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

  • question_answer
    If \[\overset{\to }{\mathop{a}}\,,\,\overset{\to }{\mathop{b}}\,,\,\overset{\to }{\mathop{c}}\,\] are three non-coplanar vectors, then the value of \[\frac{\overset{\to }{\mathop{a}}\,.(\overset{\to }{\mathop{b}}\,\times \overset{\to }{\mathop{c}}\,)}{(\overset{\to }{\mathop{c}}\,\times \overset{\to }{\mathop{a}}\,).\overset{\to }{\mathop{b}}\,}+\frac{\overset{\to }{\mathop{b}}\,.(\overset{\to }{\mathop{a}}\,\times \overset{\to }{\mathop{c}}\,)}{\overset{\to }{\mathop{c}}\,.(\overset{\to }{\mathop{a}}\,\times \overset{\to }{\mathop{b}}\,)}\] is:

    A) 0

    B) 2

    C) 1

    D) None of these

    Correct Answer: A

    Solution :

    [a] By definition of scalar triple product \[\vec{a}\cdot (\vec{b}\times \vec{c})\] can be written as \[[\vec{a}\vec{b}\vec{c}]\] \[\frac{\vec{a}.(\vec{b}\times \vec{c})}{(\vec{c}\times \vec{a}).\vec{b}}+\frac{\vec{b}.(\vec{a}\times \vec{c})}{\vec{c}.(\vec{a}\times \vec{b})}=\frac{[\vec{a}\vec{b}\vec{c}]}{[\vec{c}\vec{a}\vec{b}]}+\frac{[\vec{b}\vec{a}\vec{c}]}{[\vec{c}\vec{a}\vec{b}]}\] \[=\frac{[\vec{a}\vec{b}\vec{c}]}{[\vec{a}\vec{b}\vec{c}]}-\frac{[\vec{a}\vec{b}\vec{c}]}{[\vec{a}\vec{b}\vec{c}]}=1-1=0\] \[\because [\vec{a}\vec{b}\vec{c}]=[\vec{b}\vec{c}\vec{a}]=[\vec{c}\vec{a}\vec{b}]\] but \[[\vec{b}\vec{c}\vec{a}]=-\left[ \vec{a}\vec{b}\vec{c} \right]\]


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