I get the physical significance of vector addition & subtraction. But I don't understand what do dot & cross products mean?

More specifically,

  • Why is it that dot product of vectors $\vec{A}$ and $\vec{B}$ is defined as $AB\cos\theta$?
  • Why is it that cross product of vectors $\vec{A}$ and $\vec{B}$ is defined as $AB\sin\theta$, times a unit vector determined from the right-hand rule?

To me, both these formulae seem to be arbitrarily defined (although, I know that it definitely wouldn't be the case).

If the cross product could be defined arbitrarily, why can't we define division of vectors? What's wrong with that? Why can't vectors be divided?

  • 14
    $\begingroup$ Division is the inverse of multiplication. A vector space in which you can also multiply two vectors is called an algebra (over a field). The cross product is not a type of multiplication as it is not associative. The dot product also doesn't count as multiplication as it maps two vectors into a scalar. The Quaternions are an example of a vector space which is also an algebra. $\endgroup$
    – Olaf
    Commented Aug 29, 2011 at 11:15

9 Answers 9


I get the physical significance of vector addition & subtraction. But I don't understand what do dot & cross products mean?

Perhaps you would find the geometric interpretations of the dot and cross products more intuitive:

The dot product of A and B is the length of the projection of A onto B multiplied by the length of B (or the other way around--it's commutative).

The magnitude of the cross product is the area of the parallelogram with two sides A and B. The orientation of the cross product is orthogonal to the plane containing this parallelogram.

Why can't vectors be divided?

How would you define the inverse of a vector such that $\mathbf{v} \times \mathbf{v}^{-1} = \mathbf{1}$? What would be the "identity vector" $\mathbf{1}$?

In fact, the answer is sometimes you can. In particular, in two dimensions, you can make a correspondence between vectors and complex numbers, where the real and imaginary parts of the complex number give the (x,y) coordinates of the vector. Division is well-defined for the complex numbers.

The cross-product only exists in 3D.

Division is defined in some higher-dimensional spaces too (such as the quaternions), but only if you give up commutativity and/or associativity.

Here's an illustration of the geometric meanings of dot and cross product, from the wikipedia article for dot product and wikipedia article for cross product:

enter image description here enter image description here

  • 9
    $\begingroup$ Cross product exists in 3D and 7D only. $\endgroup$ Commented Dec 26, 2015 at 4:53
  • 1
    $\begingroup$ If you say that direction of cross product is the orthogonal direction of the area of parallelogram then how the direction of the vector changes when we commute the vectors. $\endgroup$ Commented Sep 5, 2016 at 4:50
  • 1
    $\begingroup$ Why would someone give an answer that is almost completely irrelevant to the question, I wonder. This was exactly the question in my mind as I was searching about it on Google, but this definitely is not an answer to it. I wish I could down-vote it. $\endgroup$ Commented Mar 26, 2017 at 20:41
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    $\begingroup$ @renormalizedQuanta Oh yes, I keep forgetting about 7D. I'll have to watch myself and be more careful as I'm rather fond of saying that the cross product only yields a vector because of a 3D co-incidence. Which is true, but one should also mention other special dimensions. $\endgroup$ Commented May 18, 2017 at 0:38
  • $\begingroup$ And as far as I know, it is kinda funny (and, last time I checked, not very simple) why this only works in 3 and 7 dimensions... I'm sure there is a thread in here about it. $\endgroup$
    – Vendetta
    Commented Sep 19, 2018 at 16:44

The best way is to ignore the garbage authors put in elementary physics books, and define it with tensors. A tensor is an object which transforms as a product of vectors under rotations. Equivalently, it can be defined by linear functions of (sets of vectors) and (linear functions of sets of vectors), all this is described on Wikipedia.

There are exactly two tensors which are invariant under rotations:

$\delta_{ij}$ and $\epsilon_{ijk}$

All other tensors which are invariant under rotations are products and tensor traces of these. These tensors define the "dot product" and "cross product", neither of which is a good notion of product:

$V \cdot U = V^i U^j \delta_{ij}$

and cross product

$(V \times U)_k = V^i U^j \epsilon_{ijk}$

It is pointless to try to think of the cross product as a "product", because it is not associative, $(A\times B)\times C$ does not equal $A\times(B\times C)$. It is also less than useful to think of the dot-product as a product in the usual sense, because it takes pairs of vectors to numbers, and $(A\cdot B)C$ does not equal $A(B\cdot C)$, because the first points in the C direction, and the second points in the A direction.

The best way is to get used to the invariant tensors. These generalize to arbitrary dimensions, and they are much clearer, and do not require a right-hand rule (this is taken care of by the index order convention). You will not find a single physics paper which uses the cross product, with the single exception of Feynman's 1981 paper "the qualitative behavior of Yang-Mills theory in 2+1 dimensions", and even if you do, it is trivial to translate.


You can divide vectors with clifford ("geometric") algebra.

The geometric product of vectors is associative:

$$abc = (ab)c = a(bc)$$

And the geometric product of a vector with itself is a scalar.

$$aa = |a|^2$$

These are all the properties required to define a unique product of vectors. All other properties can be derived. I'll sum them up, however: for two vectors, the geometric product marries the dot and cross products.

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

We use wedges instead of crosses because this second term is not a vector. We call it a bivector, and it represents an oriented plane. It can be instructive to introduce a basis to see this. $e_1 e_1 = e_2 e_2 = 1$ and $e_1 e_2 = -e_2 e_1$ capture the geometric product's properties for these orthonormal basis vectors. The geometric product is then,

$$ab = (a^1 e_1 + a^2 e_2) (b^1 e_1 + b^2 e_2) = (a^1 b^1 + a^2 b^2) + (a^1 b^2 - a^2 b^1) e_1 e_2$$

As I said, the geometric product of two vectors is invertible in Euclidean space. This is obvious from the associativity property: $a b b^{-1} = a(b b^{-1}) = a$. That $b b^{-1} = 1$ implies that

$$b^{-1} = b/|b|^2$$

It's informative to look at the quantity $a = (a b) b^{-1}$, using the grouping to decompose it a different way.

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

The first term is in the direction of $b$, the second is orthogonal to $b$. This decomposes $a$ into $a_\parallel$ and $a_\perp$.

What others have said is right, you can't define just the vector cross product to be invertible. This decomposition should convince you--you cannot fully reconstruct a vector without information from both the dot and cross products. And as has been said, this product is not commutative.

  • $\begingroup$ this is a great answer, except the use of superscripts for things that are not exponentiation. $\endgroup$
    – Rob
    Commented Dec 30, 2018 at 3:12

If you're going to define division of vectors, you're going to have to define over what multiplication field you're going to define the division: For ordinary numbers, I think of $\frac{x}{y}$ as being the number that, when multiplied by $y$, gives $x$. So, $\frac{\vec x}{\vec y}$ would have to be the vector that, when "multiplied" by $\vec y$, gives $\vec x$. If our multiplication field is the dot product, we're already in trouble, because the dot product of two vectors is a scalar, and the definition above would therefore require $\frac{\vec x}{\vec y}$ to simultaneously be a vector and a scalar.

Similarly, if our operation is the cross product, then we know that, for any vectors $\vec x$ and $\vec y$ and any scalar $c$, we have ${\vec x} \times {\vec y} = {\vec x}\times \left({\vec y} + c{\vec x}\right)$, so this means that there are an infinite number of vectors that satisfy the property "when crossproducted by $\vec y$, gives $\vec x$". Therefore, division over the cross product is not unique.


Another, more intuitive, take on your question (mathematicians: please look away) is to think of the dot (A.B = ABcosθ) and cross (AxB = ABsinθ) products as simply ways to measure degrees of parallelism of vectors versus perpendicularity (orthogonality) of vectors, in the sense that:

  • The dot product of parallel unit vectors, U.U yields a number U.U cos 0 = 1 (or a number of value A*B for vectors of arbitrary vector lengths) while of orthogonal (perpendicular) vectors it’s always zero, as cos 90°=0

  • Conversely, the cross product of parallel unit vectors is of magnitude 0 (as sin 0°=0) while of orthogonal unit vectors it’s 1 (or magnitude AxB for arbitrary lengths); but in this case the result is mapped into a vector, which needs to be perpendicular to the plane defined by the input vectors A and B (there’s no obvious way to assign it a direction within the plane defined by A and B). And recognise also that the cross product thereby gains a sense of handedness as AxB = -BxA, which turns out to be useful (example below).

Of course, it’s the use of sin and cos that determines the way these measures range from 0 to 1 for unit vectors; just imagine the products' values changing as you think of the A and B vectors rotating towards or away from each other for each type of product.

As far as physical significance is concerned (and ignoring the deeper insights available via Clifford Algebra etc.), these “products” turn out to be useful in so many situations that their physical significance is often taken for granted, rather than being emphasised (and maybe this underlies your question).

For the dot product: e.g. in mechanics, the scalar value of Power is the dot product of the Force and Velocity vectors (as above, if the vectors are parallel, the force is contributing fully to the power; if perpendicular to the direction of motion, the force is not contributing to the power, and it's the cos function that varies as the length of the projection of the force vector on the velocity vector varies; so it's not at all arbitrarily defined).

For the cross product: e.g. angular momentum, L = r x p (all vectors), so it seems perfectly intuitive for the vector resulting from the cross product to align with the axis of rotation involved, perpendicular to the plane defined by the radius and momentum vectors (which in this example will themselves usually be perpendicular to each other so the magnitude of rp*sin90°= rp). And if the direction of rotation changes, the sign of the momentum vector is reversed and so the cross product vector L also changes sign (hence the usefulness of mapping the cross product into a vector).

Note however, you can also calculate the number resulting from AB*sinθ (rather than mapping it into a perpendicular vector). It’s just the area of the parallelogram that is defined by the vectors A and B in the cross product.

Incidentally, there's nothing stopping you mapping the dot product into a perpendicular vector, if so desired - but it's probably not often useful to do so in physics.

As regards division, that’s a little more technical and dealt with well by the earlier replies. There’s also some approachable discussion on https://www.quora.com/Can-we-divide-a-vector-by-a-vector-and-why

I hope this is of some help for the less-expert members.


In addition to nibot's answer: division something is finding a part of something. In case of a vector, its part has the same direction but a smaller length. So it is natural to divide vectors by numbers, not by vectors.

Those dot and cross products are not simple products because they depend not only on lengths but also on orientations. They are called correspondences between a couple of vectors and numbers or vectors.


For products you have the answers. For division I recommend you to read more about quaternions. Interpretation of vectors in terms of quaternions allows for more reach algebra than vector space itself.

A little math right here. For natural definition of division you need at least division ring (one may comment that division algebra is enough, then add octonions to my answer). There is a theorem that the only finite dimensional division rings are reals, complex and quaternions. Vectors are vector space in three dimensions. So, any division for three-dimensional vectors will be "unnatural".


The question "what does $\vec{a}/\vec{b}$ equal?" is equivalent to asking "What do you multiply with $\vec b$ to get $\vec a$?" -- the answer is a matrix, assuming the multiplication involves some contraction afterwards. Equivalently, you multiply and contract a (1, 0) tensor $b^\mu$ by a (1,1) tensor $A_\mu^\nu$ to get a (0,1) tensor $b^\nu$.

But there are multiple matrices you can multiply $\vec b$ by to get $\vec a$. In two dimensions, you need two sets of "this maps to this" (and the knowledge that the mapping is linear) to pin down what the linear mapping is. In general in $n$ dimensions, you need $n$ such vectors -- so instead of dividing vectors, you divide sets of vectors -- these are called matrices.


Dividing a scalar, or a vector by a vector is a useful operation. It is an outer product (AKA, tensor product). For example, consider the Leibniz differential expressed using the basis vectors $\hat{\mathfrak{e}}^i$:

$$ d\mathit{f}=\nabla [\mathit{f}]\cdot d\mathfrak{x}=\frac{\partial \mathit{f}}{\partial x^i}dx^i$$

$$\approx\frac{\Delta \mathit{f}}{\Delta x^i}\hat{\mathfrak{e}}^i\cdot \Delta \mathfrak{x}$$

$$=\Delta \mathit{f}\left(\frac{1}{\Delta x^i}\hat{\mathfrak{e}}^i\cdot \Delta \mathfrak{x}\right)$$

$$=\frac{\Delta \mathit{f}}{\Delta \mathfrak{x}}\cdot \Delta \mathfrak{x}.$$

So $\frac{1}{\Delta x^i}\hat{\mathfrak{e}}^i=\frac{1}{\Delta \mathfrak{x}}$ is a vector. Of course we need to define our way out of $\frac{1}{\Delta x^i=0}.$

Now if we have the vector-valued function $\vec{\mathit{f}}\left[\mathfrak{x}\right]$ we can write $\frac{d\vec{\mathit{f}}}{d\mathfrak{x}},$ which is called the derivative matrix. If it's square, it is often called the Jacobean matrix.


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