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

JEE Main & Advanced Mathematics Vector Algebra Question Bank Critical Thinking

  • question_answer
    \[[(\mathbf{a}\times \mathbf{b})\times (\mathbf{b}\times \mathbf{c})\,(\mathbf{b}\times \mathbf{c})\times (\mathbf{c}\times \mathbf{a})\,(\mathbf{c}\times \mathbf{a})\times (\mathbf{a}\times \mathbf{b})]=\,\]

    A) \[{{[\mathbf{a}\,\,\mathbf{b}\,\,\mathbf{c}]}^{2}}\]

    B) \[{{[\mathbf{a}\,\,\mathbf{b}\,\,\mathbf{c}]}^{3}}\]

    C) \[{{[\mathbf{a}\,\,\mathbf{b}\,\,\mathbf{c}]}^{4}}\]

    D) None of these

    Correct Answer: C

    Solution :

    • We have \[(\mathbf{a}\times \mathbf{b})\times (\mathbf{b}\times \mathbf{c})\]          
    • \[=((\mathbf{a}\times \mathbf{b})\,.\,\mathbf{c})\mathbf{b}-((\mathbf{a}\times \mathbf{b})\,.\,\mathbf{b})\mathbf{c}=[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]\mathbf{b}\]                   
    • \[(\mathbf{b}\times \mathbf{c})\times (\mathbf{c}\times \mathbf{a})=((\mathbf{b}\times \mathbf{c})\,.\,\mathbf{a})\mathbf{c}-((\mathbf{b}\times \mathbf{c})\,.\,\mathbf{c})\mathbf{a}=[\,\mathbf{b}\,\mathbf{c}\,\mathbf{a}]\,\mathbf{c}\]                   
    • \[(\mathbf{c}\times \mathbf{a})\times (\mathbf{a}\times \mathbf{b})=((\mathbf{c}\times \mathbf{a})\,.\,\mathbf{b})\mathbf{a}-((\mathbf{c}\times \mathbf{a})\,.\,\mathbf{a})\mathbf{b}=[\mathbf{c}\,\mathbf{a}\,\mathbf{b}]\,\mathbf{a}\]                   
    • \[\therefore \,\,\,[(\mathbf{a}\times \mathbf{b})\times (\mathbf{b}\times \mathbf{c})(\mathbf{b}\times \mathbf{c})\times (\mathbf{c}\times \mathbf{a})(\mathbf{c}\times \mathbf{a})\times (\mathbf{a}\times \mathbf{b})]\]           
    • \[=[[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]\,\mathbf{a}\,[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]\,\mathbf{b}\,[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]\,\mathbf{c}]\]\[={{[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]}^{3}}[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]={{[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]}^{4}}.\]


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