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Can we divide two vector quantities? For eg., Pressure( a scalar) equals force (a vector) divided by area (a vector).

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    $\begingroup$ I think instead of saying $w=\frac{\vec u}{\vec v}$ we could say $w=\frac{\vec u}{v^2}\cdot\vec v$ $\endgroup$ – Jim May 7 '14 at 17:00
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    $\begingroup$ in the geometric algebra induced by a (positive definite) inner product, all (non-zero) vectors have an inverse; it's just not terribly interesting as it's just the same vector, rescaled... $\endgroup$ – Christoph May 7 '14 at 21:03
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    $\begingroup$ Although there is a thing called reciprocal system of vectors. Google it. $\endgroup$ – evil999man May 8 '14 at 14:15
  • $\begingroup$ This is a classic motivation for the quaternions. The quaternion system predates the vector-scalar system, and has the advantage that you can do division in it. $\endgroup$ – Ben Crowell May 15 at 15:21
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No, in general you cannot divide one vector by another. It is possible to prove that no vector multiplication on three dimensions will be well-behaved enough to have division as we understand it. (This depends on exactly what one means by 'well-behaved enough', but the core result here is Hurwitz's theorem.)

Regarding force, area and pressure, the most fruitful way is to say that force is area times pressure: $$ \vec F=P\cdot \vec A. $$ As it turns out, pressure is not actually a scalar but a matrix (or, more technically, a rank 2 tensor). This is because, in certain situations, an area with its normal vector pointing in the $z$ direction can also experience forces along $x$ and $y$, which are called shear stresses. In this case, the correct linear relation is that $$ \begin{pmatrix}F_x\\ F_y \\ F_z \end{pmatrix} = \begin{pmatrix}p_x & s_{xy} & s_{xz} \\ s_{yx} & p_y & s_{yz} \\ s_{zx} & s_{zy} & p_z\end{pmatrix} \begin{pmatrix}A_x\\ A_y \\ A_z \end{pmatrix}. $$ In a fluid, shear stresses are zero and the pressure is isotropic, so all the $p_j$s are equal, and therefore the pressure tensor $P$ is a scalar matrix. In a solid, on the other hand, shear stresses can occur even in static situations, so you need the full matrix. In this case, the matrix is referred to as the stress tensor of the solid.

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  • $\begingroup$ Could you define $$\frac{\vec{F}}{\vec{A}} := \mathbf{P}$$ such that $\vec{F} = \mathbf{P} \vec{A}$ ? $\endgroup$ – ja72 May 16 at 12:10
  • $\begingroup$ @ja72 No, you can't. $\endgroup$ – Emilio Pisanty May 16 at 12:12
  • $\begingroup$ It is akin to taking the symbolic derivative (jacobian) where $$P_{ij} = \frac{ \partial \vec{F}_i }{\partial A_j}$$ $\endgroup$ – ja72 May 16 at 12:13
  • $\begingroup$ @ja72 If you have a concrete question to ask, do so. If all you're going to do is set up a series of moving goalposts, then no thanks. $\endgroup$ – Emilio Pisanty May 16 at 13:04
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As an aside, you can actually divide two vectors. The only question is how do you want to interpret the objects and more importantly the operation.

For example, you can map the vectors to an object in a quaternion space quite simply as:

$$ \phi:V \rightarrow H: \vec{v} \mapsto (0,\vec{v}) , $$

and then division is well defined. But your answer will be, in general, quite obviously, a general quaternion $(r,\vec{u})$, and you then need a physical interpretation for this.

In the specifics of your question, you see, the objects and the operation are fixed by nature. Force and area are vectors related by a tensor called pressure as:

$$ \vec{F} = P \vec{A}, $$

where the operation of $P$ on $\vec{A}$ is defined to be the tensor action. In this setup there is no unique way to define division of two vectors to produce a tensor: the definition of the operation admits no sensible inverse.

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    $\begingroup$ note that the quaternions are a subalgebra of the geometric algebra, where vector division is essentially (up to scale) the same as (Clifford) multiplication; division of two non-parallel non-orthogonal vectors results in a mixed-grade multi-vector with scalar and pseudo-vector components $\endgroup$ – Christoph May 7 '14 at 23:45
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    $\begingroup$ @Christoph: thanks. Ignoring the impositions of nature, there's a rich structure of algebras over vector spaces for which division is well defined. Very interesting stuff, but I can't pretend I'm too well acquainted with most of it. $\endgroup$ – Zorawar May 8 '14 at 0:27
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To define vector division as the scalar result of one vector "divided" by another, where the scalar times the denominator vector would then give us the numerator vector, we can write the following: \begin{align*} \vec u&=w\vec v\\ \vec u\cdot\vec v&=w\vec v\cdot\vec v\\ \therefore w&=\frac{\vec u\cdot\vec v}{v^2} \end{align*}

The math for a scalar quotient works. That is one way to divide out a vector

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  • $\begingroup$ I see nothing wrong with your answer. $\endgroup$ – Džuris May 7 '14 at 18:05
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    $\begingroup$ In the first calculation, you don't specify what kind of mathematical object a vector quotient would be, and what kind of product would be needed to multiply two of them together. So it's not justified that you can turn $\frac{\vec{u}}{\vec{v}}\cdot\frac{\vec{v}}{\vec{v}}$ into $\frac{\vec{u}\cdot\vec{v}}{v^2}$. $\endgroup$ – David Z May 7 '14 at 20:12
  • $\begingroup$ To extend DavidZ's comment, it seems you are defining vector division by using vector division with $\vec v/\vec v\equiv1$. Rather circular logic. $\endgroup$ – Kyle Kanos May 8 '14 at 0:22
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    $\begingroup$ @Jim: Facepalm. Re-reading your post, I apparently misread/misunderstood it entirely. $\endgroup$ – Kyle Kanos May 8 '14 at 12:51
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    $\begingroup$ Now, this thing totally restricts the so called "division" in the case of linearly dependent vectors. $\endgroup$ – Subhasis Biswas Oct 6 '18 at 8:19
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It depends on the context. Division is usually defined as inverse of multiplication. If

$$x\cdot\vec{v}=\vec{u}$$

then, if there is only one $x$ that satisfies above relation, you can say that $x=\frac{\vec{u}}{\vec{v}}$.

The $x$ here can be scalar (so you multiplied vector with scalar) and it's only meaningful if you consider vectors which are pointing in the same direction.

$x$ could be a matrix and other answers have shown cases where the matrix is not unique.

$x$ could also be a vector and you could consider either dot or cross product. Again there are cases when this works and when it does not.

So you cannot divide by anything, there can be some divisions that cannot be defined, but that's fine - you cannot divide by zero in reals aswell. You just have to understand what you are doing and whether inverse is unique and if it's definable at all. There are cases where vector division makes sense and is useful.

For example, let's consider Lorentz force on charge that's moving in magnetic field. $$\vec{F}=q \vec{v}\times\vec{B}$$

If you can measure the force and one of the quantities on the right hand side, the other is the division (however, beware if it's inverse of right side or left side multiplication :)) of force and the measured right hand side quantity.

It could be written as $$\vec{v}=\frac{\vec{F}}{\vec{B}}(left)$$ where "left" and "right" is a matter of convention.

However, as Jerry pointed out, the solution is not unique.

So whenever you can multiply, you can check if there exists inverse. There are cases when there is no unique inverse, but if there is one, you can call it the division. Vectors are not totally on one side or the other - you can usually find a set of vectors for which certain division is meaningful.

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    $\begingroup$ Your statement about the cross product is not quite right -- if ${\vec F} = q{\vec v} \times {\vec B}$ for some $F,v,B$, then, for every choice of $c$, you also have ${\vec F} = q\left(c{\vec B} + {\vec v}\right)\times {\vec B}$, so the division will not be unique. $\endgroup$ – Jerry Schirmer May 7 '14 at 18:23
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    $\begingroup$ @JerrySchirmer oops, right. You can only determine the component that's normal to $\vec{B}$. Thanks! $\endgroup$ – Džuris May 7 '14 at 18:28
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As per the Wolfram Mathworld page

In general, there is no unique matrix solution to the matrix equation, $$\mathbf y=\mathbb A\mathbf x$$

An example is then given for $\mathbf y=2\mathbf x=(2,4)$ in which there are 3 different solutions.

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    $\begingroup$ are there 3 solutions, or infinite solutions of which they give three examples? $\endgroup$ – DavePhD May 7 '14 at 16:39
  • $\begingroup$ @DavePhD: While the website does not expressly say so, I believe that there are an infinite number of solutions (or at least extremely large number) and they give 3 examples. $\endgroup$ – Kyle Kanos May 7 '14 at 16:42
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    $\begingroup$ If you're interested in how many solutions are there, you can treat it as a particular linear system for the coefficients of $\mathbb A$. $\endgroup$ – fqq May 7 '14 at 17:25
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    $\begingroup$ I don't think it's possible for a system of linear equations to have only three solutions. Geometrically, the solution space corresponds to an intersection of hyperplanes, which can only ever be a point, line, plane, etc. depending on how many equations there are. You'd need to have some kind of curve involved (i.e. a higher-order polynomial) to get a finite number of solutions greater than 1. $\endgroup$ – David Z May 7 '14 at 22:30
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    $\begingroup$ My comment about the three solutions was not to stay that there are only three solutions, just that there are at least three solutions to show the non-uniqueness of the division. $\endgroup$ – Kyle Kanos May 7 '14 at 22:36
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Suppose we take $A = TB$, where $A$ and $B$ are vectors and $T$ is a tensor. Now if $A$ and $B$ given and vector division possible we can find the value of $T$. If we take a simple example $A = (a_1,a_2,a_3)$, $B = (b_1,b_2,b_3)$ and $T$ is a 3x3 matrix: $$T=\left(\begin{matrix}t_{11} & t_{12} & t_{13} \\ t_{21} & t_{22} & t_{23}\\t_{31} & t_{32} & t_{33}\end{matrix}\right)$$ Now from above relation we get three equations with nine unknowns, which never give a unique solution, so we can say vector division impossible unless we take $A$ and $B$ as parallel and $T$ as a scalar.

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    $\begingroup$ Welcome to PhysicsSE! Note that this site supports MathJax. You can click the link to learn the basics. I've taken the liberty of typesetting this post for you. $\endgroup$ – Chris May 15 at 15:08

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