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

  • question_answer
    The direction ratios (d,r,'s) of the normal to the plane throuth (1, 0, 0) and (0, 1, 0) which makes an angle \[\pi /4\]with the plane \[x+y=3\]are

    A) \[1,\,\sqrt{2,}\,1\]          

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

    C) \[1,\,\,1\,,\,\,2\]  

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

    Correct Answer: B

    Solution :

    [b] The equation of the plane through (1, 0, 0) is \[a(x-1)+by+cz=0\] (i) Passes through, (0, 1, 0) \[-a+b=0\Rightarrow b=a\] Also, \[\cos 45{}^\circ =\frac{a+a}{\sqrt{2(2{{a}^{2}}+{{c}^{2}})}}\] \[\Rightarrow 2a=\sqrt{2{{a}^{2}}+{{c}^{2}}}\] \[\Rightarrow 2{{a}^{2}}={{c}^{2}}\] \[\Rightarrow c=\sqrt{2}a\] So the d.r.'s of the normal are \[a,a,\sqrt{2}a,\]i.e., 1, 1,\[\sqrt{2}\].


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