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The equation of the plane containing the line $$2x-5y+z=3; x+y+4z=5,$$ and parallel to the plane, $$x+3y+6z=1,$$ is :

JEE · Math · previous-year question

  1. A.$$x+3y+6z=7$$correct
  2. B.$$2x+6y+12z=-13$$
  3. C.$$2x+6y+12z=13$$
  4. D.$$x+3y+6z=-7$$

Answer

A. $$x+3y+6z=7$$

Explanation

Equation of the plane containing the lines $$2x - 5y + z = 3$$ and $$x + y + 4z = 5$$ is $$2x - 5y + z - 3 + \lambda \left( {x + y + 4z - 5} \right) = 0$$ $$ \Rightarrow \left( {2 + \lambda } \right)x + \left( { - 5 + \lambda } \right)y + \left( {1 + 4\lambda } \right)z + $$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( { - 3 - 5\lambda } \right) = 0....\left( i \right)$$ Since the plane $$(i)$$ parallel to the given plane $$x + 3y + 6z = 1$$ $$\therefore$$ $$\,\,\,{{2 + \lambda } \over 1} = {{ - 5 + \lambda } \over 3} = {{1 + 4\lambda } \over 6}$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \lambda = - {{11} \over 2}$$ Hence equation of the required plane is $$\left( {2 - {{11} \over 2}} \right)x + \left( { - 5 - {{11} \over 2}} \right)y + \left( {1 - {{44} \over 2}} \right)z + $$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( { - 3 + {{55} \over 2}} \right) = 0$$ $$ \Rightarrow x + 3y + 6z = 7$$

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