JEE Main & Advanced JEE Main Paper (Held On 9 April 2016)

  • question_answer
    The distance of the point (1, .2, 4) from the plane passing through the point (1, 2, 2) and perpendicular to the planes \[x-y+2z=3\] and \[2x-2y+z+12=0\], is     JEE Main Online Paper (Held On 09 April 2016)

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

    B) 2

    C) \[\sqrt{2}\]                                        

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

    Correct Answer: D

    Solution :

    Equation of plane\[\bot \] to the planes. x . y + 2z = 3 & 2x . 2y + z + 12 = 0 and passes through (1, 2, 2) is\[\left| \begin{matrix}    x-1 & y-2 & z-2  \\    1 & -1 & 2  \\    2 & -2 & 1  \\ \end{matrix} \right|=0\] 3(x . 1) + 3(y . 2) = 0 x + y = 3    ..... (1) distance of plane x + y . 3 = 0 from (1, . 2, 4) is\[\left| \frac{1-2-3}{\sqrt{1+1}} \right|=2\sqrt{2}\]


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