JEE Main & Advanced Mathematics Vector Algebra Question Bank Self Evaluation Test - Vector Algebra

  • question_answer
    Resolved part of vector \[\vec{a}\] along vector \[\vec{b}\] is \[{{\vec{a}}_{1}}\] and that perpendicular to \[\vec{b}\] is \[{{\vec{a}}_{2}}\] then \[{{\vec{a}}_{1}}\times {{\vec{a}}_{2}}\] is equal to

    A) \[\frac{(\vec{a}\times \vec{b})\cdot \vec{b}}{{{\left| {\vec{b}} \right|}^{2}}}\]

    B) \[\frac{(\vec{a}\cdot \vec{b})\vec{a}}{{{\left| {\vec{a}} \right|}^{2}}}\]

    C) \[\frac{(\vec{a}\cdot \vec{b})(\vec{b}\times \vec{a})}{{{\left| {\vec{b}} \right|}^{2}}}\]

    D) \[\frac{(\vec{a}\cdot \vec{b})(\vec{b}\times \vec{a})}{\left| \vec{b}\times \vec{a} \right|}\]

    Correct Answer: C

    Solution :

    [c] \[{{\vec{a}}_{1}}=(\vec{a}\cdot \hat{b})\hat{b}=\frac{(\vec{a}\cdot \vec{b})\vec{b}}{|\vec{b}{{|}^{2}}}\] \[\Rightarrow {{\vec{a}}_{2}}=\vec{a}-{{\vec{a}}_{1}}=\vec{a}-\frac{(\vec{a}\cdot \vec{b})\vec{b}}{|\vec{b}{{|}^{2}}}\] Thus, \[{{\vec{a}}_{1}}\times {{\vec{a}}_{2}}=\frac{(\vec{a}\cdot \vec{b})\vec{b}}{|\vec{b}{{|}^{2}}}\times \left( \vec{a}-\frac{(\vec{a}\cdot \vec{b})\vec{b}}{|\vec{b}{{|}^{2}}} \right)\] \[=\frac{(\vec{a}\cdot \vec{b})(\vec{b}\times a)}{|\vec{b}{{|}^{2}}}\]


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