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The Equation of the Plane Parallel to the Lines X − 1 = 2y − 5 = 2z and 3x = 4y − 11 = 3z − 4 and Passing Through the Point (2, 3, 3) is (A) X − 4y + 2z + 4 = 0 (B) X + 4y + 2z + 4 = 0 - Mathematics

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Question

The equation of the plane parallel to the lines x − 1 = 2y − 5 = 2z and 3x = 4y − 11 = 3z − 4 and passing through the point (2, 3, 3) is

Options

  •  x − 4y + 2z + 4 = 0

  • x + 4y + 2z + 4 = 0

  •  x − 4y + 2z − 4 = 0

  • None of these

     
MCQ
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Solution

 x − 4y + 2z + 4 = 0

\[\text{ Let a, b, c be the direction ratios of the required plane } .\]
\[\text{ The given line equations can be rewritten as } \]
\[\frac{x - 1}{1} = \frac{y - \frac{5}{2}}{\frac{1}{2}} = \frac{z - 0}{\frac{1}{2}} . . . \left( 1 \right)\]
\[\frac{x - 0}{\frac{1}{3}} = \frac{y - \frac{11}{4}}{\frac{1}{4}} = \frac{z - \frac{4}{3}}{\frac{1}{3}} . . . \left( 2 \right)\]
\[\text{ Since the required plane is parallel to the lines (1) and  } (2),\]
\[a + \frac{b}{2} + \frac{c}{2} = 0 \Rightarrow 2a + b + c = 0 . . . \left( 3 \right)\]
\[\frac{a}{3} + \frac{b}{4} + \frac{c}{3} = 0 \Rightarrow 4a + 3b + 4c = 0 . . . \left( 4 \right)\]
\[\text{ Solving (3) and (4) using cross-multiplication method, we get } \]
\[\frac{a}{1} = \frac{b}{- 4} = \frac{c}{2} = \lambda (\text{say} )\]
\[ \Rightarrow a = \lambda, b = - 4\lambda, c = 2\lambda\]
\[\text{ Now, the equation of the plane whose direction ratios are } \lambda, -4\lambda, 2\lambda \text{ and passing through the point (2, 3, 3) is } \]
\[\lambda \left( x - 2 \right) + \left( - 4\lambda \right)\left( y - 3 \right) + 2\lambda \left( z - 3 \right) = 0\]
\[ \Rightarrow x - 4y + 2z + 4 = 0\]

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Chapter 29: The Plane - MCQ [Page 85]

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RD Sharma Mathematics [English] Class 12
Chapter 29 The Plane
MCQ | Q 15 | Page 85

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