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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2000

  • question_answer
    The shortest distance of the point (2,10,1) from the plane\[\overrightarrow{r}.(3\hat{i}-\hat{j}+4\hat{k})=2\sqrt{26}\]is:

    A)  \[2\sqrt{26}\]                  

    B)  \[\sqrt{26}\]

    C)  \[\frac{1}{\sqrt{26}}\]  

    D)         \[4\]

    E)  \[2\]

    Correct Answer: E

    Solution :

    Point \[\overrightarrow{a}=2\hat{i}+10\hat{j}+\hat{k}\] Equation of plane\[\overrightarrow{r}.(3\hat{i}-\hat{j}+4\hat{k})=2\sqrt{26}\] In cartesion form, it is written as \[3x-y+4z=2\sqrt{26}\] \[\therefore \]Distance\[\left| \frac{6-10+4-2\sqrt{26}}{\sqrt{9+1+16}} \right|\]                 \[=\frac{2\sqrt{26}}{\sqrt{26}}=2\]


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