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

4

We have,

$\vec a×\vec b+ \vec b×\vec c =p\vec a+q\vec b+r \vec c$   ...(i)

Taking dot product with $\vec a$, we obtain

$\vec a.(\vec a×\vec b) + \vec a. (\vec b×\vec c) = p(\vec a· \vec a)+q(\vec a·\vec b) +r (\vec a.\vec c)$

$⇒[\vec a\,\,\vec b\,\,\vec c]=p+\frac{q}{2}+\frac{r}{2}$   ...(ii)

Taking dot product on both sides of (i) by $\vec b$, we get

$\vec b.(\vec a×\vec b)+\vec b. (\vec b×\vec c) = p(\vec b. \vec a)+q(\vec b. \vec b) +r(\vec c.\vec c)$

$⇒0=\frac{p}{2}+q+\frac{r}{2}$   ...(iii)

Taking dot product on both sides of (i) by $\vec c$, we get

$\vec c.(\vec a×\vec b)+\vec c. (\vec b×\vec c) = p(\vec c · \vec a) +q(\vec c.\vec b) +r (\vec c. \vec c)$

$⇒[\vec c\,\,\vec a\,\,\vec b]=\frac{p}{2}+\frac{q}{2}+r$

$⇒[\vec a\,\,\vec b\,\,\vec c]=\frac{p}{2}+\frac{q}{2}+r$   ...(iv)

From (ii) and (iv), we get

$p+\frac{q}{2}+\frac{r}{2}=\frac{p}{2}+\frac{q}{2}+r⇒p=r$

Putting $p=r$ in (iii), we get $q=-r$. Thus, we obtain

$p=r=-q$

$∴\frac{p^2+2q^2 + r^2}{q^2}=\frac{(-q)^2+2q^2 +(-q)^2}{q^2}=4$