2. Questions Bank
  3. JEE Main & Advanced
  4. Mathematics
  5. Vector Algebra

JEE Main & Advanced Mathematics Vector Algebra Question Bank Critical Thinking

  • question_answer
    If b and c are any two non-collinear unit vectors and a is any vector, then \[(\mathbf{a}\,.\,\mathbf{b})\,\mathbf{b}+(\mathbf{a}\,.\,\mathbf{c})\,\mathbf{c}+\frac{\mathbf{a}\,.\,(\mathbf{b}\times \mathbf{c})}{|\mathbf{b}\times \mathbf{c}|}\,(\mathbf{b}\times \mathbf{c})=\] [IIT 1996]

    A) a

    B) b

    C) c

    D) 0

    Correct Answer: A

    Solution :

    • Let \[\mathbf{i}\] be a unit vector in the direction of \[\mathbf{b},\,\mathbf{j}\] in the direction of \[\mathbf{c}.\] Note that \[\mathbf{b}=\mathbf{i}\] and \[\mathbf{c}=\mathbf{j}\]                   
    • We have \[\mathbf{b}\times \mathbf{c}=\,|\mathbf{b}||\mathbf{c}|\sin \alpha \,\mathbf{k}=\sin \alpha \,\mathbf{k}\], where \[\mathbf{k}\] is a unit vector perpendicular to \[\mathbf{b}\] and \[\mathbf{c}.\]                   
    • \[\Rightarrow \,|\mathbf{b}\times \mathbf{c}|\,=\sin \alpha \Rightarrow \mathbf{k}=\frac{\mathbf{b}\times \mathbf{c}}{|\mathbf{b}\times \mathbf{c}|}\]          Any vector \[\mathbf{a}\] can be written as a linear combination of \[\mathbf{i},\,\,\mathbf{j}\] and \[\mathbf{k}.\]                   
    • Let \[\mathbf{a}={{a}_{1}}\mathbf{i}+{{a}_{2}}\mathbf{j}+{{a}_{3}}\mathbf{k}.\]                   
    • Now   \[\mathbf{a}\,.\,\mathbf{b}=\mathbf{a}\,.\,\mathbf{i}={{a}_{1}},\] \[\mathbf{a}\,.\,\mathbf{c}=\mathbf{a}\,.\,\mathbf{j}={{a}_{2}}\]                                 and \[\mathbf{a}\,.\frac{\mathbf{b}\times \mathbf{c}}{|\mathbf{b}\times \mathbf{c}|}=\mathbf{a}\,.\,\mathbf{k}={{a}_{3}}\]                   
    • Thus \[(\mathbf{a}\,.\,\mathbf{b})\mathbf{b}+(\mathbf{a}\,.\,\mathbf{c})\mathbf{c}+\frac{\mathbf{a}\,.\,(\mathbf{b}\times \mathbf{c})}{|\mathbf{b}\times \mathbf{c}|}(\mathbf{b}\times \mathbf{c}|\]                   
    • \[={{a}_{1}}\mathbf{b}+{{a}_{2}}\mathbf{c}+{{a}_{3}}\frac{(\mathbf{b}\times \mathbf{c})}{|\mathbf{b}\times \mathbf{c}|}={{a}_{1}}\mathbf{i}+{{a}_{2}}\mathbf{j}+{{a}_{3}}\mathbf{k}=\mathbf{a}\].


You need to login to perform this action.
You will be redirected in 3 sec spinner