A) 0
B) 1
C) 2
D) 3
Correct Answer: D
Solution :
We have \[\mathbf{p}\,.\,(\mathbf{a}+\mathbf{b})=\mathbf{p}\,.\,\mathbf{a}+\mathbf{p}\,.\,\mathbf{b}\] \[=\frac{(\mathbf{b}\times \mathbf{c})\,.\,\mathbf{a}}{[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]}+\frac{(\mathbf{b}\times \mathbf{c})\,.\,\mathbf{b}}{[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]}=\frac{[\mathbf{b}\,\mathbf{c}\,\mathbf{a}]}{[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]}+\frac{[\mathbf{b}\,\,\mathbf{c}\,\,\mathbf{b}]}{[\mathbf{a}\,\,\mathbf{b}\,\,\mathbf{c}]}\] \[=1+0=1\], \[\left\{ \because \,[\mathbf{b}\,\mathbf{c}\,\mathbf{a}]=[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]\,\text{and}\,[\mathbf{b}\,\mathbf{c}\,\mathbf{b}]=0 \right\}\] Similarly, \[\mathbf{q}.(\mathbf{b}+\mathbf{c})=1\] and \[\mathbf{r}\,.\,(\mathbf{a}+\mathbf{c})=1\] Thus, required result is 1+1+1=3.
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