JEE Main & Advanced Mathematics Three Dimensional Geometry Question Bank Plane

  • question_answer
    A plane meets the co-ordinate axes in \[A,B,C\] and \[(\alpha ,\beta ,\gamma )\] is the centered of the triangle \[ABC\]. Then  the equation of the plane is [MP PET 2004]

    A)            \[\frac{x}{\alpha }+\frac{y}{\beta }+\frac{z}{\gamma }=3\]

    B)            \[\frac{x}{\alpha }+\frac{y}{\beta }+\frac{z}{\gamma }=1\]

    C)            \[\frac{3x}{\alpha }+\frac{3y}{\beta }+\frac{3z}{\gamma }=1\]

    D)            \[\alpha x+\beta y+\gamma z=1\]

    Correct Answer: A

    Solution :

               Let the co-ordinates of the points where the plane cuts the axes are (a, 0, 0), (0, b, 0), (0, 0, c). Since centroid is \[(\alpha ,\,\,\beta ,\,\,\gamma ),\] therefore \[a=3\alpha ,\,\] \[b=3\beta ,\,\,c=3\gamma .\]            Equation of the plane will be \[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\]                    \[\Rightarrow \,\,\frac{x}{3\alpha }+\frac{y}{3\beta }+\frac{z}{3\gamma }=1\,\,\Rightarrow \,\,\frac{x}{\alpha }+\frac{y}{\beta }+\frac{z}{\gamma }=3.\]

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