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प्रश्न
The equation of the plane through the line x + y + z + 3 = 0 = 2x − y + 3z + 1 and parallel to the line \[\frac{x}{1} = \frac{y}{2} = \frac{z}{3}\] is
विकल्प
x − 5y + 3z = 7
x − 5y + 3z = −7
x + 5y + 3z = 7
x + 5y + 3z = −7
उत्तर
x − 5y + 3z = 7
\[\text{ The equation of the plane passing through the line of intersection of the given planes is } \]
\[x + y + z + 3 + \lambda \left( 2x - y + 3z + 1 \right) = 0 \]
\[\left( 1 + 2\lambda \right)x + \left( 1 - \lambda \right)y + \left( 1 + 3\lambda \right)z + 3 + \lambda = 0 . . . \left( 1 \right)\]
\[\text{ This plane is parallel to the line} \frac{x}{1}=\frac{y}{2}=\frac{z}{3} . \text{ It means that this line is perpendicular to the normal of the plane (1) } .\]
\[ \Rightarrow 1 \left( 1 + 2\lambda \right) + 2 \left( 1 - \lambda \right) + 3 \left( 1 + 3\lambda \right) = 0.................... (\text{ Because }a_1 a_2 + b_1 b_2 + c_1 c_2 = 0)\]
\[ \Rightarrow 1 + 2\lambda + 2 - 2\lambda + 3 + 9\lambda = 0\]
\[ \Rightarrow 9\lambda + 6 = 0\]
\[ \Rightarrow \lambda = \frac{- 2}{3}\]
\[\text{ Substituting this in (1), we get } \]
\[\left( 1 + 2 \left( \frac{- 2}{3} \right) \right)x + \left( 1 - \left( \frac{- 2}{3} \right) \right)y + \left( 1 + 3 \left( \frac{- 2}{3} \right) \right)z + 3 + \left( \frac{- 2}{3} \right) = 0\]
\[ \Rightarrow - x + 5y - 3z + 7 = 0\]
\[ \Rightarrow x - 5y + 3z = 7\]
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