Question

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

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

Answer 1

The correct option is C $\frac{1}{2}(\overrightarrow{p}+\overrightarrow{q}+\overrightarrow{r})$

Consider the problem

Let

$|p|=|q|=|r|=c$

And

$p.q=0=p.r=q.r$

$\begin{array}{c}p\times |(x-q)\times p|+q\times |(x-r)\times q|+r\times |(x-p)\times r|=0\\ \Rightarrow (p.p)(x-q)-\{p.(x-q\left)\right\}p+.........=0\\ \Rightarrow c^2(x-q+x-r+x-p)-(p.x)p-(q.x)q-(r.x)r=0\\ \Rightarrow c^2\{3x-(p+q+r\left)\right\}-\left[\right(p.x)p+(q.x)q+(r.x\left)r\right]=0\end{array}$

which is satisfied by $x=\frac{1}{2}(p+q+r)$

Hence, Option $C$ is the correct answer.