Question

Find the equation of the plane through the intersection of the planes 3x − 4y + 5z = 10 and 2x + 2y − 3z = 4 and parallel to the line x = 2y = 3z.

 

Answer 1

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

\[\left( 3x - 4y + 5z - 10 \right) + \lambda \left( 2x + 2y - 3z - 4 \right) = 0\]

\[ \Rightarrow \left( 3 + 2\lambda \right) x + \left( - 4 + 2\lambda \right) y + \left( 5 - 3\lambda \right) z - 10 - 4\lambda = 0 . . . \left( 1 \right)\]

\[\text{ The given line is} \]

\[x = 2y = 3z\]

\[\text{ Dividing this equation by 6, we get} \]

\[\frac{x}{6} = \frac{y}{3} = \frac{z}{2}\]

\[\text{ The direction ratios of this line are proportional to 6, 3, 2} .\]

\[\text{ So, the normal to the plane is perpendicular to the line whose direction ratios are proportional to6, 3, 2 .}  \]

\[ \Rightarrow \left( 3 + 2\lambda \right) 6 + \left( - 4 + 2\lambda \right) 3 + \left( 5 - 3\lambda \right) 2 = 0\]

\[ \Rightarrow 18 + 12\lambda - 12 + 6\lambda + 10 - 6\lambda = 0\]

\[ \Rightarrow 12\lambda + 16 = 0\]

\[ \Rightarrow \lambda = \left( \frac{- 4}{3} \right)\]

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

\[\left( 3 + 2\left( \frac{- 4}{3} \right) \right) x + \left( - 4 + 2 \left( \frac{- 4}{3} \right) \right) y + \left( 5 - 3 \left( \frac{- 4}{3} \right) \right) z - 10 - 4 \left( \frac{- 4}{3} \right) = 0\]

\[ \Rightarrow x - 20y + 27z = 14\]