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JEE Main & Advanced Sample Paper JEE Main Sample Paper-16

  • question_answer
    Let A (1, 1,1), B (2, 3, 5) and C (-1, 0. 2) be three points then equation of plane parallel to plane ABC and at distance 2 unit from origin is

    A) \[2x-3y+z+2\sqrt{14}=0\]

    B) \[2x-3y+z-\sqrt{14}=0\]

    C) \[2x-3y+z+2=0\]              

    D) \[2x-3y+z-2=0\]

    Correct Answer: A

    Solution :

    Dr?s of AB are (1, 2, 4) and dr?s of AC are (-2,-1,1) dr?s of normal to the plane ABC are (2,-3,1) Equation of plane ABC is \[2(x-1)-3(y-1)+1(z-1)=0\] or\[2x-3y+z=0\] Let required plane is \[2x-3y+z+k=0\] then\[\left| \frac{k}{\sqrt{4}+9+1} \right|=2\]or\[k=\pm 2\sqrt{14}\]


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