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 \[1+0+1+0+1+0=3\].You need to login to perform this action.
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