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JEE Main & Advanced Mathematics Three Dimensional Geometry Question Bank Critical Thinking

  • question_answer
    The plane \[lx+my=0\] is rotated an angle \[\alpha \] about its line of intersection with the plane \[z=0\], then the equation to the plane in its new position is

    A) \[lx+my\pm z\sqrt{({{l}^{2}}+{{m}^{2}})}\tan \alpha =0\]

    B) \[lx-my\pm z\sqrt{({{l}^{2}}+{{m}^{2}})}\tan \alpha =0\]

    C) \[lx+my\pm z\sqrt{({{l}^{2}}+{{m}^{2}})}\cos \alpha =0\]

    D) \[lx-my\pm z\sqrt{({{l}^{2}}+{{m}^{2}})}\cos \alpha =0\]

    Correct Answer: A

    Solution :

    • Equation of any plane through the line of intersection of plane \[lx+my=0\]and \[z=0\] is \[lx+my+\lambda z=0\]. Given that angle between plane is
    • \['\alpha '\] so, angle between their normals is \['\pi -\alpha '\], also direction cosines of their normals are l, m, 0 and \[l,m,\lambda \]                           
    • \[\tan \,(\pi -\alpha )=\pm \frac{\sqrt{\Sigma ({{m}_{1}}{{n}_{2}}-{{m}_{2}}{{n}_{1}})}}{\Sigma {{l}_{1}}{{l}_{2}}}\]                                
    • \[-\tan \alpha =\pm \frac{\lambda }{\sqrt{{{l}^{2}}+{{m}^{2}}}}\]                                                              
    • Required equation is \[lx+my\pm z\sqrt{({{l}^{2}}+{{m}^{2}})}\tan \alpha =0\].


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