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]