1
$\begingroup$

Question:

If $$\vec p \times (\vec u \times \vec p) + \vec q \times (\vec v \times \vec q) + \vec r \times (\vec w \times \vec r) = 0$$ Where $\vec p, \vec q, \vec r$ are mutually perpendicular vectors with the same magnitude and: $$\vec u = \vec X - \vec q$$ $$\vec v = \vec X - \vec r$$ $$\vec w = \vec X - \vec p$$ Then $\vec X$ can be expressed as?

Not sure how I should proceed with this question. As it is given that $p,q,r$ are perpendicular vectors with the same magnitude, I tried to suppose they are unit vectors along the three axes, however, that didn't help. I attempted to use the fact that: $$\vec a \times (\vec b \times \vec c) = (\vec a \cdot \vec c)\vec b - (\vec a \cdot \vec b) \vec c$$

However, I was unable to proceed from there as well. What should I do?

$\endgroup$
1
$\begingroup$

I think this works: we have that

$$ 0 = \vec p \times (\vec X \times \vec p - \vec q \times \vec p) + \vec q \times (\vec X \times \vec q - \vec r \times \vec q) + \vec r \times (\vec X \times \vec r - \vec p \times \vec r) $$ instead of $\vec p, \vec q, \vec r$, lets call them $p_1 , p_2 , p_3$; in fact, we are going to be using cyclical notation, so lets just say that $p_{3k+1} = \vec p$, $p_{3k+2} = \vec q$, and $p_{3k} = \vec r$ for any $k$. Then, $$ \sum _i p_i \times ( p_{i+1} \times p_i ) = \sum _i p_i \times ( X \times p_i ). $$ Now, since three mutually orthogonal vectors are a frame for $\mathbb{R}^3$, we know that $p_i \times p_{i+1} = \left( \pm \frac{|p_i||p_{i+1}|}{|p_{i+2}|} \right) p_{i+2}$. So, the sum on the left becomes $$ \sum _i p_i \times ( p_{i+1} \times p_i ) = \sum _i p_i \times ( \left( \mp \frac{|p_i||p_{i+1}|}{|p_{i+2}|} \right) p_{i+2} ) = \sum _i \left( \mp \frac{|p_i||p_{i+1}|}{|p_{i+2}|} \right) p_i \times ( p_{i+2} ) = \sum _i \left( \mp \frac{|p_i||p_{i+1}|}{|p_{i+2}|} \right) p_i \times ( p_{i+{(3=0)} - 1} ) = \sum _i \left( \mp \frac{|p_i||p_{i+1}|}{|p_{i+2}|} \right) \left( \mp \frac{|p_i||p_{i-1}|}{|p_{i+1}|} \right) p_{i+1} = \sum _i |p_i|^2 p_{i+1}, $$ that is to say, merely $$ | \vec p |^2 \vec q + | \vec q |^2 \vec r + | \vec r |^2 \vec p. $$ Then, by the lagrange (?) formula, the right hand sum becomes $$ \sum _i p_i \times ( X \times p_i ) = \sum_i | p_i |^2 X - \sum_i ( p_i \cdot X ) p_i. $$ So, $$ X = \frac{ \sum _i |p_i|^2 p_{i+1} + \sum_i ( p_i \cdot X ) p_i }{ \sum_i | p_i |^2 } = \frac{ \sum _i (|p_{i+2}|^2 + p_i \cdot X ) p_{i} }{ \sum_i | p_i |^2 } . $$ Unfortunately, off the top of my head, I am not sure how to go much farther...

EDIT: I thought about how to go further

Since the $p_i$ are a frame, we can write down the coordinate for $X$ relative to that frame: $X = x_{p_1} p_1 + x_{p_2} p_2 + x_{p_3} p_3 = \sum_i x_{p_i} p_i$. Then, we must have from the last equation that $$ x_{p_i} = \frac{ (|p_{i+2}|^2 + x_{p_i} )}{ \sum_i | p_i |^2 }, $$ so that $$ x_{p_i} = \frac{|p_{i+2}|^2}{\sum_i | p_i |^2 - 1}. $$ hence, $X = \frac{1}{ |p|^2 + |q|^2 + |r|^2 - 1 } ( |r|^2 , |p|^2 , |q|^2 )$.

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.