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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    The distance between the lines\[\overrightarrow{r}=(4\hat{i}-7\hat{j}-9\hat{k})+t(3\hat{i}-7\hat{j}+4\hat{k})\]and\[\overrightarrow{r}=(7\hat{i}-14\hat{j}-5\hat{k})+s(3\hat{i}+7\hat{j}-4\hat{k})\]is equal to

    A)  1                                            

    B)  \[\frac{1}{2}\]

    C)  \[\frac{3}{4}\]                  

    D)         \[17\]

    E)  \[0\]

    Correct Answer: E

    Solution :

    On comparing the given lines with \[\overrightarrow{r}=\overrightarrow{a}+t\overrightarrow{c}\]and\[\overrightarrow{r}=\overrightarrow{b}+s\overrightarrow{d}\] We get, \[\overrightarrow{a}=(4\hat{i}-7\hat{i}-9\hat{k}),\overrightarrow{c}=3\hat{i}-7\hat{j}+4\hat{k}\]                 \[\overrightarrow{b}=7\hat{i}-14\hat{j}-5\hat{k},\overrightarrow{d}=-3\hat{i}+7\hat{j}-4\hat{k}\] Now, \[[\overrightarrow{a}-\overrightarrow{b}\overrightarrow{c}\overrightarrow{d}]=\left| \begin{matrix}    -3 & 7 & -4  \\    3 & -7 & 4  \\    -3 & 7 & -4  \\ \end{matrix} \right|\] \[=0\]            (\[\because \]two rows are identical) \[\therefore \]  Shortest distance\[=\frac{[\overrightarrow{a}-\overrightarrow{b}\overrightarrow{c}\overrightarrow{d}]}{[\overrightarrow{c}\times \overrightarrow{d}]}\]                 \[=0\]


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