JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2004

  • question_answer
        If the points whose position vectors are\[\frac{x}{1}=\frac{y}{0}=\frac{z}{-1}\] and \[\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\]lie on a plane, then \[{{\cos }^{-1}}\left( \frac{1}{5} \right)\]is equal to :

    A)  \[{{\cos }^{-1}}\left( \frac{1}{3} \right)\]                              

    B)  \[{{\cos }^{-1}}\left( \frac{1}{2} \right)\]

    C)  \[{{\cos }^{-1}}\left( \frac{1}{4} \right)\]                              

    D)  \[(a-b).(b-c)\times (c-a)\]

    Correct Answer: A

    Solution :

                    Let \[\overrightarrow{a}=3\hat{i}-2\hat{j}-\hat{k},\] \[\overrightarrow{b}=2\hat{i}+3\hat{j}-4\hat{k},\]\[\overrightarrow{c}=-\hat{i}+\hat{j}+2\hat{k}\]and \[\overrightarrow{d}=4\hat{i}+5\hat{j}+\lambda \hat{k}\]are coplanar so, \[[d\,b\,c]+[d\,c\,a]+[d\,a\,b]=[a\,b\,c]\] \[\Rightarrow \]\[\left| \begin{matrix}    4 & 5 & \lambda   \\    2 & 3 & -4  \\    2 & 3 & -4  \\ \end{matrix} \right|+\left| \begin{matrix}    4 & 5 & \lambda   \\    -1 & 1 & 2  \\    3 & -2 & -1  \\ \end{matrix} \right|\]                                 \[\left| \begin{matrix}    4 & 5 & \lambda   \\    3 & -2 & -1  \\    2 & 3 & -4  \\ \end{matrix} \right|=\left| \begin{matrix}    3 & -2 & -1  \\    2 & 3 & -4  \\    -1 & 1 & 2  \\ \end{matrix} \right|\] \[\Rightarrow \]               \[40+5\lambda +37-\lambda +94+13\lambda =25\] \[\Rightarrow \]               \[\lambda =-\frac{146}{17}\]


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