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

  • question_answer
    The d.r-s of normal to the plane through \[(1,\,0,\,0),\,\,(0,\,1,\,0)\] which makes an angle \[\frac{\pi }{4}\] with plane \[x+y=3\], are [AIEEE 2002]

    A) \[1,\sqrt{2},1\]

    B) 1,1, \[\sqrt{2}\]

    C) 1, 1, 2

    D) \[\sqrt{2},\,1,\,1\]

    Correct Answer: B

    Solution :

    • The plane by intercept form is \[\frac{x}{1}+\frac{y}{1}+\frac{z}{c}=1\].                   
    • D.r-s of normal are 1,1,\[\frac{1}{c}\]and of given plane are 1,1, 0. Now, \[\cos \frac{\pi }{4}=\frac{1.1+1.1+\frac{1}{c}.0}{\left( \sqrt{\frac{1}{{{c}^{2}}}+2} \right)\,\sqrt{2}}\]Þ \[\frac{1}{\sqrt{2}}=\frac{2}{\left( \sqrt{\frac{1}{{{c}^{2}}}+2} \right)\,\,\sqrt{2}}\]           
    • Þ\[\frac{1}{{{c}^{2}}}+2=4\Rightarrow {{c}^{2}}=\frac{1}{2}\]Þ \[c=\frac{1}{\sqrt{2}}\]                   
    • \  D.r-s of required normal are 1, 1, \[\sqrt{2}\].


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