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Manipal Engineering Manipal Engineering Solved Paper-2009

  • question_answer
    The volume of the tetrahedron having the edges\[\widehat{\mathbf{i}}+2\widehat{\mathbf{j}}-\widehat{\mathbf{k}},\,\,\widehat{\mathbf{i}}+\widehat{\mathbf{j}}+\widehat{\mathbf{k}},\,\,\widehat{\mathbf{i}}-\widehat{\mathbf{j}}+\lambda \widehat{\mathbf{k}}\]as coterminous, is\[\frac{2}{3}\]cubic unit. Then\[\lambda \]equals

    A)  1                                            

    B)  2

    C)  3                                            

    D)  4

    Correct Answer: A

    Solution :

    Let\[\overrightarrow{\mathbf{a}}=\widehat{\mathbf{i}}+2\widehat{\mathbf{j}}-\widehat{\mathbf{k}},\,\,\overrightarrow{\mathbf{b}}=\widehat{\mathbf{i}}+\widehat{\mathbf{j}}+\widehat{\mathbf{k}}\] and\[\overrightarrow{\mathbf{c}}=\widehat{\mathbf{i}}-\widehat{\mathbf{j}}+\lambda \widehat{\mathbf{k}}\] Since, volume of tetrahedron\[=\frac{1}{6}[\overrightarrow{\mathbf{a}}\overrightarrow{\mathbf{b}}\overrightarrow{\mathbf{c}}]\] \[\Rightarrow \]               \[\frac{2}{3}=\frac{1}{6}\left| \begin{matrix}    1 & 2 & -1  \\    1 & 1 & 1  \\    1 & -1 & \lambda   \\ \end{matrix} \right|\] \[\Rightarrow \]               \[\frac{2}{3}=\frac{1}{6}[1(\lambda +1)-2(\lambda -1)-1(1-1)]\] \[\Rightarrow \]               \[4=[-\lambda +5]\] \[\Rightarrow \]               \[\lambda =1\]


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