Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-coplanar unit vectors such that the angle between

Question

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

Professor PC Thomas Classes · Answer

(4) : As $\vec{a} 	imes \vec{b}+\vec{b} 	imes \vec{c}=p \vec{a}+q \vec{b}+r \vec{c}$ Taking dot product of vectors with $\vec{a}, \vec{b}$ and $\vec{c}$\ $[\vec{a} \vec{b} \vec{c}]=p a^{2}+q(\vec{b} \cdot \vec{a})+r(\vec{c} \cdot \vec{a})=p+\frac{q}{2}+\frac{r}{2} ;$\ $0=\frac{p}{2}+q+\frac{r}{2}$ and $[\vec{a} \vec{b} \vec{c}]=\frac{p}{2}+\frac{q}{2}+r$\ Also, $\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}ight]^{2}=\left|\begin{array}{lll}\vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} & \vec{a} \cdot \vec{c} \ \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} & \vec{b} \cdot \vec{c} \ \vec{c} \cdot \vec{a} & \vec{c} \cdot \vec{b} & \vec{c} \cdot \vec{c}\end{array}ight|$\ $=\left|\begin{array}{ccc}1 & 1 / 2 & 1 / 2 \ 1 / 2 & 1 & 1 / 2 \ 1 / 2 & 1 / 2 & 1\end{array}ight|=\frac{1}{2}$\ $	herefore \quad[\vec{a} \vec{b} \vec{c}]=\frac{1}{\sqrt{2}}$\ Thus, $p+\frac{q}{2}+\frac{r}{2}=\frac{1}{\sqrt{2}} ; \frac{p}{2}+q+\frac{r}{2}=0$ $$ \frac{p}{2}+\frac{q}{2}+r=\frac{1}{\sqrt{2}} $$ On solving, we get $p=-q=r$\ Hence, $\frac{p^{2}+2 p^{2}+r^{2}}{q^{2}}=\frac{p^{2}+2 p^{2}+p^{2}}{p^{2}}=4$\