What is the intuition behind the fact that the cross product of two vectors is orthogonal? Every video I've seen just says it is orthogonal but they do not explain why. Since I have terrible memory, I need to understand things intuitively or I will just forget them.

By the way, I do not understand the intuition behind the calculation for the cross product either, so if that is necessary to understand why it is orthogonal, please explain.

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    $\begingroup$ Do any of the answers here help you? $\endgroup$ – Casteels Mar 13 '15 at 11:32
  • $\begingroup$ Two vectors define a plane. The most symmetrical way to leave that plane is the perpendicular direction. Another choice could have been a direction such that all three angles are equal. $\endgroup$ – user65203 Mar 13 '15 at 20:46
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    $\begingroup$ "Orthogonal" takes as input two vectors: the grammar is "a is orthogonal to b," not "a is orthogonal." $\endgroup$ – Qiaochu Yuan Mar 22 '15 at 8:19
  • $\begingroup$ This video is really all you need. $\endgroup$ – Daniel W. Farlow Mar 24 '15 at 20:32
  • $\begingroup$ Which answer is closest to your intuition? $\endgroup$ – user48672 Mar 31 '15 at 11:19

Its not really an "intuition" thing, the cross product is defined that way. The following may help you understand why.

The short answer is that the cross product appears in many physical systems - most famously in calculating the force produced by an electrical current in a magnetic field. But this begs the question of why it appears so often in physics.

In physics, you often get linear relationships - in the example above doubling the current will double the force, as will doubling the magnetic field. So the current vector and magnetic vector are somehow multiplied together to find the force vector. You could potentially define vector multiplication in lots of ways. But there is an additional constraint in physics, that you must get the same answer for the force vector however you orient your co-ordinate system. Nature doesn't have a coordinate system, so the answer must be the same however you define what is the "x axis" and what is the "y axis". If you set up a coordinate system with the x axis along the direction of the magnetic field and the y axis along the current vector, then you should get the same answer for force as if you set up the co-ordinate system with x axis in the current direction and the y axis along the magnetic field direction.

All physical laws must follow this rule; nature doesn't have a preferred direction for x and y. The dot product (which presumably you have just learned about) follows this rule.

If you want to multiply two vectors in 3D space to form another vector, up to a constant term the only definition of vector multiplication which has this property of being independent of the coordinate system is the cross product. It is the unique definition of vector multiplication which is independent of the choice of coordinate system.

So nature follows this rule whenever two (3D) vectors are multiplied together to form a 3rd vector. Any other rule would be depend on the choice of coordinate system. So natural processes in three dimensions are in a sense forced to use this rule, because its the only rule which is independent of the co-ordinate system. In 2D, there is no possible rule which is independent of the co-ordinate system. In higher dimensions there are more choices. But we live in a 3D Universe (spatial dimensions), so the cross product rule is forced onto nature and hence onto physics and maths.

  • $\begingroup$ I'm not sure what you mean here, since (1) there is always a possible and unique definition of product of vectors via Clifford algebra that matches the cross product (the Hodge dual of the wedge product of the vectors.) So, there is one for 2D, but it's just a scalar; there is one for >3D, but its not a vector. You're right that 3D is special, but not for the existence and uniqueness of such product, but the fact that it is still a vector. $\endgroup$ – Artur Araujo Mar 27 '15 at 15:20
  • $\begingroup$ (2) It's not clear why frame invariance leads to the vector product (in fact in metric free geometry, not such product is defined at all, and you still have physics - with differential forms.) In fact, many frame invariant physical quantities are not cross products... From a frame-invariance point of view, the vector product is only understood in terms of rotational (and spin) symmetries, not general frame invariance. $\endgroup$ – Artur Araujo Mar 27 '15 at 15:20

The answer depends on the definition of the cross product. One possibility of the definition is to use that the map

$$E=\Bbb R^3\to E^*,\quad x\mapsto \langle x,\cdot\rangle$$ is an isomorphism (Riesz representation) and for $x,y\in E$ the map

$$\det: E\to \Bbb R,\quad z\mapsto \det(x,y,z)$$ is a linear form on $E$ so there is a unique element of $\Bbb R^3$ denoted $x\wedge y$ such that $$\det(x,y,z)=\langle x\wedge y,z\rangle,\quad \forall z$$ so using the properties of the determinant we see that

$$(x\wedge y)\perp x\quad\text{and}\quad (x\wedge y)\perp y$$

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    $\begingroup$ I am only just learning about the cross product so do not understand most of the symbols and terminology you used, can use more basic terminology? $\endgroup$ – Ray Kay Mar 13 '15 at 11:50

The best way to intuitively understand the vector product is by thinking of it in terms of an area form. By this I mean something to which I can give a patch of a plane, and it will tell me the area. This is important, for example, to define area/surface/double integrals. Usually, one intuitively motivates the double integral by cutting the plane by little squares with sides parallel to the coordinate axes, and computing their area as $dxdy$ (you will then multiply by whatever you're integrating.

This procedure, however, doesn't quite cut it. The first thing is that when you go on to surfaces in three dimensions, you not only need to know how large the area is, but you need to know how the little patches of surfaces stand inside three-dimensional space. The easiest way to do this is by giving a direction that is perpendicular to the plane (you can think of the patch being so small that it approximates a patch of a plane.) This is where the vector nature of the vector product kicks in. Another thing is that I need to have some orientation for the patch, for the integration (this is a complicated statement, but remember that even in one dimension, integrating forwards or backwards gives different results; more importantly I might be computing the flux of something, so I need to know what way is in, what way is out.) This orientation requirement gives the vector product its anti-symmetry.

So, here's the input we can give to the vector product: take a small rhombus, I can consider the two vectors defining its sides. Here's what I want from the vector product: that its norm equal the area of the rhombus; that its direction tell me how the rhombus is in space; that it also point "up" or "down" according to which way I'm going. The formula for the vector product accomplishes all these things, as I'm sure you have, or you can check.

As an extra, thinking of the vector product in this way will open you the door to later on understand the definition of important objects called differential forms. Once you study differential geometry, these objects are hugely important, and they pretty much generalize this notion. A further way to generalize it is precisely what user48672 mentioned, by defining quaternions, and, more generally, Clifford algebras.

  • $\begingroup$ I like this explanation and I have a question about it. Is the history of the development of the vector product similar to what you have here? By that I mean, was the definition of the vector product crafted specifically to generate one vector that indicates the area, orientation, and order of traversal of a plane, or was the vector product originally motivated by something else and it also works for this? $\endgroup$ – Todd Wilcox Mar 27 '15 at 14:58
  • $\begingroup$ To be completely honest, I cannot accurately describe the historical development. What I can say is the following: the vector product first shows up as a quaternion product, and it was quickly integrated into electromagnetism, where surface integrals are prominent and where the cross product in the differentatial form of Maxwell's laws is counterpart to those integrals. I'm not sure what the precise motivation was, but already early on it was understood that this product is an area form: Weyl himself mentions this in an essay in this book <store.doverpublications.com/0486489035.html> $\endgroup$ – Artur Araujo Mar 27 '15 at 15:08

You got tons of answers but Sami Ben Romdhane's post is worth expanding upon, in my opinion, for if you've just started studying linear algebra, this approach will be relatively new.

I don't know how familiar you are with terms, so I will assume you know what $\mathbb{R}^3$ is and what matrices and determinants are, and of course the standard scalar and cross product in $\mathbb{R}^3$.

If $A$ is some sort of function between vector spaces, for example between $\mathbb{R}^n$ and $\mathbb{R}^m$, in this case, we call this function linear, if for every $x,y\in\mathbb{R}^n$ and every $\alpha,\beta\in\mathbb{R}$, it is true that $$ A(\alpha x+\beta y)=\alpha A(x)+\beta A(y). $$

As you can see, this is rather similiar to distributivity.

Let us define volume before area. Let $u,v,w\in\mathbb{R}^3$ be vectors, and let $\mathrm{vol}$ be a function that takes in three vectors and returns the (oriented) volume of the paralelepiped they span, so $$ \mathrm{vol}:\mathbb{R}^3\times\mathbb{R}^3\times\mathbb{R}^3\rightarrow\mathbb{R},\ (u,v,w)\mapsto\mathrm{vol}(u,v,w). $$ What sort of properties does this function have?

You can check by elementary geometric methods, that the volume of a paralelepiped must be linear in its three independent edges separately, if you allow the volume to be oriented. Here oriented means that it can take on negative values, and if you exchange two of its sides, it must change signs.

This determines most properties of this $\mathrm{vol}$ function's properties. The volume function, $\mathrm{vol}(u,v,w)$ must be linear in all three of its variables separately, and must be so-called "skew-symmetric", which means that if you swap any two of its variables, it will change sign, so $$ \mathrm{vol}(u,v,w)=-\mathrm{vol}(v,u,w). $$

This alone does not determine the volume function exactly, but if $$ e_1=(1,0,0),\ e_2=(0,1,0),\ e_3=(0,0,1) $$ are the standard basis vectors of $\mathbb{R}^3$, you know that this is an orthonormal basis, meaning that all of these are of unit length and any different two of them are orthogonal to one another, plus we take this $\{e_1,e_2,e_3\}$ set to be of positive orientation in this order. This means that $e_1,e_2,e_3$ basically span the "unit cube", so we want $$ \mathrm{vol}(e_1,e_2,e_3)=1. $$

These properties now uniquely determine the volume function. If $$ u=u_1e_1+u_2e_2+u_3e_3 \\ v=v_1e_1+v_2e_2+v_3e_3 \\ w=w_1e_1+w_2e_2+w_3e_3,$$ then the value of $\mathrm{vol}$ on them is given by $$ \mathrm{vol}(u,v,w)=\mathrm{vol}\left(\sum_{i=1}^3u_ie_i,\sum_{j=1}^3v_je_j,\sum_{k=1}^3w_ke_k\right), $$ and by linearity, this is $$ \sum_{i,j,k=1}^3u_iv_jw_k\mathrm{vol}(e_i,e_j,e_k), $$ and we know that $\mathrm{vol}(e_1,e_2,e_3)=1$, but we also know that by the skew-symmetric property, $\mathrm{vol}(e_2,e_1,e_3)=-1$ and $\mathrm{vol}(e_1,e_1,e_3)=0$, well, basically, the value of $\mathrm{vol}$ on $e_i,e_j,e_k$ is $1$ if $i,j,k$ is an even permutation of $1,2,3$, 0 if any of the indices are repeated, and -1 if $i,j,k$ is an odd permutation of $1,2,3$.

(If you have ever encountered the so-called Levi-Civita symbol, $\epsilon_{ijk}$, then this is its definition, so $\mathrm{vol}(e_i,e_j,e_k)=\epsilon_{ijk}$.)

Because of this, instead of summing on indices, we can sum on permutations, let $S_3$ be the symmetric group of $(1,2,3)$, so the set of all permutations of $(1,2,3)$, then $$ \sum_{i,j,k=1}^3u_iv_jw_k\mathrm{vol}(e_i,e_j,e_k)=\sum_{\pi\in S_3}\mathrm{sgn}(\pi)u_{\pi(1)}v_{\pi(2)}w_{\pi(3)}\mathrm{vol}(e_1,e_2,e_3)= \\ =\sum_{\pi\in S_3}\mathrm{sgn}(\pi)u_{\pi(1)}v_{\pi(2)}w_{\pi(3)}, $$ where $\mathrm{sgn}(\pi)$ is the sign of the permutation $\pi$, and as you can see what we got here is exactly the definition of the determinant whose rows (or columns) are the vectors $u,v$ and $w$.

So $$ \mathrm{vol}(u,v,w)=\det(u,v,w)=\begin{vmatrix}u_1&u_2&u_3\\v_1&v_2&v_3\\w_1&w_2&w_3\end{vmatrix} .$$

Now, as we have noted, the volume function, thus the determinant is linear in all of its variables separately, so let $u,v\in\mathbb{R}^3$ fixed vectors and let $x$ be a variable vector, then $$ \det(u,v,x) $$ is linear in $x$, therefore the rule $x\mapsto \det(u,v,x)$ is a linear function from $\mathbb{R}^3$ to $\mathbb{R}$, because it maps the vector $x$, to a number $\det(u,v,x)$.

It is a linear algebraic theorem that you are probably unfamiliar with, that if $V$ is a (finite-dimensional) vector space that has a scalar product on it (like $\mathbb{R}^3$), then the set of all number-valued linear functions that are defined on $V$ can be identified with the set $V$ itself.

In our example, that means that if $\omega:\mathbb{R}^3\rightarrow\mathbb{R}$ is a linear function, that maps a vector to a number, then there is also such an $y_\omega\in\mathbb{R}^3$ vector, that $\omega(x)=y_\omega\cdot x$ for any $x$ vector.

Our linear function in this case is the map $x\mapsto\det(u,v,x)$, which is also dependant on $u$ and $v$. Let us denote the unique vector which corresponds to this linear map as $u\times v$. Then by our prior theorem $$ \det(u,v,x)=(u\times v)\cdot x, $$ and since the determinant is linear and skew-symmetric in $u$ and $v$ too, this $u\times v$ vector must be linear and skew-symmetric in both vectors as well.

This $u\times v$ vector is what we call the cross product of $u$ and $v$.

Now, why is this orthogonal to $u$ and $v$?

We see that the cross product of $u$ and $v$ is that vector, that when scalar producted with an arbitrary vector, the result of the scalar product is the volume of the three vectors ($u,v$ and the arbitrary vector). We know that the scalar product is the largest when the two vectors you take the scalar product of are parallel, and the scalar product of two orthogonal vectors is null.

We also know that the volume of a paralelepiped is the largest if a designated "third" vector is orthogonal to the base (which is spanned by the "first two" vectors), and zero, if the "third" vector is parallel with the base.

Putting the two together, the $\det(u,v,x)$ expression for equal lengths, but different angles will be the largest, when $x$ is orthogonal to the plane spanned by $u$ and $v$, but $(u\times v)\cdot x$ will be the largest, if $u\times v$ is parallel with $x$, but the two expressions are the same, so the cross product vector is orthogonal to the two vectors that it results from.


You can use the triple product: $A\cdot(B\times C)$. This is equal to the signed volume of the paralellepiped defined by the three vectors. So in particular, if A is in the plane of B and C, the result will be zero.


Try this. In 3 dimensions, the cross product of (a1, a2, a3) and (b1, b2, b3) is (a2*b3- a3*b2, a3*b1- a1*b3, a1*b2- a2*b1) Now try the inner product of a and this vector, the inner product of b and this vector. This cross product is orthogonal (perpendicular) to each of these. Since two vectors in 3 dimensions define a plane (unless the two are collinear), the cross product is perpendicular to the plane.

  • $\begingroup$ I think you know how to take the inner product of two 3 dimensional vectors. In sticking to 3 for purposes of intuition. So just take the inner product of the vectors as above. Now about why the cross product has the form or takes. $\endgroup$ – waltzingmonkey Mar 24 '15 at 10:30
  • $\begingroup$ I think you know how to take the inner product of two 3 dimensional vectors. In sticking to 3 for purposes of intuition. So just take the inner product of the vectors as above. Now about why the cross product has the form or takes. You know that (a•b) is |a||b|cos(theta) where theta is the angle between the two vectors a and b. You can express this in terms of the components of a and b. Now try and compute |a||b| sine(theta) in terms of the components of a and b. You may find its a familiar expression. You will also see this doesn't work in dimensions other than 3. $\endgroup$ – waltzingmonkey Mar 24 '15 at 10:37

Let's say you want to take the cross-product of the two vectors shown coordinate-wise below:

$$v_1 \times v_2 = \begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix} \times \begin{pmatrix}x_2\\y_2\\z_2\end{pmatrix}$$

If you know about the determinant of a $3\times3$ matrix, let $x, y, z$ be unknown and write

$$\Delta = \begin{vmatrix} x & y & z \\ x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \end{vmatrix}$$

Then by the usual formula for the determinant of a $3\times3$ matrix (I think of it as adding the products along all the "right diagonals" and subtracting the products along all the "left diagonals"), $$\begin{eqnarray} \Delta &=& x y_1 z_2 - x z_1 y_2 + y z_1 x_2 - y x_1 z_2 + z x_1 y_2 - z y_1 x_2 \\ &=& (y_1 z_2 - z_1 y_2)x + (z_1 x_2 - x_1 z_2)y + (x_1 y_2 - y_1 x_2)z \end{eqnarray}$$

Now we know that $ax + by + cz$ is the dot product of the vectors $\begin{pmatrix}a\\b\\c\end{pmatrix}$ and $\begin{pmatrix}x\\y\\z\end{pmatrix},$ and that if the dot product is zero these two vectors are orthogonal. But in fact this is exactly the formula we have just written, if we let $$\begin{pmatrix}a\\b\\c\end{pmatrix} =\begin{pmatrix}y_1 z_2 - z_1 y_2 \\ z_1 x_2 - x_1 z_2 \\ x_1 y_2 - y_1 x_2\end{pmatrix} = v_1 \times v_2.$$ So if we replace $x,y,z$ in the matrix with the coordinates of $v_1$, which are $x_1,y_1,z_1$, the determinant of the matrix will be the dot product of $v_1 \times v_2$ with $v_1$. But the matrix written this way has two rows the same, so its determinant is zero, ergo that dot product is zero, ergo $v_1 \times v_2$ is orthogonal to $v_1$. A similar argument shows that $v_1 \times v_2$ is orthogonal to $v_2$.

As far as remembering the formula goes, I too have a bad memory for such things. I usually can remember that each coordinate of the cross-product involves two products of coordinates of the two given vectors, and that the $i$th coordinate of the cross-product does not involve the $i$th coordinate of either given vector. For example, for the $z$-coordinate of the cross product I can use $x_1, x_2, y_1,$ and $y_2$ but not $z_1$ or $z_2$. This reminds me that the $z$-coordinate of the cross product is either $x_1 y_2 - y_1 x_2$ or $y_1 x_2 - x_1 y_2$. To remember which one it is, I then try an easy problem that will give me a non-zero $z$-coordinate, for example I try to compute $\begin{pmatrix}1\\0\\0\end{pmatrix} \times \begin{pmatrix}0\\1\\0\end{pmatrix},$ which I know should come out to $\begin{pmatrix}0\\0\\1\end{pmatrix}.$ Only one of the two formulas works, so that's the one I use.

Another mnemonic I have seen uses the matrix determinant again, that is, you write

$$ \begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix} \times \begin{pmatrix}x_2\\y_2\\z_2\end{pmatrix} = \begin{vmatrix} e_1 & e_2 & e_3 \\ x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \end{vmatrix} $$

and then write out the determinant as I did before, but with $e_1, e_2, e_3$ instead of $x,y,z$. Then you say that instead of numbers, $e_1,$ $e_2,$ and $e_3$ are the basis vectors of $\mathbb R^3$, so the things we multiply them by are the coordinates of a vector.


The cross-product was defined by Gibbs to mimic the properties of quaternion multiplication without the negativity of the square. Quaternions are very economic way to represent rotations in 3D and therefore many physical equations.

See for example here.




  • $\begingroup$ Now, please CHECK the sources before down-voting. $\endgroup$ – user48672 Mar 23 '15 at 1:10
  • $\begingroup$ id just like to point out that the asker said many time he is just beginning to learn about these things. so it is quite clear he would not know about quaternions $\endgroup$ – Quality Mar 23 '15 at 1:12
  • $\begingroup$ Quite the contrary, if he reads about Geometric algebra many obscure conventions will be come intuitively clear. $\endgroup$ – user48672 Mar 23 '15 at 1:13
  • $\begingroup$ I think that here the only qualified person to rate the answers is the one who set the bounty. So spare useless efforts :) $\endgroup$ – user48672 Mar 23 '15 at 21:32

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