JEE Main & Advanced Mathematics Three Dimensional Geometry Question Bank Self Evaluation Test - Three Dimensional Geometry

  • question_answer
    The angle between the pair of planes represented by equation\[2{{x}^{2}}-2{{y}^{2}}+4{{z}^{2}}+6xz+2yz+3xy=0\]is  

    A) \[{{\cos }^{-1}}\left( \frac{1}{3} \right)\]

    B) \[{{\cos }^{-1}}\left( \frac{4}{21} \right)\]

    C) \[{{\cos }^{-1}}\left( \frac{4}{9} \right)\]

    D) \[{{\cos }^{-1}}\left( \frac{7}{\sqrt{84}} \right)\]

    Correct Answer: C

    Solution :

    [c] \[2{{x}^{2}}-2{{y}^{2}}+4{{z}^{2}}+(6xz+2yz+3xy)=0\] or     \[2{{x}^{2}}+x(6z+3y)-2{{y}^{2}}+4{{z}^{2}}+2yz=0\] \[x=\frac{-(6z+3y)\pm \sqrt{36{{z}^{2}}+9{{y}^{2}}+36yz-8(-2{{y}^{2}}+4{{z}^{2}}+2yz)}}{4}\]\[x=\frac{-(6z+3y)\pm \sqrt{{{(2z+5y)}^{2}}}}{4}\] \[\Rightarrow x=\frac{-(6z+3y)\pm (2z+5y)}{4}\] or         \[2x-y+2z=0,x+2y+2z=0\] \[\therefore \] Angle between planes is \[{{\cos }^{-1}}\left( \frac{4}{9} \right).\]


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