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Manipal Engineering Manipal Engineering Solved Paper-2010

  • question_answer
    Consider a tetrahedron with faces\[{{F}_{1}},\,\,{{F}_{2}},\,\,{{F}_{3}},\,\,{{F}_{4}}\]. Let\[{{V}_{1}},\,\,{{V}_{2}},\,\,{{V}_{3}},\,\,{{V}_{4}}\]be the vectors whose magnitudes are respectively equal to areas of\[{{F}_{1}},\,\,{{F}_{2}},\,\,{{F}_{3}},\,\,{{F}_{4}}\] and whose directions are perpendicular to these faces in outward direction, then\[|{{\overset{\to }{\mathop{\mathbf{V}}}\,}_{\mathbf{1}}}\mathbf{+}{{\overset{\to }{\mathop{\mathbf{V}}}\,}_{\mathbf{2}}}\mathbf{+}{{\overset{\to }{\mathop{\mathbf{V}}}\,}_{\mathbf{3}}}\mathbf{+}{{\overset{\to }{\mathop{\mathbf{V}}}\,}_{\mathbf{4}}}|\]equals

    A)  1                                            

    B)  4

    C)  \[\vec{0}\]                        

    D)         None of these

    Correct Answer: C

    Solution :

    We have, \[{{\mathbf{\vec{V}}}_{1}}=\frac{1}{2}(\mathbf{\vec{a}}\times \mathbf{\vec{b}}),\,\,{{\mathbf{\vec{V}}}_{2}}=\frac{1}{2}(\mathbf{\vec{b}}\times \mathbf{\vec{c}}),\]\[{{\mathbf{\vec{V}}}_{3}}=\frac{1}{2}(\mathbf{\vec{c}}\times \mathbf{\vec{a}})\]and \[{{\mathbf{\vec{V}}}_{4}}=\frac{1}{2}\{(\mathbf{\vec{c}}-\mathbf{\vec{a}})\times (\mathbf{\vec{b}}-\mathbf{\vec{a}})\}\] \[\therefore \]\[{{\mathbf{\vec{V}}}_{1}}+{{\mathbf{\vec{V}}}_{2}}+{{\mathbf{\vec{V}}}_{3}}+{{\mathbf{\vec{V}}}_{4}}=\frac{1}{2}(\mathbf{\vec{a}}\times \mathbf{\vec{b}}+\mathbf{\vec{b}}\times \mathbf{\vec{c}}+\mathbf{\vec{c}}\times \mathbf{\vec{a}}+\]                                                                         \[+\mathbf{\vec{c}}\times \mathbf{\vec{b}}-\mathbf{\vec{c}}\times \mathbf{\vec{a}}-\mathbf{\vec{a}}\times \mathbf{\vec{b}})=0\] \[|{{\mathbf{\vec{V}}}_{1}}+{{\mathbf{\vec{V}}}_{\mathbf{2}}}+{{\mathbf{\vec{V}}}_{3}}+{{\mathbf{\vec{V}}}_{4}}|\,\,=0\]


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