JEE Main & Advanced Mathematics Vector Algebra Question Bank Vector or Cross product of two vectors and its application

  • question_answer
    Three forces \[\mathbf{i}+2\,\mathbf{j}-3\,\mathbf{k},\,\,2\,\mathbf{i}+3\,\mathbf{j}+4\,\mathbf{k}\] and \[\mathbf{i}-\mathbf{j}+\mathbf{k}\] are acting on a particle at the point (0, 1, 2). The magnitude of the moment of the forces about the point \[(1,\,-2,\,0)\] is                 [MNR 1983]

    A)                 \[2\sqrt{35}\]

    B)                 \[6\sqrt{10}\]

    C)                 \[4\sqrt{17}\]

    D)                 None of these

    Correct Answer: B

    Solution :

               Let \[{{\mathbf{F}}_{1}}=\mathbf{i}+2\mathbf{j}-3\mathbf{k},\]\[{{\mathbf{F}}_{2}}=2\mathbf{i}+3\mathbf{j}+4\mathbf{k}\], \[{{\mathbf{F}}_{3}}=\mathbf{i}-\mathbf{j}+\mathbf{k}\].                    \[O\,(0,\,1,\,2)\] and \[P\,(1,\,-2,\,0)\Rightarrow \overrightarrow{OP}=\mathbf{i}-3\mathbf{j}-2\mathbf{k}\]                    Resultant force \[(\mathbf{F})={{\mathbf{F}}_{1}}+{{\mathbf{F}}_{2}}+{{\mathbf{F}}_{3}}=4\mathbf{i}+4\mathbf{j}+2\mathbf{k}\]            Hence moment of force is \[=\overrightarrow{OP}\times \mathbf{F}\]            \[=\left| \begin{matrix}    \mathbf{i} & \mathbf{j} & \mathbf{k}  \\    1 & -3 & -2  \\    4 & 4 & 2  \\ \end{matrix} \right|=2\mathbf{i}-10\mathbf{j}+16\mathbf{k}\]                                 Magnitude of moment of force is  \[|\overrightarrow{OP}\times \mathbf{F}|=\sqrt{4+100+256}=6\sqrt{10}.\]


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