JEE Advance - Mathematics (2014 - Paper 1 Offline - No. 11)
Let $$\overrightarrow a \,\,,\,\,\overrightarrow b $$ and $$\overrightarrow c $$ be three non-coplanar unit vectors such that the angle between every pair of them is $${\pi \over 3}.$$ 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 $${{{p^2} + 2{q^2} + {r^2}} \over {{q^2}}}$$ is
Answer
4
Explanation
Given $\vec{a} \times \vec{b}+\vec{b} \times \vec{c}=p \vec{a}+q \vec{b}+r \vec{c}$ ..........(1)
Taking dot product with $\vec{a}$ :
Hence,
$$ 0+\vec{a} \cdot \vec{b} \times \vec{c}=p(1 \cdot 1 \cdot \cos 0)+q\left(1 \cdot 1 \cdot \cos \frac{\pi}{3}\right)+r\left(1 \cdot 1 \cdot \cos \frac{\pi}{3}\right) $$
$\Rightarrow \vec{a} \cdot \vec{b} \times \vec{c}=p+\frac{q}{2}+\frac{r}{2}$ ..........(2)
Taking the dot product of (1) with $\vec{b}$ :
$$ 0+0=\frac{p}{2}+q+\frac{r}{2} $$ ............(3)
Taking the dot product of (1) with $\vec{c}$ :
$$ \vec{c} \cdot \vec{a} \times \vec{b}+0=\frac{p}{2}+\frac{q}{2}+r$$ .............(4)
From (2) and (4), we get
$$ \begin{aligned} p+\frac{q}{2}+\frac{r}{2} & =\frac{p}{2}+\frac{q}{2}+r \\\\ \frac{p}{2} & =\frac{r}{2} \Rightarrow p=r \end{aligned} $$
Now, from Eq. (3), we get $0=\frac{r}{2}+q+\frac{r}{2} \Rightarrow q=-r$.
Now, $\frac{p^2+2 q^2+r^2}{q^2}=\frac{r^2+2(-r)^2+r^2}{(-r)^2}=\frac{4 r^2}{r^2}=4$.
Taking dot product with $\vec{a}$ :
Hence,
$$ 0+\vec{a} \cdot \vec{b} \times \vec{c}=p(1 \cdot 1 \cdot \cos 0)+q\left(1 \cdot 1 \cdot \cos \frac{\pi}{3}\right)+r\left(1 \cdot 1 \cdot \cos \frac{\pi}{3}\right) $$
$\Rightarrow \vec{a} \cdot \vec{b} \times \vec{c}=p+\frac{q}{2}+\frac{r}{2}$ ..........(2)
Taking the dot product of (1) with $\vec{b}$ :
$$ 0+0=\frac{p}{2}+q+\frac{r}{2} $$ ............(3)
Taking the dot product of (1) with $\vec{c}$ :
$$ \vec{c} \cdot \vec{a} \times \vec{b}+0=\frac{p}{2}+\frac{q}{2}+r$$ .............(4)
From (2) and (4), we get
$$ \begin{aligned} p+\frac{q}{2}+\frac{r}{2} & =\frac{p}{2}+\frac{q}{2}+r \\\\ \frac{p}{2} & =\frac{r}{2} \Rightarrow p=r \end{aligned} $$
Now, from Eq. (3), we get $0=\frac{r}{2}+q+\frac{r}{2} \Rightarrow q=-r$.
Now, $\frac{p^2+2 q^2+r^2}{q^2}=\frac{r^2+2(-r)^2+r^2}{(-r)^2}=\frac{4 r^2}{r^2}=4$.
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