A) 0
B) -3
C) 3
D) -9
Correct Answer: C
Solution :
[c] As given \[\overset{\to }{\mathop{p}}\,=\frac{\overset{\to }{\mathop{b}}\,\times \overset{\to }{\mathop{c}}\,}{[\overset{\to }{\mathop{a}}\,\overset{\to }{\mathop{b}}\,\overset{\to }{\mathop{c}}\,]},\overset{\to }{\mathop{q}}\,=\frac{\overset{\to }{\mathop{c}}\,\times \overset{\to }{\mathop{a}}\,}{[\overset{\to }{\mathop{a}}\,\overset{\to }{\mathop{b}}\,\overset{\to }{\mathop{c}}\,]},\] and |
\[\overset{\to }{\mathop{r}}\,=\frac{\overset{\to }{\mathop{a}}\,\times \overset{\to }{\mathop{b}}\,}{[\overset{\to }{\mathop{a}}\,\overset{\to }{\mathop{b}}\,\overset{\to }{\mathop{c}}\,]}\] |
\[\therefore (\overset{\to }{\mathop{a}}\,-\overset{\to }{\mathop{b}}\,-\overset{\to }{\mathop{c}}\,).\overset{\to }{\mathop{p}}\,+(\overset{\to }{\mathop{b}}\,-\overset{\to }{\mathop{c}}\,-\overset{\to }{\mathop{a}}\,).\overset{\to }{\mathop{q}}\,+(\overset{\to }{\mathop{c}}\,-\overset{\to }{\mathop{a}}\,-\overset{\to }{\mathop{b}}\,).\overset{\to }{\mathop{r}}\,\] |
\[=\frac{a.(\overset{\to }{\mathop{b}}\,\times \overset{\to }{\mathop{c}}\,)}{[\overset{\to }{\mathop{a}}\,\overset{\to }{\mathop{b}}\,\overset{\to }{\mathop{c}}\,]}+\frac{\overset{\to }{\mathop{b}}\,.(\overset{\to }{\mathop{c}}\,\times \overset{\to }{\mathop{a}}\,)}{[\overset{\to }{\mathop{a}}\,\overset{\to }{\mathop{b}}\,\overset{\to }{\mathop{c}}\,]}+\frac{\overset{\to }{\mathop{c}}\,.(\overset{\to }{\mathop{a}}\,\times \overset{\to }{\mathop{b}}\,)}{[\overset{\to }{\mathop{a}}\,\overset{\to }{\mathop{b}}\,\overset{\to }{\mathop{c}}\,]}\] |
[Since \[\overset{\to }{\mathop{b}}\,.(\overset{\to }{\mathop{b}}\,\times \overset{\to }{\mathop{c}}\,)=0,\] |
\[\overset{\to }{\mathop{c}}\,.(\overset{\to }{\mathop{b}}\,\times \overset{\to }{\mathop{c}}\,),=0,\overset{\to }{\mathop{c}}\,.(\overset{\to }{\mathop{c}}\,\times \overset{\to }{\mathop{a}}\,)=0\overset{\to }{\mathop{a}}\,.(\overset{\to }{\mathop{c}}\,\times \overset{\to }{\mathop{a}}\,)]\] |
\[=0,\overset{\to }{\mathop{a}}\,.(\overset{\to }{\mathop{a}}\,\times \overset{\to }{\mathop{b}}\,)=0\] and \[\overset{\to }{\mathop{b}}\,.(\overset{\to }{\mathop{a}}\,\times \overset{\to }{\mathop{b}}\,)=0\] |
\[=\frac{[\overset{\to }{\mathop{a}}\,\overset{\to }{\mathop{b}}\,\overset{\to }{\mathop{c}}\,]}{[\overset{\to }{\mathop{a}}\,\overset{\to }{\mathop{b}}\,\overset{\to }{\mathop{c}}\,]}+\frac{[\overset{\to }{\mathop{a}}\,\overset{\to }{\mathop{b}}\,\overset{c}{\mathop{c}}\,]}{[\overset{\to }{\mathop{a}}\,\overset{\to }{\mathop{b}}\,\overset{\to }{\mathop{c}}\,]}+\frac{[\overset{\to }{\mathop{a}}\,\overset{\to }{\mathop{b}}\,\overset{\to }{\mathop{c}}\,]}{[\overset{\to }{\mathop{a}}\,\overset{\to }{\mathop{b}}\,\overset{\to }{\mathop{c}}\,]}=3\] |
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