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

  • question_answer
    The d. r. of normal to the plane through (1, 0, 0), (0, 1, 0) which makes an angle \[\pi /4\] with 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] Equation of plane through (1, 0, 0) is \[a(x-1)+by+cz=0\]                      (i) (i) Passes through (0, 1, 0). \[-a+b=0\] \[\Rightarrow b=a;\] Also, \[\cos 45{}^\circ =\frac{a+a}{\sqrt{2(2{{a}^{2}}+{{c}^{2}})}}\Rightarrow 2a=\sqrt{2{{a}^{2}}+{{c}^{2}}}\] \[\Rightarrow 2{{a}^{2}}={{c}^{2}}\Rightarrow c=\sqrt{2a}.\] So. d. r of normal area \[a,a\sqrt{2a}\] i.e., \[1,1,\sqrt{2}.\]


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