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JEE Main & Advanced Mathematics Three Dimensional Geometry Question Bank Self Evaluation Test - Three Dimensional Geometry

  • 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) \[\frac{10}{3\sqrt{3}}\]

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

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

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

    Correct Answer: A

    Solution :

    [a] It is obvious that the given line and plane are parallel. Given point on the lie is \[A(2,-2,3).\] \[B(0,0,5)\] is a point the plane. Therefore, \[\xrightarrow[AB]{}=(2-0)\hat{i}+(-2-0)\hat{j}+(3-5)\hat{k}\] Then distance of B from the plane = Projection of \[\xrightarrow[AB]{}\]on vector \[\hat{i}+5\hat{j}+\hat{k}\] \[P=\left| \frac{(2\hat{i}-2\hat{j}-2\hat{k}).(\hat{i}+5\hat{j}+\hat{k})}{\sqrt{1+25+1}} \right|=\left| \frac{2-10-2}{\sqrt{27}} \right|=\frac{10}{3\sqrt{3}}\]


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