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

  • question_answer
    The area of a triangle whose vertices are \[A\,(1,\,-1,\,2),\] \[B\,(2,\,1,\,-1)\] and \[C\,(3,\,-1,\,2)\] is [MNR 1983; IIT 1983]

    A)                 13

    B)                 \[\sqrt{13}\]

    C)                 6

    D)                 \[\sqrt{6}\]

    Correct Answer: B

    Solution :

               Here \[\overrightarrow{OA}=\mathbf{i}-\mathbf{j}+2\mathbf{k}\] and \[\overrightarrow{OB}=2\mathbf{i}+\mathbf{j}-\mathbf{k}\] and \[\overrightarrow{OC}=3\mathbf{i}-\mathbf{j}+2\mathbf{k}\]            These implies \[\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\mathbf{i}+2\mathbf{j}-3\mathbf{k}\]            and                             \[\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=2\mathbf{i}\]            Hence required area is given by \[=\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|\]            \[\overrightarrow{AB}\times \overrightarrow{AC}=\left| \begin{matrix}    \mathbf{i} & \mathbf{j} & \mathbf{k}  \\    1 & 2 & -3  \\    2 & 0 & 0  \\ \end{matrix} \right|=-2(3\mathbf{j}+2\mathbf{k})\]                                 \[\Rightarrow \]Area of triangle \[=\frac{1}{2}\times 2|3\mathbf{j}+2\mathbf{k}|=\sqrt{13}\].


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