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

  • question_answer
    \[\mathbf{a},\,\mathbf{b}\] and c are three vectors with magnitude \[|\mathbf{a}|\,=4,\] \[|\mathbf{b}|\,=4,\] \[|\mathbf{c}|\,=2\] and such that \[\mathbf{a}\] is perpendicular to \[(\mathbf{b}+\mathbf{c}),\,\mathbf{b}\] is perpendicular to \[(\mathbf{c}+\mathbf{a})\] and \[\mathbf{c}\] is perpendicular to \[(\mathbf{a}+\mathbf{b}).\] It follows that  \[|\mathbf{a}+\mathbf{b}+\mathbf{c}|\] is equal to            [UPSEAT 2004]

    A)             9

    B)             6

    C)             5

    D)             4

    Correct Answer: B

    Solution :

               Here |a|=4; \[|\mathbf{b}|=4;\,\,\,|\mathbf{c}|=2\]                    and   \[\mathbf{a}.(\mathbf{b}+\mathbf{c})=0\Rightarrow \mathbf{a}.\mathbf{b}+\mathbf{a}\mathbf{.c}=0\]                       .....(i)                             \[\mathbf{b}.(\mathbf{c}+\mathbf{a})=0\Rightarrow \mathbf{b}\mathbf{.c}+\mathbf{b}\mathbf{.a}=0\]              .....(ii)                           \[\mathbf{c}.(\mathbf{a}+\mathbf{b})=0\Rightarrow \mathbf{c}\mathbf{.a}+\mathbf{c}\mathbf{.b}=0\]                .....(iii)                                 Adding (i), (ii) and (iii), we get,  \[2[\mathbf{a}\mathbf{.b}+\mathbf{b}\mathbf{.c}+\mathbf{c}\mathbf{.a}]=0\]                                 \ \[|\mathbf{a}+\mathbf{b}+\mathbf{c}|=\sqrt{|\mathbf{a}{{|}^{2}}+|\mathbf{b}{{|}^{2}}+|\mathbf{c}{{|}^{2}}+2(\mathbf{a}.\mathbf{b}+\mathbf{b}.\mathbf{c}+\mathbf{c}.\mathbf{a})}\]                                        \[=\sqrt{|\mathbf{a}{{|}^{2}}+|\mathbf{b}{{|}^{2}}+|\mathbf{c}{{|}^{2}}}\]=\[\sqrt{16+16+4}\]                                 Þ \[|\mathbf{a}+\mathbf{b}+\mathbf{c}|=6\].


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