A) 0
B) 1
C) 2
D) 3
Correct Answer: D
Solution :
Given, \[\vec{a},\vec{b},\vec{c}\]are three non-coplanar vectors and \[\vec{p},\vec{q},\vec{r}\]defined by the relations \[\vec{p}=\frac{\vec{b}\times \vec{c}}{[\vec{a}\vec{b}\vec{c}]},\vec{q}=\frac{\vec{c}\times \vec{a}}{[\vec{a}\vec{b}\vec{c}]}\] and \[\vec{r}=\frac{\vec{a}\times \vec{b}}{[\vec{a}\vec{b}\vec{c}]}\] \[\therefore \] \[\vec{a}.\vec{p}=\frac{\vec{a}.\vec{b}\times \vec{c}}{[\vec{a}\vec{b}\vec{c}]}\] \[=\frac{\vec{a}.(\vec{b}\times \vec{c})}{[\vec{a}\vec{b}\vec{c}]}=1\] and \[\vec{a}.\vec{q}=\vec{a}.\frac{\vec{c}\times \vec{a}}{[\vec{a}\vec{b}\vec{c}]}\] \[=\frac{\vec{a}.(\vec{c}\times \vec{a})}{[\vec{a}\vec{b}\vec{c}]}=0\] Similarly, \[\vec{b}.\vec{q}=\vec{c}.\vec{r}=1\] and \[\vec{a}.\vec{r}=\vec{b}.\vec{p}=\vec{b}.\vec{r}=\vec{c}.\vec{q}=\vec{c}.\vec{p}=0\] \[\therefore \]\[(\vec{a}+\vec{b}).\vec{p}+(\vec{b}+\vec{c}).\vec{q}+(\vec{c}+\vec{a}).\vec{r}\] \[=\vec{a}.\vec{p}+\vec{b}.\vec{p}+\vec{b}.\vec{q}+\vec{c}.\vec{q}\] \[+\,\vec{c}.\vec{r}+\vec{a}.\vec{r}\] \[=1+1+1=3.\]
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