JEE Main & Advanced AIEEE Solved Paper-2005

  • question_answer
    The distance between the line\[r=2\hat{i}-2\hat{j}+3\hat{k}+\lambda (\hat{i}-\hat{j}+4\hat{k})\]and the plane\[r.(\hat{i}+5\hat{j}+\hat{k})=5\]is     AIEEE  Solved  Paper-2005

    A) \[\frac{10}{3}\]                                

    B) \[\frac{3}{10}\]                

    C)        \[\frac{10}{3\sqrt{3}}\]                 

    D) \[\frac{10}{9}\]

    Correct Answer: C

    Solution :

    Line is parallel to plane as \[(\hat{i}-\hat{j}+4\hat{k}).(\hat{i}+5\hat{j}+\hat{k})=0\] General point on the line is\[(\lambda +2,-\lambda -2,4\lambda +3)\].For\[\lambda =0,\]a point on this line is (2,-2, 3) and distance from \[r.(\hat{i}+5\hat{j}+\hat{k})=5\]or\[x+5y+z=5\] \[d=\left| \frac{2+5(-2)+3-5}{\sqrt{1+25+1}} \right|\] \[\Rightarrow \]               \[d=\left| \frac{-10}{3\sqrt{3}} \right|=\frac{10}{3\sqrt{3}}\]


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