Why is Price Vector Orthogonal to Vector connecting two bundles on Budget Hyperplane Here is my revised version/understanding why price vector is orthogonal to any vector from a bundle on the budget hyperplane to another bundle on the hyperplane: (see below for original question)


  Want to show geometrically that budget hyperplane is the relative terms of exchange. This is a fancy way of saying it represents the 'ratio' of the prices between any two commodities. For simplicity sake, look at $L=2$. We have two options for a ratio: either $\frac{p_1}{p_2}$ or $\frac{p_2}{p_1}$. What would it be? The way the price vector is constructed is p$=(p_1,p_2)$, so when you draw this vector from any point on the budget line that is negatively sloped in the case of $L=2$, the vector is essentially the slope (e.g. rise over run) of $\frac{p_2}{p_1}$. So obviously, if the negatively sloped budget line is supposed to represent the price ratio, then it has to be the case $\frac{-p_1}{p_2}$. In fact, this is so from the Walrasian budget set definition where $w=x\cdot p$.


When we look at a typical Walrasian budget set in $\mathbb{R^+_2}$, why is the price vector orthogonal to the consumption vector (e.g. any two on the slope) on the budget hyperplane? This goes back to Chapter 2 of MWG.
 I understand the analytical explanation using dot product. 


  $p\cdot x=p\cdot x'=w$ for $x,x'\in\{x\in\mathbb{R^+_2}:p\cdot x=w\}$.
   Hence, $p\cdot\Delta x=0$.


But I am having a tough time understanding two things:


  (Q) Why does this orthogonality between price and consumption vector on hyperplane relate back to the slope of the budget line determining the relative rate of exchange between two commodities? How do you make sense between intuition and geometric interpretation?


Thanks for your 2 cent!
Note that $p$ is not orthogonal to consumption vectors on the budget line, but it is orthogonal to a any vector $v$ that satisfies $x+v=x'$ with $x,x'$ in the budget line. MWG are drawing vectors starting from some $x$.

About the slope: In the budget line you can see that the slope for any $x$ is $D_x (p.x)=p$ (that is the line that is drawn from $x$). That line or vector can be represented with the function $x_2=(p_2/p_1)*x_1$ (note that if $x_1=p_1$, $x_2=p_2$). The slope is $p_2/p_1$.

On the other hand, all vectors on the budget line satisfy $x_2=(-p_1/p_2)*x_1+w/p_2$. The slope here is $(-p_1/p_2)$. This slope is the rate of exchange. Both slopes imply that the two functions (the vector $p$ and the budget line) are orthogonal. Or maybe it's better to say that the orthogonality between the budget line and the vector $p$ imply that $dx_2/dx_1=-p_1/p_2$, that is, the last term captures the rate of exchange.