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JCECE Engineering JCECE Engineering Solved Paper-2014

  • question_answer
    The distance of the point \[(-1,\,\,5,\,\,-10)\] from the point   of   intersection   of   the   line \[\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}\] and the plane\[x-y+z=5\], is

    A) \[10\]                                   

    B) \[11\]

    C)  \[12\]                                  

    D)  \[13\]

    Correct Answer: D

    Solution :

    Given line is\[\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}=t\](say) Then, any point on this line is \[(3t+2,\,\,4t-1,\,\,12t+2)\]. This point lies on the plane\[x-y+z=5\]. \[\therefore \]\[3t+2-4t+1+12t+2=5\] \[\Rightarrow \]               \[11t=0\Rightarrow t=0\] The point is\[(2,\,\,-1,\,\,2)\]. Its distance from\[(-1,\,\,-5,\,\,-10)\].                 \[=\sqrt{{{(2+1)}^{2}}+{{(-1+5)}^{2}}+{{(2+10)}^{2}}}\]                 \[=\sqrt{9+16+144}=13\]


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