J & K CET Engineering J and K - CET Engineering Solved Paper-2005

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

    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 o planes as \[(\hat{i}-\hat{j}+4\hat{k}).(\hat{i}+5\hat{j}+\hat{k})=0\] Any point on the given line is \[(\lambda +2,\,-\lambda -2,\,4\lambda +3).\] For \[\lambda =0\] a point on this line is \[(2,-2,3)\] and the distance from \[\vec{r}.(\hat{i}+5\hat{j}+\hat{k})=5\] or \[x+5y+z=5\] is \[d=\frac{|2+5(-2)+3-5|}{\sqrt{1+25+1}}=\frac{|-10|}{3\sqrt{3}}=\frac{10}{3\sqrt{3}}\]


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