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JEE Main & Advanced Mathematics Vector Algebra Question Bank Scalar triple product and their applications

  • question_answer
    If a,b,c are three non-zero, non-coplanar vectors and \[{{b}_{1}}=b-\frac{b.a}{|a{{|}^{2}}}a,\,{{b}_{2}}=b+\frac{b.a}{|a{{|}^{2}}}a\],\[{{c}_{1}}=c-\frac{c.a}{|a{{|}^{2}}}a-\frac{c.b}{|b{{|}^{2}}}b\], \[{{c}_{2}}=c-\frac{c.a}{|a{{|}^{2}}}a\frac{c.{{b}_{1}}}{|{{b}_{1}}{{|}^{2}}}{{b}_{1}}\], \[{{c}_{3}}=c-\frac{c.a}{|a{{|}^{2}}}a\frac{c.{{b}_{2}}}{|{{b}_{2}}{{|}^{2}}}{{b}_{2}}\], \[{{c}_{4}}=a-\frac{c.a}{|a{{|}^{2}}}a\]. Then which of the following is a set of mutually orthogonal vectors is                        [IIT Screening 2005]

    A)             \[\{\mathbf{a},\,{{\mathbf{b}}_{1}},\,{{c}_{1}}\}\]

    B)             \[\{\mathbf{a},\,{{\mathbf{b}}_{1}},\,{{c}_{2}}\}\]

    C)             \[\{\mathbf{a},\,{{\mathbf{b}}_{2}},\,{{c}_{3}}\}\]

    D)             \[\{\mathbf{a},\,{{\mathbf{b}}_{2}},\,{{c}_{4}}\}\]

    Correct Answer: B

    Solution :

                    We have, \[\mathbf{a}\,.\,{{\mathbf{b}}_{1}}=0\], \[{{\mathbf{b}}_{1}}\,.\,{{\mathbf{c}}_{2}}=0\], \[\mathbf{a}\,.\,{{\mathbf{c}}_{2}}=0\]                                 \[\therefore \] Set of orthogonal vectors, \[[\mathbf{a}\,\,{{\mathbf{b}}_{1}}\,\,{{\mathbf{c}}_{2}}]=0\]                                 \[\therefore \] Option  is the correct answer.


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