Question

Let $\vec{a},\vec{b}$ and $\vec{c}$ be three non-coplanar vectors, and let $\vec{p},\vec{q}$ and $\vec{r}$ be the vectors defined by the relations $\vec{P}=\frac{\vec{b}\times \vec{c}}{\left[\vec{a}\vec{b}\vec{c}\right]},\vec{q}=\frac{\vec{c}\times \vec{a}}{\left[\vec{a}\vec{b}\vec{c}\right]}and\vec{r}=\frac{\vec{a}\times \vec{b}}{\left[\vec{a}\vec{b}\vec{c}\right]}.$
Then the value of the expression
$\left(\vec{a}+\vec{b}\right).\vec{p}+\left(\vec{b}+\vec{c}\right).\vec{q}+\left(\vec{c}+\vec{a}\right).\vec{r}$ is equal to

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (D)

$\vec{a}.\vec{p}=\frac{\vec{a}.\left(\vec{b}\times \vec{c}\right)}{\left[\vec{a}\vec{b}\vec{c}\right]}=\frac{\left[\vec{a}\vec{b}\vec{c}\right]}{\left[\vec{a}\vec{b}\vec{c}\right]}=1=\vec{b}.\vec{q}=\vec{c}.\vec{r}$
$\vec{b}.\vec{p}=\frac{\vec{b}.\left(\vec{b}\times \vec{c}\right)}{\left[\vec{a}\vec{b}\vec{c}\right]}=\frac{0}{\left[\vec{a}\vec{b}\vec{c}\right]}=0=\vec{c}.\vec{p}=\vec{a}.\vec{r}$
Therefore, the given expression is equal to
1 + 0 + 1 + 0 + 1 + 0 = 3.
[Also see the system of reciprocal vectors]