Answer:
Option B
Explanation:
Plan
(i)a.b= |a|.|b| cosθ
(ii) [a b c]2= $\begin{bmatrix}a.a & a.b &a.c\\b.a & b.b &b.c \\ c.a &c.b&c.c\end{bmatrix}$
a.a=|a|2 =1,similarly b.b=c.c=1
a.b=|a||b|cos $\left(\frac{\pi}{3}\right)=\left(\frac{1}{2}\right)$similarly
b.c=c.a=$\frac{1}{2}$
[a b c]2= $\begin{bmatrix}a.a & a.b &a.c\\b.a & b.b &b.c \\ c.a &c.b&c.c\end{bmatrix}$
= $\begin{bmatrix}1 & \frac{1}{2} &\frac{1}{2}\\\frac{1}{2} & 1 &\frac{1}{2} \\ \frac{1}{2} &\frac{1}{2}&1\end{bmatrix}=\frac{3}{4}-\frac{1}{2}=\frac{1}{2}$
$\therefore$ $[ a.b.c]=\frac{1}{\sqrt{2}}$ .........(i)
As , given a x b+b x c=pa+qb+rc
Take dot product with a
a.(axb)+a.(bxc) = p a2+q b2+r c.a
$\Rightarrow$ $ 0+\frac{1}{\sqrt{2}}=p+\frac{q}{2}+\frac{r}{2}$
$\therefore$ [a a b]=0
Now, take dot product with b and c
$0=\frac{p}{2}+q+\frac{r}{2}$ ......(iii)
and $\frac{1}{\sqrt{2}}=\frac{p}{2}+\frac{q}{2}+r$ .......(iv)
On subtracting Eq(ii) from Eq. (iv), we get
$\frac{p}{2}-\frac{r}{2}=0\Rightarrow p=r$
$\Rightarrow$ p+r=0 [By Eq.(iii)]
$\therefore$ $\frac{p^{2}+2q^{2}+r^{2}}{q^{2}}$
$=\frac{p^{2}+2p^{2}+p^{2}}{p^{2}}=4$
