# 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?

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.
• Alvaro, thanks for the answer. I follow what you have written. But I am trying to flesh out the intuition behind the orthogonality and relative terms of exchange, say between two products. I am trying to think hard to see what kind of absurdity it would result in if we have, say they are not orthogonal, meaning the dot product isn't zero. I suppose that would probably lead to a contradiction that $x^0$ or $x'$ is not on the budget hyperplane... What would be the situation when it is NOT orthogonal? – Frank Swanton Jun 27 '16 at 21:54
• If the budget line and the vector p are not orthogonal, then $x_2=-\alpha∗x_1+w/p_2$ with $\alpha$ different from $p_1/p_2$. Then $dx_1/dx_2=-\alpha$ (relative prices are not the rate of exchange). The consequence would be that the budget line can be written as $\alpha∗p_2*x_1+p_2*x_2=w$. Hence, for any $x'>>0$ in the original budget line we have $\alpha∗p_2*x'_1+p_2*x'_2=p_1*x'_1+p_2*x'_2+(\alpha∗p_2-p_1)*x'_1=w+(\alpha∗p_2-p_1)*x'_1\neq w$. $x'$ is not in the budget line. – Belisario Jun 27 '16 at 22:35
• Also, the way I think about the relative terms of exchange is with the indifference curves. So given say a certain utility level, the tangency on each point of that indifference curve represents the marginal rate of substitution between $x_1,x_2$ for $L=2$. The budget hyperplane would be I suppose tangent to some indifference curve given we have a quasiconcave utility function $u(.)$. So, that does represent, given some price and wealth, a particular rate of marginal substitution then? Again, it's hard to see why this has to do with the price vectors being orthogonal :( – Frank Swanton Jun 28 '16 at 3:40
• Actually I think I am getting confused with the jargon "the relative terms of exchange" meaning it is just a ratio of prices. So it has to be either $p_1$ over $p_2$ or the other way around. So in that sense, I get both graphically and algebraically when $L=2$, the slope is the ratio. – Frank Swanton Jun 28 '16 at 3:57
• Yes, the (market) rate of exchange is the ratio at which people exchange products in the market. So, a unit of $x_2$ (with price $p_2$) can be exchanged for $p_2/p_1$ units of $x_1$ (think $p_2$ as the income you get when you sell one unit of $x_2$). This ratio is equal to the marginal rate of substitution when the consumer is maximizing, but MWG are not talking about it. Your answer is correct. Note that you can always get the slope between $x_1$ and $x_2$ using the implicit function theorem: $dx_2/dx_1=-p_1/p_2$. This turns out to be the rate of exchange and also the orthogonal slope. – Belisario Jun 28 '16 at 19:09