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

  • question_answer
    Let \[\overset{\to }{\mathop{a}}\,,\text{ }\overset{\to }{\mathop{b}}\,\] and \[\overset{\to }{\mathop{c}}\,\] be three non-coplanar vectors, and let \[\overset{\to }{\mathop{p}}\,,\text{ }\overset{\to }{\mathop{q}}\,\] and \[\overset{\to }{\mathop{r}}\,\] be the vectors defined by  the relations \[\overset{\to }{\mathop{p}}\,=\frac{\overset{\to }{\mathop{b}}\,\times \overset{\to }{\mathop{c}}\,}{[\overset{\to }{\mathop{a}}\,\,\overset{\to }{\mathop{b}}\,\,\overset{\to }{\mathop{c}}\,]},\overset{\to }{\mathop{q}}\,=\frac{\overset{\to }{\mathop{c}}\,\times \overset{\to }{\mathop{a}}\,}{[\overset{\to }{\mathop{a}}\,\,\overset{\to }{\mathop{b}}\,\,\overset{\to }{\mathop{c}}\,]}\] and \[\overset{\to }{\mathop{r}}\,=\frac{\overset{\to }{\mathop{a}}\,\times \overset{\to }{\mathop{b}}\,}{[\overset{\to }{\mathop{a}}\,\,\overset{\to }{\mathop{b}}\,\,\overset{\to }{\mathop{c}}\,]}.\] Then the value of the expression \[(\overset{\to }{\mathop{a}}\,+\overset{\to }{\mathop{b}}\,).\overset{\to }{\mathop{p}}\,+(\overset{\to }{\mathop{b}}\,+\overset{\to }{\mathop{c}}\,).\overset{\to }{\mathop{q}}\,+(\overset{\to }{\mathop{c}}\,+\overset{\to }{\mathop{a}}\,).\overset{\to }{\mathop{r}}\,\] is equal to

    A) 0

    B) 1

    C) 2

    D) 3

    Correct Answer: D

    Solution :

    [d] \[\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{p}}\,=\frac{\overset{\to }{\mathop{a}}\,.(\overset{\to }{\mathop{b}}\,\times \overset{\to }{\mathop{c}}\,)}{[\overset{\to }{\mathop{a}}\,\,\overset{\to }{\mathop{b}}\,\,\overset{\to }{\mathop{c}}\,]}=\frac{[\overset{\to }{\mathop{a}}\,\overset{\to }{\mathop{b}}\,\overset{\to }{\mathop{c}}\,]}{[\overset{\to }{\mathop{a}}\,\overset{\to }{\mathop{b}}\,\overset{\to }{\mathop{c}}\,]}=1=\overset{\to }{\mathop{b}}\,.\overset{\to }{\mathop{q}}\,=\overset{\to }{\mathop{c}}\,.\overset{\to }{\mathop{r}}\,\] \[\overset{\to }{\mathop{b}}\,.\overset{\to }{\mathop{p}}\,=\frac{\overset{\to }{\mathop{b}}\,.(\overset{\to }{\mathop{b}}\,\times \overset{\to }{\mathop{c}}\,)}{[\overset{\to }{\mathop{a}}\,\,\overset{\to }{\mathop{b}}\,\,\overset{\to }{\mathop{c}}\,]}=\frac{0}{[\overset{\to }{\mathop{a}}\,\overset{\to }{\mathop{b}}\,\overset{\to }{\mathop{c}}\,]}=0=\overset{\to }{\mathop{c}}\,.\overset{\to }{\mathop{p}}\,=\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{r}}\,\] Therefore, the given expression is equal to \[1+0+1+0+1+0=3\].


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