Question

Let $\overrightarrow{a},\overrightarrow{b},$ and $\overrightarrow{c}$ be three non-coplanar unit vectors such that the angle between every pair of them is $\frac{\pi }{3}$. If $\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}=p\overrightarrow{a}+q\overrightarrow{b}+r\overrightarrow{c}$, where $p,q$ and $r$ are scalars, then the value of $\frac{p^2+2q^2+r^2}{q^2}$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is D $4$

$|\overrightarrow{a}|=|\overrightarrow{b}|=|\overrightarrow{c}|=1$
$\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}=p\overrightarrow{a}+q\overrightarrow{b}+r\overrightarrow{c}$
$\Rightarrow \overrightarrow{a}\cdot (\overrightarrow{b}\times \overrightarrow{c})=p+q(\overrightarrow{a}\cdot \overrightarrow{b})+r(\overrightarrow{a}\cdot \overrightarrow{c})$

$\Rightarrow p+\frac{q}{2}+\frac{r}{2}=\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]$ ........ $\left(1\right)$
Similarly,

$\frac{p}{2}+q+\frac{r}{2}=0$ ....... $\left(2\right)$

$\frac{p}{2}+\frac{q}{2}+r=\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]$ ..... $\left(3\right)$

Solving these equations, we get

$\Rightarrow p=r=-q$

$\frac{p^2+2q^2+r^2}{q^2}=4$

Hence, option $D$.