If $\vec p\times((\vec x-\vec q)\times\vec p)+\vec q\times((\vec x-\vec r)\times\vec q)+\vec r\times((\vec x-\vec p)\times\vec r)=0$, find $\vec x$

Question

Let $\vec p, \vec q$ and $\vec r$ are three mutually perpendicular vectors of the same magnitude. If a vector $\vec x$ satisfies the equation $\begin{aligned} \vec p \times ((\vec x - \vec q) \times \vec p)+\vec q \times ((\vec x - \vec r) \times \vec q)+\vec r \times ((\vec x - \vec p) \times \vec r)=0\end{aligned}$, then find $\vec x$.

I considered $p^2 = q^2 = r^2 = a^2$ (say) and since $\vec p, \vec q$ and $\vec r$ are three mutually perpendicular vectors, therefore $\vec p \cdot \vec q = \vec q \cdot \vec r = \vec p \cdot \vec r = 0$. Then I solved the given equation using these data and got:

$3a^2 \vec x +((\vec p\cdot\vec x)- a^2)\vec p +((\vec q\cdot\vec x)- a^2)\vec q+((\vec r\cdot\vec x)- a^2)\vec r= 0$

How should I proceed after this to find $\vec x$ ? Thanks.

Answer 1

Take the dot product of your equation with $\vec{p}$, and you get $$3a^2(\vec{x} \cdot \vec p) + ((\vec p \cdot \vec{x}) - a^2)a^2 = 0$$ This solves to $\vec{x} \cdot \vec p = {a^2 \over 2}$. By symmetry you must also have $\vec{x} \cdot \vec q = \vec{x} \cdot \vec r = {a^2 \over 2}$.

$\vec p, \vec q, \vec r$ are an orthogonal basis so you therefore have $\vec {x} = {1 \over 4} (\vec p + \vec q + \vec r)$.

Answer 2

$\bf{My\; Solution::}$ Let $$\left|\vec{p}\right|=\left|\vec{q}\right|=\left|\vec{p}\right|=a\;,$$ Then Given $$\vec{p}\cdot \vec{q}=\vec{q}\cdot \vec{r}=\vec{r}\cdot \vec{p} = 0.$$

And Given $$\begin{aligned} \vec p \times ((\vec x - \vec q) \times \vec p)+\vec q \times ((\vec x - \vec r) \times \vec q)+\vec r \times ((\vec x - \vec p) \times \vec r)=0\end{aligned}$$

Using Vector Triple Product Property, We Get

$$\Rightarrow \displaystyle \vec{p}\times \left(\vec{x}\times \vec{p}-\vec{q}\times \vec{p}\right)+\vec{q}\times \left(\vec{x}\times \vec{q}-\vec{r}\times \vec{q}\right)+\vec{r}\times \left(\vec{x}\times \vec{r}-\vec{p}\times \vec{r}\right)=0$$

$$\Rightarrow \displaystyle \vec{p}\times (\vec{x}\times \vec{p})- \vec{p}\times (\vec{q}\times \vec{p})+ \vec{q}\times (\vec{x}\times \vec{q}) - \vec{q}\times (\vec{r}\times \vec{q}) + \vec{r}\times (\vec{x}\times \vec{r}) - \vec{r}\times (\vec{p}\times \vec{r}) = 0$$.

$$\Rightarrow \displaystyle \left(\left|\vec{p}\right|^2+\left|\vec{q}\right|^2+\left|\vec{r}\right|^2\right)\vec{x}-\left(\vec{p}\cdot \vec{x}\right)\vec{p}- \left(\vec{q}\cdot \vec{x}\right)\vec{q}-\left(\vec{r}\cdot \vec{x}\right)\vec{r}-\left|\vec{p}\right|^2\vec{q}-\left|\vec{q}\right|^2\vec{r}-\left|\vec{r}\right|^2\vec{p}=0$$

Now above Given $$\vec{p}\;,\vec{q}\;,\vec{r}$$ are Three Perpendicular vector of same magnitude.

So We can write $$\vec{p}=a\vec{i}$$ and $$\vec{q}=a\vec{j}$$ and $$\vec{r}=a\vec{k}.$$ and $$\vec{x}=x_{1}\vec{i}+y_{1}\vec{j}+z_{1}\vec{k}$$

So Our equation convert into......

$$\displaystyle \Rightarrow 3a^2\vec{x}-a^2\left(x_{1}\vec{i}+y_{1}\vec{j}+z_{1}\vec{k}\right)-a^2(a\vec{i}+a\vec{j}+a\vec{k})=0$$

$$\displaystyle \Rightarrow 3a^2\vec{x}-a^2\vec{x}-a^2(\vec{p}+\vec{q}+\vec{r})=\vec{0}$$

So we get $$\displaystyle \vec{x} = \frac{1}{2}\left(\vec{p}+\vec{q}+\vec{r}\right)$$