Question

Let $\overrightarrow{p},\overrightarrow{q},\overrightarrow{r}$ be three mutually perpendicular vectors of the same magnitude. If a vector $\vec{x}$ satisfies the equation
$\overrightarrow{p}\times [(\overrightarrow{x}-\overrightarrow{q})\times \overrightarrow{p}]+\overrightarrow{q}\times [(\overrightarrow{x}-\overrightarrow{r})\times \overrightarrow{q}]$
$+\overrightarrow{r}\times [(\overrightarrow{x}-\overrightarrow{p})\times \overrightarrow{r}]=\overrightarrow{0}$, then $\overrightarrow{x}$ is given by

• A. $\frac{1}{2}(\overrightarrow{p}+\overrightarrow{q}-2\overrightarrow{r})$
• B. $\frac{1}{2}(\overrightarrow{p}+\overrightarrow{q}+\overrightarrow{r})$
• C. $\frac{1}{3}(\overrightarrow{p}+\overrightarrow{q}+\overrightarrow{r})$
• D. $\frac{1}{3}(2\overrightarrow{p}+\overrightarrow{q}-\overrightarrow{r})$

Answer 1

Answer: (B)

Since, $\overrightarrow{p},\overrightarrow{q},\overrightarrow{r}$ are mutually perpendicular vectors of same magnitude, so let us consider
$|\overrightarrow{p}|=|\overrightarrow{q}|=|\overrightarrow{r}|=\lambda$ and
$\overrightarrow{p}\cdot \overrightarrow{q}=\overrightarrow{q}\cdot \overrightarrow{r}=\overrightarrow{r}\cdot \overrightarrow{p}=\mathrm{0.....}$ (i)
Given, $\overrightarrow{p}\times \{(\overrightarrow{x}-\overrightarrow{q})\times \overrightarrow{p}\}+\overrightarrow{q}\times \{(\overrightarrow{x}-\overrightarrow{r})\times \overrightarrow{q}\}$
$+\overrightarrow{r}\times \{(\overrightarrow{x}-\overrightarrow{p})\times \overrightarrow{r}\}=\overrightarrow{0}$
$\Rightarrow (\overrightarrow{p}\cdot \overrightarrow{p})(\overrightarrow{x}-\overrightarrow{q})-\{\overrightarrow{p}\cdot (\overrightarrow{x}-\overrightarrow{q})\}\overrightarrow{p}+(\overrightarrow{q}\cdot \overrightarrow{q})(\overrightarrow{x}-\overrightarrow{r})$
$-\{\overrightarrow{q}\cdot (\overrightarrow{x}-\overrightarrow{r})\}\overrightarrow{q}+(\overrightarrow{r}\cdot \overrightarrow{r})(\overrightarrow{x}-\overrightarrow{p})-\{\overrightarrow{r}\cdot (\overrightarrow{x}-\overrightarrow{p})\}\overrightarrow{r}=0$
$\Rightarrow \overrightarrow{x}(\overrightarrow{p}\cdot \overrightarrow{p})+(\overrightarrow{q}\cdot \overrightarrow{q})+(\overrightarrow{r}\cdot \overrightarrow{r})\}-(\overrightarrow{p}\cdot \overrightarrow{p})\overrightarrow{q}$
$-(\overrightarrow{q}\cdot \overrightarrow{q})r-(\overrightarrow{r}\cdot \overrightarrow{r})\overrightarrow{p}=(\overrightarrow{x}\cdot \overrightarrow{p})\overrightarrow{p}+(\overrightarrow{x}\cdot \overrightarrow{q})\overrightarrow{q}+(\overrightarrow{x}\cdot \overrightarrow{r})\overrightarrow{r}$
$\Rightarrow 3\overrightarrow{x}|\lambda {|}^2-(\overrightarrow{p}+\overrightarrow{q}+\overrightarrow{r})|\lambda {|}^2=(\overrightarrow{x}\cdot \overrightarrow{p})\overrightarrow{p}$
$+(\overrightarrow{x}\cdot \overrightarrow{q})\overrightarrow{q}+(\overrightarrow{x}\cdot \overrightarrow{r})\overrightarrow{r}....$(ii)
Taking dot of Eq. (ii) with $\overrightarrow{p}$, we get
$3(\overrightarrow{x}\cdot \overrightarrow{p})|\lambda {|}^2-|\lambda {|}^4=(\overrightarrow{x}\cdot \overrightarrow{p})|\lambda {|}^2\Rightarrow \overrightarrow{x}\cdot \overrightarrow{p}=\frac{1}{2}|\lambda {|}^2$
Similarly, taking dot of Eq. (ii) with $\overrightarrow{q}$ and $\overrightarrow{r}$ respectively, we get
$\overrightarrow{x}\cdot \overrightarrow{q}=\frac{|\lambda {|}^2}{2}=\overrightarrow{x}\cdot \overrightarrow{r}$
$\therefore$ Eq. (ii) becomes
$3\overrightarrow{x}|\lambda {|}^2-(\overrightarrow{p}+\overrightarrow{q}+\overrightarrow{r})|\lambda {|}^2=\frac{|\lambda {|}^2}{2}(\overrightarrow{p}+\overrightarrow{q}+\overrightarrow{r})$
$\Rightarrow 3\overrightarrow{x}=\frac{1}{2}(\overrightarrow{p}+\overrightarrow{q}+\overrightarrow{r})+(\overrightarrow{p}+\overrightarrow{q}+\overrightarrow{r})\Rightarrow \vec{x}=\frac{1}{2}(\overrightarrow{p}+\overrightarrow{q}+\overrightarrow{r})$