2. Solved Papers
  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2005

  • question_answer
    The equation of the plane through the point \[(2,-1,-3)\]and parallel to the lines\[\frac{x-1}{3}=\frac{y+2}{2}=\frac{z}{-4}\]and\[\frac{x}{2}=\frac{y-1}{-3}=\frac{z-2}{2}\]is:

    A)  \[8x+14y+13z+37=0\]

    B)  \[8x-14y+13z+37=0\]

    C)  \[8x+14y-13z+37=0\]

    D)  \[8x+14y+13z-37=0\]

    E)  \[8x-14y-13z-37=0\]

    Correct Answer: A

    Solution :

    Given equation of lines are\[\frac{x-1}{3}=\frac{y+2}{2}=\frac{z}{-4}\]. and\[\frac{x}{2}=\frac{y-1}{-3}=\frac{z-2}{2}\] Equation of plane is \[a(x-2)+b(y+1)+c(z+3)=0\]            ...(i) Now, given lines are parallel to it. \[\therefore \]  \[3a+2b-4c=0\]           ...(ii) and        \[2a-3b+2c=0\]            ...(iii) Elimination of a, b and c gives \[\left| \begin{matrix}    x-2 & y+1 & z+3  \\    3 & 2 & -4  \\    2 & -3 & 2  \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[[(x-2)(4-12)-(y+1)(6+8)+\]\[(z+3)(-9-4)]=0\] \[\Rightarrow \]\[-8x+16-14y-14-13z-39=0\] \[\Rightarrow \]               \[8x+14y+13z=-37\] \[\Rightarrow \]               \[8x+14y+13z+37=0\]


You need to login to perform this action.
You will be redirected in 3 sec spinner