Can we divide two vectors?

Question

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

Answer 1

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.

Answer 2

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

Answer 3

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.

Answer 4

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.

Answer 5

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.

Answer 6

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.