My question is: We have addition, subtraction and muliplication of vectors. Why cannot we define vector division? What is division of vectors?

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    $\begingroup$ What multiplication are you talking about? An inner product? Because that's not an operation over the vectors. $\endgroup$
    – Stefan
    Nov 28, 2012 at 16:34
  • $\begingroup$ Division is the inverse of multiplication. If that inverse does not exist, then the statement "we have multiplication" may not be true in the sense you may be thinking. $\endgroup$
    – adam W
    Nov 28, 2012 at 16:35
  • $\begingroup$ I am thinking, that dot product and cross product are multiplications of vectors. $\endgroup$
    – Reader
    Nov 28, 2012 at 16:38
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    $\begingroup$ To prove that, and therewith to really explain why you cannot "have" something is in general very hard. Think of the trisection of an angle with ruler and compass. - By the way: If you define multiplication of two-dimensional vectors by imitating multiplication of complex numbers you can divide. $\endgroup$ Nov 28, 2012 at 16:44
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    $\begingroup$ @Reader Cross products are well-defined only for vectors of 3 and 7 dimensions. There is a generalization of a cross product called the wedge product, but this is also not a "multiplication" of vectors in a sense analogous to multiplication of the reals. $\endgroup$
    – Emily
    Nov 28, 2012 at 17:51

10 Answers 10


As already mentioned in the comments you have two ways of "multiplying" vectors. You have the dot product and the cross product. However, the dot product isn't a product.

When you multiply two rational numbers, you get a rational number. When you multiply two matrices, you get a matrix. When you multiply two complex numbers, you get a complex number. So you would want your product to satisfy that the multiplication of two vectors gives a new vector. However, the dot product of two vectors gives a scalar (a number) and not a vector.

But you do have the cross product. The cross product of two (3 dimensional) vectors is indeed a new vector. So you actually have a product. It is still a bit of a strange product in that it is not commutative. $\vec{x}\times\vec{y}$ isn't the same as $\vec{y}\times\vec{x}$. $\vec{x}\times\vec{y}$ = -$\vec{y}\times\vec{x}$

Now about division. If you have two real numbers $x$ and $y\neq 0$, we say that $\frac{x}{y} = z$ exactly when $x = yz$. So in that sense you could define a type of division of vectors.

However, again there are some problems with vectors. When we divide by a real number $y$, we can also consider this as multiplying by the inverse of $y$, that is, $y^{-1}$. The inverse of $y$ is that unique number $y^{-1}$ such that $yy^{-1} = 1$. The number $1$ is that "special" number that satisfies that $1x = x$ for all real numbers $x$. And you see that any (nonzero) number divided is $1$. The question is: what would the equivalent of $1$ be for vectors?

With vectors, you don't have such a "unit". There is no vector $\vec{1}$ such that the cross product of $\vec{1}$ with any other vector $\vec{x}$ is $\vec{x}$, that is, $\vec{1}\times \vec{x} = \vec{x}$.

So that is way we don't really have a division of vectors that "works" just like division of real numbers do.

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    $\begingroup$ Excellent answer! I want to add that the reason $\vec{1}\times \vec{x} = \vec{x}$ is not true is because the resulting vector by definition would be a vector perpendicular to the two vectors in the other side. you can't have $\vec{1}=\vec{1}$ such that the RHS vector is perpendicular to the vector on the LHS $\endgroup$ Dec 20, 2021 at 16:24
  • $\begingroup$ In particular, if we take $\vec{c} = \vec{a} \times \vec{b}$ then we say, in simple presentations, that, "the cross product gives a vector". But, in more advanced treatments we see that $\vec{c}$ is another "kind" of vector, which transforms differently under coordinate transforms. Some people call $\vec{c}$ an "axial vector". To really see what's going on we have to look at tensor character, which is complicated... But anyway, if we could divide and say $\vec{a} = \vec{c}/\vec{b}$ we would have a strange equation where the two sides of the equation transform differently. $\endgroup$ Jan 12, 2022 at 21:58

You can define division of vectors, but as multiplication and division are related operations, you can only do so by choosing a definition of multiplication that allows it.

As has been pointed out, in vector algebra we typically just define the dot and cross products. For two vectors $a$ and $b$, the dot product $a\cdot b$ tells us how much the two vectors are parallel. The cross product $a \times b$ tells us how perpendicular the vectors are, and moreover, it tells us something about their relative orientation--about the plane that the two vectors lie in. I submit to you without proof that the dot and cross products contain all the relevant information possible from two vectors. In other words, if one knows $a$ and $a \cdot b$ and $a \times b$, then one can reconstruct $b$.

Indeed, the formula for doing so is something like

$$b = (a \cdot b) a/|a|^2 + (a \times b) \times a/|a|^2$$

It should be intuitive that $a/|a|^2$ somehow "undoes" these two products. If there were a candidate for $a^{-1}$, then $a/|a|^2$ would be it.

But how can we go about this in a formalized way? The answer is to define a new product, one that combines the properties of the dot and cross products into a single operation. This operation is called the geometric product.

Let $e_1, e_2, \ldots, e_n$ be an orthonormal basis for $\mathbb R^n$. The geometric product of vectors is defined as follows:

$$e_i e_j = \begin{cases} 1, & i = j \\ -e_j e_i, & i \neq j\end{cases}$$

When two basis vectors are the same and multiplied via the geometric product, the result is a scalar, and so we capture the behavior of the dot product. When the two basis vectors are orthogonal, the result is antisymmetric, and we capture the behavior of the cross product. It's important to note, though, that this antisymmetric part does not result in a vector--rather, it results in a new object we call a bivector. Think of it as an oriented planar subspace, just as vectors are oriented line-like subspaces through $\mathbb R^n$.

The geometric product is linear on its arguments, so we can find the geometric product of $a$ and $b$ just by breaking them down into components. Furthermore, the geometric product is associative, so we can find $ab$ and then multiply (on either the left or right) by another vector $c$, and so on. For now, though, we can restrict ourselves to the case of two vectors. The geometric product is often written as

$$ab = a \cdot b + a \wedge b$$

This wedge neatly avoids one problem with the cross product--it doesn't exist in dimensions outside of 3 or 7. The wedge (which produces the bivector part mentioned earlier) does exist in any number of dimensions, though.

Now then, the geometric product admits multiplicative inverses (essentially, division). See that $aa^{-1} = 1 \implies a^{-1} = a/a^2$, just as I observed before. Because the geometric product is associative, it is meaningful to say that

$$a^{-1} a b = (a^{-1} a) b = \frac{1}{a^2}aab = \frac{a^2}{a^2} b = b$$

where on the other hand, associativity gives us the freedom to group the products differently, like so:

$$b=a^{-1} a b = a^{-1} (ab) = a^{-1} (a \cdot b + a \wedge b) = \frac{1}{a^2} [a(a \cdot b) + a (a \wedge b)]$$

which is just the geometric algebra form of the decomposition I wrote earlier. Here, it follows just from the freedom to group products as one sees fit. This is a powerful technique in geometric algebra, useful for proving many identities (even up to vector calculus and beyond).

All that aside, what is the product of two vectors then, under the geometric product? It is a scalar and a bivector, as we've established. One name for the set of such objects is spinors. Spinors are useful for representing rotations, and indeed, the product $ab$ gives us a spinor corresponding to a rotation from $b$ into the direction of $a$. In 2 dimensions, such spinors have only two components, and these correspond to complex numbers. In 3 dimensions, such spinors have 4 components (1 scalar, 3 bivector--for the 3 planes of 3d space), and these spinors correspond to quaternions and so on. Thus, the geometric product gives great insight into the nature of rotations and how they can be built from vectors.

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    $\begingroup$ Summary. We we want to divide the given vector $c$ with the given vector $a$, i.e. we are looking for the quotient $b=c/a$ for which $ba=c$. This can be done if the product $ba$ is the Clifford (or geometric) product. In this case, $b=c/a=ca^{-1}=ca/|a|^2$, since with this $b$, $$ba = (ca/|a|^2)a = c|a|^2/|a|^2=c$$ Here the product $ca$ is $$ca = c\cdot a + c\wedge a$$ which is generally not a vector, so neither is the quotient $b=c/a=ca/|a|^2$. Generally, the quotient of two vectors is a sum of a scalar and a bivector, just as the (Clifford) product of them. $\endgroup$
    – mma
    Sep 11, 2021 at 7:06

Given any vector $b$, you can find some nonzero $c,d$ with $b \cdot c=0$ and $b \times d=0$ (just take $c$ perpendicular to $b$, and $d$ parallel).

To say that $x=a / b$, where $/$ is a division operation that corresponds to dot product, should be equivalent to saying that $a = b \cdot x$. But if this is true, then also $a = b \cdot (x+c)$ where $b \cdot c=0$, so we should also say that $x+c=a/b$. That is, "dot-product division" is never uniquely defined, no matter the choice of $a$ and $b$. So it's not really a useful concept. Similar remarks apply to "cross-product division" — just replace $c$ by $d$.

On the other hand, there is (sort of) a definition of vector division based on scalar multiplication: if $a$ and $b$ are parallel vectors, then you can divide $a$ by $b$ to get a real number. Of course, this isn't defined for general pairs of vectors. But it is unique whenever it exists, which means it's occasionally a useful concept...


The quotient of two vectors is a quaternion by definition. (The product of two vectors can also be regarded as a quaternion, according to the choice of a unit of space.) A quaternion is a relative factor between two vectors that acts respectively on the vector's two characteristics length and direction; through its tensor or modulus, the ratio of lengths taken as a positive number; and the versor or radial quotient, the ratio of orientations in space, taken as being equal to an angle in a certain plane. The versor has analogues in the $+$ and $-$ signs of the real numbers, and in the argument or phase of the complex numbers; in space, a versor is described by three numbers: two to identify a point on the unit-sphere which is the axis of positive rotation, and one to identify the angle around that axis. (The angle is canonically taken to be positive and less than a straight angle, so the axis of positive rotation is reversed when the two vectors are exchanged in their plane.) The tensor and versor which describe a vector quotient together have four numbers in their specification (therefore a quaternion).

Two quaternions are multiplied or divided by multiplying or dividing their respective tensors and versors. A versor has a representation as a great circle arc, connecting the points where a sphere is pierced by the dividend and divisor rays going from its center; these arcs have the same condition for equality as vectors do, viz. equal magnitude, and parallel direction. But on a sphere, no two arcs are parallel unless they are part of the same great circle. So, the vector-arcs are compared or compounded by moving one end of each to the line of intersection of their two planes, then taking the third side of the spherical triangle as the arc to be the product or the quotient of the versors.


Let vector $A = 2i+4j+8k$ and vector B is unknown but the cross product $C= 4i+6j=16k$

Let $B=xi+yj+zk$


$(4z-8y)i+(2z+8x)j+(2y-4x)k=4i+8j+16k $

$z-2y=1$ gives $z=1+2y$

$z+4x=4$ gives $2y+4x=3$


On solving those equations $y=\frac{19}{4}$, hence $x= -\frac{23}{8}$ and $z=\frac{21}{2}$ and vector $B= (-23/4)i+(19/4)j+(21/2)k$


One way to give a meaning to the division between vectors could be the following:

let $\vec{x}$ and $\vec{y}$ be two vectors of $R^n$, then finding the "division of two vectors $\vec{x}$ and $\vec{y}$" can mean finding all matrices A, such that:

$A\vec{x} = \vec{y}$

It is clear that in general the result will not be unique, but given a solution $A_1$ another solution $A_2$ can be found by adding $A_1$ with a certain matrix that has $x$ in its kernel, i.e.:


So we could say that if $A\vec{x} = \vec{y}$, then $A$ belongs to the set $\vec{y}/\vec{x}$, i.e. : the "division of vector $\vec{y}$ by $\vec{x}$"

For example:

if $\vec{y}$ and $\vec{x}$ are collinear, the solution could be written: as $y_k/x_k I+N_x$, where :

  • $y_k$ and $x_k$ are the $k$-th component for some k of $\vec{y}$, $\vec{x}$, respectively; because of collinearity the ratio is the same for any $k$;
  • $I$ is the identity matrix
  • $N$ are all matrices whose kernel contains $\vec{x}$.

In the two-dimensional case, assuming in the following example that the components $x_k$ are all non-zero, one can write:

$A_{11} x_1+A_{12} x_1=y_1$

$A_{21} x_1+A_{22} x_2=y_2$

this gives us two equations in 4 unknowns. We can think of writing a particular solution like this:

$A_{11}=y_1/x_1, A_{22}=y_2/x_2, A_{12}=A_{21}=0$


$$A=\left[\begin{array}{cc}y_1/x_1&0\\0&{y_2/x_2}\end{array}\right] $$

which provides us with a particular solution to which to add the matrices, parameterized in the form (u,v are the parameters): $$N=\left[\begin{array}{cc}u x_2&-y x_1\\v x_2&-v x_1\end{array}\right] $$ We can easily see that :$N\vec{x}=0$

Therefore in $R^2$ the solution $\vec{y}/\vec{x}$, under the abovementioned hipothesys, can be parametrized as follows:

$$\vec{y}/\vec{x}=A(u,v)=\left[\begin{array}{cc}y_1/x_1&0\\0&{y_2/x_2}\end{array}\right] + \left[\begin{array}{cc}u x_2&-u x_1\\v x_2&-v x_1\end{array}\right] $$

PS: This concept seems linked with the "vector inversion" discussed here:

Inverse of a vector


The definition of a Vector space allows us to add two vectors, subtract two vectors, and multiply a vector by a scalar.

When it comes to multiplying two vectors you have the dot product and the cross product. However, the dot product gives a scalar as output.

This leaves us with the cross product whose result is a vector.

Still,the division of vectors cannot be defined because of the uniqueness issue.

Let $\vec A,\vec B,\vec C$ be three vectors such that $$\vec A \times \vec B =\vec C$$

We want vector division $\frac {\vec C}{\vec A}$ so that $$ \frac {\vec C}{\vec A}=\vec B $$

But Vectors other than $\vec B$ can be cross multiplied with $\vec A$ to obtain the same cross product.

Let $\vec V$ be any vector $ (\vec B \not=\vec V)$ such that $\vec B - \vec V $ is parallel to $\vec A$.Therefore $$ \vec A \times (\vec B - \vec V)=0 $$

$$\implies \vec A \times \vec B-\vec A \times \vec V =0 \text{ (Distributive Law) } $$

$$\implies \vec A \times \vec B=\vec A \times \vec V=\vec C $$

$$ \frac {\vec C}{\vec A}=\vec B \text{ or } \vec V \text{??}$$

So we cannot uniquely define vector division.

We can also see this algebraically in terms of components.

$\vec A$ and $\vec C$ are known and $\vec B$ is unknown.


$$\vec A = a_1\hat i+a_2\hat j+a_3\hat k $$

$$\vec C= c_1\hat i+c_2\hat j+c_2\hat k $$

$$ \vec B = x\hat i + y\hat j + z\hat k $$

$\vec C$ is perpendicular to $\vec A $ and $\vec B$. $$\vec A . \vec C = (a_1c_1+a_2c_2+a_3c_3) = 0 $$ $$ \vec A \times \vec B = (a_1\hat i+a_2\hat j+a_3\hat k)\times(x\hat i+y\hat j+z\hat k)= c_1\hat i+c_2\hat j+c_2\hat k $$

$$ \implies \begin{vmatrix} \hat i & \hat j &\hat k \\ a_1 & a_2 & a_3 \\ x & y & z \end{vmatrix} =c_1\hat i+c_2\hat j+c_2\hat k $$

$$ \implies (a_2z-a_3y)\hat i+(a_3x-a_1z)\hat j+(a_1y-a_2x)\hat k=c_1\hat i+c_2\hat j+c_2\hat k $$

  • $a_2z-a_3y=c_1 \implies 0.x+(-a_3)y+a_2z=c_1 $

  • $a_3x-a_1z=c_2 \implies a_3x+0.y+(-a_1)z=c_2$

  • $a_1y-a_2x=c_3 \implies (-a_2)x+a_1y+0.z=c_3$

$$ D=\begin{vmatrix} 0 & -a_3 & a_2 \\ a_3 & 0 & -a_1 \\ -a_2 & a_1 & 0\end{vmatrix}(\text {Determinant of a }3\times 3 \text { skew symmetric matrix}) $$


$$ D_x=\begin{vmatrix} c_1 & -a_3 & a_2 \\ c_2 & 0 & -a_1 \\ c_3 & a_1 & 0\end{vmatrix}=c_3(a_1a_3)-a_1(-c_1a_1-c_2a_2)$$

$$= a_1(a_1c_1+a_2c_2+a_3c_3)$$


Similarly $$ D_y=a_2(a_1c_1+a_2c_2+a_3c_3)=0 $$


From Cramer's Rule we know that this implies there are infinitely many solutions for $x , y \text{ and } z$.Meaning $\vec B$ is not unique.Thus vector division cannot be defined.


$ a/b = D^{-1}.a $ , where D is the diagonal matrix with $ D_{ij} = \delta_{ij}b_i $. This operation produces a vector whose entries are the divisions of the corresponding entries for a and b.


Introduction. There are very good answers and mine will only supplement them. About "inverse" of dot product, imagine you have $\vec a \cdot \vec b=c$ and you know $\vec b$ and $c$ and want to know what is $\vec a$. What this problem means? It means that you have parallelogram of known area witch lies on known vector and you need to determine other vector. It turns out that there are many such $\vec a$, so you have a set of vectors, more precisely a curve. Every curve can be expressed as vector function. Now we will try to find this vector function.


  1. $\vec a \cdot \vec b=c$
  2. $(a_x i+a_y j )\cdot (b_x i+b_y j)=c$
  3. Let us call $a_x=x$ and $a_y=y$
  4. $(x i+y j )\cdot (b_x i+b_y j)=c$
  5. $x b_x+y b_y=c$
  6. $y=\frac{c- x b_x}{b_y}$
  7. $b_x= \vec b \cdot i$ and $b_y=\vec b \cdot j$
  8. $y=\frac{c- x \vec b \cdot i}{\vec b \cdot j}$
  9. Vector function of this function is: $\vec a=x i+(\frac{c- x \vec b \cdot i}{\vec b \cdot j})j$
  10. It is equivalent to: $\vec a = x i+\left (\frac{c- x |b|cos(\alpha)}{|b| sin(\alpha)} \right)j$, where $\alpha=\angle(\vec a;\vec b)$

Summery. These derivations can be carried out in both directions so: $\vec a \cdot \vec b=c$ is equivalent to $\vec a=x i+(\frac{c- x \vec b \cdot i}{\vec b \cdot j})j$ and it is equivalent to $\vec a = x i+\left (\frac{c- x |b|cos(\alpha)}{|b| sin(\alpha)} \right)j$, where $\alpha=\angle(\vec a;\vec b)$

Similar answer. You can find similar answer about finding "inverse" of cross product here: Finding $\vec b$ from $\vec a \times \vec b$ , $\vec a$ and $\alpha=\angle(\vec a;\vec b)$ using only vector algebra..

  • $\begingroup$ Although this answer is interesting, it doesn't talk about division between vectors, but between a scalar and a vector! $\endgroup$ Oct 28 at 16:45

In three dimensions:

$$\cos \theta = \frac{\boldsymbol a \cdot \boldsymbol b}{|\boldsymbol a| |\boldsymbol b|}, \sin \theta = \frac{|\boldsymbol a \times \boldsymbol b|}{|\boldsymbol a| |\boldsymbol b|}$$

hence $\frac{|\boldsymbol a \times \boldsymbol b|}{\boldsymbol a \cdot \boldsymbol b}$, the magnitude of the cross product divided by the dot product is $\tan \theta$, where $\theta$ is the angle between the two vectors.


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