x + y - z + 5 \right) = 0 \]
\[\left( 1 + 2\lambda \right)x + \left( 2 + \lambda \right)y + \left( 3 - \lambda \right)z - 4 + 5\lambda = 0 . . . \left( 1 \right)\]
\[\text{ This plane is perpendicular to 5x + 3y - 6z + 8 = 0 . So } ,\]
\[5 \left( 1 + 2\lambda \right) + 3\left( 2 + \lambda \right) - 6 \left( 3 - \lambda \right) =\text{  0 }(\text{ Because } a_1 a_2 + b_1 b_2 + c_1 c_2 = 0)\]
\[ \Rightarrow 5 + 10\lambda + 6 + 3\lambda - 18 + 6\lambda = 0\]
\[ \Rightarrow 19\lambda - 7 = 0\]
\[ \Rightarrow \lambda = \frac{7}{19}\]
\[\text{ Substituting this in (1), we get } \]
\[\left( 1 + 2 \left( \frac{7}{19} \right) \right)x + \left( 2 + \frac{7}{19} \right)y + \left( 3 - \frac{7}{19} \right)z - 4 + 5 \left( \frac{7}{19} \right) = 0\]
\[ \Rightarrow 33x + 45y + 50z - 41 = 0\]