$\mathbf{a}\times \mathbf{b}$ follows the right hand rule? Why not left hand rule? Why is it $a b \sin (x)$ times the perpendicular vector? Why is $\sin (x)$ used with the vectors but $\cos(x)$ is a scalar product?

So why is cross product defined in the way that it is? I am mainly interested in the right hand rule defintion too as it is out of reach?

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    $\begingroup$ The simplest answer is probably "because physicists historically have found that definition useful". $\endgroup$
    – BHT
    Sep 25, 2016 at 17:18
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    $\begingroup$ At a really basic level we want the cross product to take two vectors and give us a vector perpendicular to both. Clearly we have two choices in non-degenerate cases, and choosing the "right hand rule" is just a matter of convention. You can equally well take the left hand rule, but this is just multiplying the usual cross product by $-1$. $\endgroup$
    – 3-in-441
    Sep 25, 2016 at 17:20
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    $\begingroup$ The cake is a lie. Cross products don't give real vectors. I avoid them as a rule except as a useful way to get an orthogonal vector to two others when I happen to need one in $\Bbb R^3$. $\endgroup$
    – user137731
    Sep 25, 2016 at 18:15
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    $\begingroup$ Why the right-hand rule as opposed to the left-hand rule? Because it had to be one or the other, and if it was the left-hand rule, you'd be asking why it wasn't the right-hand rule. $\endgroup$
    – user253751
    Sep 25, 2016 at 23:11
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    $\begingroup$ @koe: no this is not just a math question. It is a question about why it was defined that way. So it is a question about human convention used in mathematics and that has often to do with convenience for physical purpose. You could, however, still ask why were physical quantities (magnetic field, electrical current/charge and force) defined that way that we use the right hand rule (and not the left hand rule). Answer: Probably because most humans are right handed. $\endgroup$
    – Curd
    Sep 26, 2016 at 10:23

13 Answers 13


The cross product originally came from the quaternions, which extend the complex numbers with two other 'imaginary units' $j$ and $k$, that have noncommutative multiplication (i.e. you can have $uv \neq vu$), but satisfy the relations

$$ i^2 = j^2 = k^2 = ijk = -1 $$

AFAIK, this is the exact form that Hamilton originally conceived them. Presumably the choice that $ijk = -1$ is simply due to the convenience in writing this formula compactly, although it could have just as easily been an artifact of how he arrived at them.

Vector algebra comes from separating the quaternions into scalars (the real multiples of $1$) and vectors (the real linear combinations of $i$, $j$, and $k$). The cross product is literally just the vector component of the ordinary product of two vector quaternions. (the scalar component is the negative of the dot product)

The association of $i$, $j$, and $k$ to the unit vectors along the $x$, $y$, and $z$ axes is just lexicographic convenience; you're just associating them in alphabetic order.

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    $\begingroup$ "Presumably the choice that $ijk=−1$ is simply due to the convenience in writing this formula compactly" Do you mean that $ijk=1$ would have also worked? $\endgroup$
    – PyRulez
    Sep 25, 2016 at 23:12
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    $\begingroup$ @PyRulez: Yes. And that convention gives the opposite algebra whose operation I will denote $\odot$. From $i \odot j \odot k = 1$ (along with $i \odot i = -1$ et cetera) you get, for example, $i \odot j = -k$ and $j \odot i = k$. In fact, there is an identity $uv = v \odot u$. $\endgroup$
    – user14972
    Sep 26, 2016 at 0:01
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    $\begingroup$ The Cross Product is older than quarternions. Wikipedia $\endgroup$ Sep 26, 2016 at 7:12
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    $\begingroup$ @StigHemmer - Read carefully: "[Lagrange] introduced the component form of both the dot and cross products". I.e., he first wrote out the numeric formulas that we now consider to be the expressions the dot product and the cross-product components in terms of $x, y, z$. However, he did not consider them to be operations on vectors, nor that the cross-product produced a vector. The concept of vector did not exist yet. Nor did he come up with the definitions given in the OP. All of these came forth from the study of quaternions. $\endgroup$ Sep 26, 2016 at 15:54
  • $\begingroup$ Ordinary vector calculus, as in electromagnetic theory, mechanics and other branches of physics, is limited to three dimensions per the above discussion of quats. However, a vector space can have any number of dimensions, or even a countably infinite number of dimensions (called a Banach space, if I remember correctly). And the elements of a vector need not be real; they only need to be members of rings (which might be finite fields like integers under multiplication modulo a prime). $\endgroup$ Oct 1, 2016 at 2:54

If the right hand rule seems too arbitrary to you, use a definition of the cross product that doesn't make use of it (explicitly). Here's one way to construct the cross product:

Recall that the (signed) volume of a parallelepiped in $\Bbb R^3$ with sides $a, b, c$ is given by

$$\textrm{Vol} = \det(a,b,c)$$

where $\det(a,b,c) := \begin{vmatrix}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{vmatrix}$.

Now let's fix $b$ and $c$ and allow $a$ to vary. Then what is the volume in terms of $a = (a_1, a_2, a_3)$? Let's see:

$$\begin{align}\textrm{Vol} = \begin{vmatrix}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{vmatrix} &= a_1\begin{vmatrix} b_2 & c_2 \\ b_3 & c_3\end{vmatrix} - a_2\begin{vmatrix} b_1 & c_1 \\ b_3 & c_3\end{vmatrix} + a_3\begin{vmatrix} b_1 & c_1 \\ b_2 & c_2\end{vmatrix} \\ &= a_1(b_2c_3-b_3c_2)+a_2(b_3c_1-b_1c_3)+a_3(b_1c_2-b_2c_1) \\ &= (a_1, a_2, a_3)\cdot (b_2c_3-b_3c_2,b_3c_1-b_1c_3,b_1c_2-b_2c_1)\end{align}$$

So apparently the volume of a parallelopiped will always be the vector $a$ dotted with this interesting vector $(b_2c_3-b_3c_2,b_3c_1-b_1c_3,b_1c_2-b_2c_1)$. We call that vector the cross product and denote it $b\times c$.

From the above construction we can define the cross product in either of two equivalent ways:

Implicit Definition
Let $b,c\in \Bbb R^3$. Then define the vector $d = b\times c$ by $$a\cdot d = \det(a,b,c),\qquad \forall a\in\Bbb R^3$$

Explicit Definition
Let $b=(b_1,b_2,b_3)$, $c=(c_1,c_2,c_3)$. Then define the vector $b\times c$ by $$b\times c = (b_2c_3-b_3c_2,b_3c_1-b_1c_3,b_1c_2-b_2c_1)$$

Now you're probably wondering where that arbitrary right-handedness went. Surely it must be hidden in there somewhere. It is. It's in the ordered basis I'm implicitly using to give the coordinates of each of my vectors. If you choose a right-handed coordinate system, then you'll get a right-handed cross product. If you choose a left-handed coordinate system, then you'll get a left-handed cross product. So this definition essentially shifts the choice of chirality onto the basis for the space. This is actually rather pleasing (at least to me).

The other properties of the cross product are readily verified from this definition. For instance, try checking that $b\times c$ is orthogonal to both $b$ and $c$. If you know the properties of determinants it should be immediately clear. Another property of the cross product, $\|b\times c\| = \|b\|\|c\|\sin(\theta)$, is easily determined by the geometry of our construction. Draw a picture and see if you can verify this one.

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    $\begingroup$ This answer captures the fundamental geometric motivation which is missing in most others. Volumes of parallelopipeds are something which one clearly wants to be able to calculate; determinants arise naturally as the formula for that. Then if you ask “Fix two sides of the parallelopiped; how will the volume change as you vary the third side?” the definition of the cross product immediately falls out, as the coefficients of the resulting formula. $\endgroup$ Sep 26, 2016 at 15:48
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    $\begingroup$ determinants can be negative - in fact you will get a negative "Volume" for a cube with the determinant if you instruct someone to mark the edges with a right handed coordinate system (=standard basis) and then tell them to keep track of the signs when multiplying as they follow a left handed order of edge measurements hence my posts' claim that the Choice of standard basis vector order, orientation says all that can/needs be said in 3-D about "why the right hand rule" $\endgroup$
    – f5r5e5d
    Sep 26, 2016 at 17:51

Well, one can see the rule as is in order to agree with the standard orientation of $\mathbb{R}^3$ $(e_1 \times e_2=e_3)$. Or just convention.

There is a way to see how this can come up "naturally", though.

Given $a,b \in \mathbb{R}^3$, $a \times b$ is the vector which comes out from Riesz representation theorem (a fancy way to say that $V \simeq V^*$ using the isomorphism given by the inner product) applied to the linear functional

$$\det(\cdot, a,b).$$

Computing the vector yields both the fact that it has its norm being what it should be and its direction also being what it should be.

To get to the other direction, we would have to invert $a,b$, considering $a \times b$ to come from the vector corresponding to $\det(\cdot, b,a)$, an "unnatural" thing.

  • 1
    $\begingroup$ Equivalently, $a\times b = \operatorname{Riesz}(\det(a,b,\cdot))$, which is IMO an even more natural way to write it. Actually I'd consider $\det(a,\cdot,b)$ pretty natural as well though, but that would yield the flipped convention. $\endgroup$ Sep 25, 2016 at 20:00
  • $\begingroup$ As I noted in a comment to another answer, this has nothing to do with using your right hand rather than the left. Unless you already chose to draw your system of coordinate axes following the right-hand rule, but then it is circular reasoning. $\endgroup$ Sep 26, 2016 at 12:28

The current Gibbs Vector Algebra System was invented by "Physicists" of the day - not "Pure Mathematicians" and its explicit purpose was describing Physical Phenomina in 3-D space - so it has to have a Convention for Handedness

http://worrydream.com/refs/Crowe-HistoryOfVectorAnalysis.pdf outlines the history, the book even more thoroughly, but I gave a try below

The Gibbs Vector Algebra was the winner of the "Quaternion Wars" in the late 19th century among Physicists as a useful system to describe Physical phenomena, in particular Electromagnetism, Heavisde essentially co-invented the same system.

At the beginning of the 19th century there were no Vector Algebras to choose from, Mathematics was not separate from Physics the way it may look today.
Math was being invented/made up/discovered by Natural Philosophers as they tried to describe the world in formal systems.

But what we now distinguish as Physics was advancing to describe more complicated phenomena like fluid flow in 3-D and writing everything in 3 sets of Cartesian component equations was limiting.
The patterns we now encode as Vector Dot and Cross Products can be seen in the Cartesian x, y, z equation systems of the early formulations of Fluid Mechanics and Maxwell's Equations.

Hamilton taking inspiration from the Gauss/Wessel/Argand Complex Number plane representation of 2-D space came up with Quaternions in 1843 and coined the term "Vector" for his "pure quaternion" 3 "imaginary" components.
Hamilton's influence from his many discovery's in Physics, Optics, Matrix Theory lead many to try to understand and use his Quaternion Algebra. Maxwell's Equations in particular became more compact compared to the original ~20 Cartesian component version.

But many found reasoning in the Hamilton Quaternion Algebra very unintuitive with his i, j, k basis unit elements squaring to -1.

This lead to Gibbs in the early 1880's "stealing" Hamilton's "Vector" name, pulling out the Dot and Cross Product from the Quaternion Multiplication rules and giving his system our current interpretation with x, y, z basis unit vectors each squaring to +1. "Closing" Gibbs new Vector Algebra required the Vector Cross Product we now know and use to give another Vector - but it really isn't mathematically correct as the Polar/Axial or Vector/PseudoVector distinction is needed in Gibbs Vector Algebra.

But both Hamilton and Gibbs had to choose an Orientation, basis metric and Order, impose a Convention on their systems to give agreement with the inherent Chirality of Euclidean 3-D space.


the Gibbs Vector Algebra Cross Product Right Hand Rule is a Convention

Any description of the (nominally) Euclidean 3 dimensional world we perceive runs into Chirality, the distinction between Right and Left handedness, the difference in a Mirror Reflection. https://en.wikipedia.org/wiki/Chirality

A Mathematical System describing this world has to have a Convention for coordinate order, orientation.
Justifications for the Right Hand Rule for the Vector Cross Product are a consequence of the choice of Chirality in the the "Standard Basis" and are circular, not deep - you just have to choose.

In fact the other choice has been used: http://web.stanford.edu/class/me331b/documents/VectorBasisIndependent.pdf

The right-hand rule is a recently accepted universal convention, much like driving on the right-hand side of the road in North America. Until 1965, the Soviet Union used the left-hand rule, logically reasoning that the left-hand rule is more convenient because a right-handed person can simultaneously write while performing cross products.

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    $\begingroup$ The Soviets were not wrong... $\endgroup$ Sep 26, 2016 at 12:25
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    $\begingroup$ If to write with the right hand were wrong, then what's left is to write with the left, right? $\endgroup$
    – LarsH
    Sep 27, 2016 at 17:44
  • $\begingroup$ Is there anything like chirality in four dimensional space? Higher dimensions? Would a vector cross product in four dimensions be a product of THREE vectors? $\endgroup$ Oct 1, 2016 at 3:03

This may be a bit too deep but let $V$ be a finite dimensional vector space with basis $v_1,...,v_n$. We say $(v_1,...,v_n)$ is an oriented basis for $V$. We can define an equivalence class on orientations of $V$ by $[v_1,...,v_n] \sim [b_1,...,b_n] \iff [v_1,...,v_n] = A[b_1,...,b_n]$ (where $A$ is the transition matrix) and $\textbf{det}(A)>0$. Therefore, orientations are broken up into two classes, positive and negative. If we let $\textbf{e}^1,...,\textbf{e}^n$ be the standard basis for $\mathbb{R}^n$ then if we wish to check that $[b_1,...,b_n] \sim [\textbf{e}^1,...,\textbf{e}^n]$ then we simply have to look at;

$$\textbf{det}\left(\begin{bmatrix} b_1 & b_2 & \cdots & b_n \end{bmatrix}\right) >0$$

Since $A = [b_i]$ is the change of basis matrix.


Edit (1): Above gives a generalization of determining orientation. In your case you would have $E=[\textbf{e}^1,\textbf{e}^2, \textbf{e}^3]$ which gives positive orientation and the ''right hand rule'' should be thought of as a geometric property that one can check without knowing all the information I presented above.

Edit (2): In regards to your question about why $\sin \theta, \cos \theta$ are used in certain situations is also because of geometry. The area of a parallelogram is base $\times$ height $= |\vec{a} \times \vec{b}| \sin \theta$ where $\vec{a}, \vec{b}$ span the parallelogram.

  • $\begingroup$ A yes for ixj the det assumes it is going to be k,not -k, so that's my question, as det is based on that definition,and therefore doesnt answer the question $\endgroup$
    – koe
    Sep 25, 2016 at 17:30
  • $\begingroup$ You should really read into the edit I wrote as well. Your question starts to far into the story to understand what is going on. The standard from $E$ obeys the right hand rule which is a geometric property and we have $\{\textbf{e}^1, \textbf{e}^2, \textbf{e}^1 \times \textbf{e}^2 = \textbf{e}^3\}$. The right hand rule is a property observed from this case. To know if $\{\vec{a}, \vec{b}, \vec{a} \times \vec{b}\}$ also satisfies this rule, you with check geometrically or do exactly what I explained above. So again, why doesn't this answer your question? $\endgroup$ Sep 25, 2016 at 17:36
  • $\begingroup$ A i think ixj definition is more fundamental than using determinants $\endgroup$
    – koe
    Sep 25, 2016 at 17:36
  • $\begingroup$ yeah i dont get what geometric property u are talking, as the math too you use is too advanced for a simple question, which i dont understand $\endgroup$
    – koe
    Sep 25, 2016 at 17:38
  • $\begingroup$ how can i "check" this geometric property? $\endgroup$
    – koe
    Sep 25, 2016 at 17:41

That the cross product has been defined as a vector at all is perhaps an accident of the fact that we perceive three spatial dimensions. It does not make sense in any other number of dimensions (e.g. 2 or 4, though some unnatural version can be constructed in 7). The exterior product (also called the wedge product) makes sense in any number of dimensions. It may be envisaged as an oriented surface with a direction of rotation, the area (or magnitude) of which is equal to the parallelogram defined by the vectors. These can be added and scaled, just like vectors. The existence of the cross product owes its existence to the isomorphism between the space of these products and the 3-dimensional vector space. It is simply the choice of considering these products (examples include torque, oriented area, angular momentum) to be normal vectors, though they are not (witness the fact that they are called pseudo-vectors). Their components do not scale with a change of basis (or units) as do vectors, but rather as their square. The need to choose of left- or right-handedness results entirely from the choice to map the product onto "normal" vectors, made possible by their isomorphism.

When the vectors are two-dimensional, the product inhabits a one dimensional space (torque in a two dimensions makes sense, so why insist on a third dimension to accommodate the normal vector?). In four dimensions, the equivalent of torque (the oriented product of a force and the length of a lever arm) inhabits a space of six dimensions, not four.

So, the short version is: it is a convention needed to deal with the unfortunate representation of the product of vectors as normal vectors. It would be completely (and more intuitively) avoided entirely if these products were thought of as oriented areas rather than as being classical vectors.

  • $\begingroup$ would you elaborate on what "unnatural version can be constructed in 7"? $\endgroup$
    – hyportnex
    Sep 26, 2016 at 21:54
  • $\begingroup$ Wikipedia's Seven-dimensional cross product gives such a beast that shares several properties with the three-dimensional cross product. I label it as unnatural because there are many (I suspect an infinity, not just 480) equivalent choices. $\endgroup$
    – qman
    Sep 27, 2016 at 2:40
  • $\begingroup$ The reason there are 480 7D cross products is due to the combinatorics of the Fano plane used to define the multiplication tables, there cannot be any others without breaking some other rule of cross products. As it stands, all 480 of them already break the Jacobi identity - which is a better reason to call them "unnatural" then the idea that there are more than 1 of them. For that matter, there are, in fact, two 3D cross products - the right handed one and the left handed one - hence the OP's question. $\endgroup$
    – No Name
    Oct 19, 2020 at 16:05

There's a whole algebra ("Geometric Algebra", or Clifford algebra) behind the concept of multiplying vectors. Look there to find out where the handedness comes from, why $ \cos \theta $ for $a \cdot b$ but $\sin \theta$ for $a \times b$, and many other gems. Here's a great reference.

Essentially, the cross product is a vector that describes a plane. A plane is described by two vectors (in any number of dimensions), but in $\mathbb{R}^3$ we have the handy situation that a plane can also be described by a vector. So, in 3-d only, vector $c = a \times b$ describes the plane described by $a$ and $b$. Since a plane has two sides, there's a two-fold arbitrariness ("orientation") that the choice of sign handles.

The $\sin$ and $\cos$ bit is slightly more complicated, but not much. Check out the reference.


As far as $\sin \theta$ and $\cos \theta$ are concerned,

Using the law of cosines,

\begin{align} \|\overrightarrow{v_2}\|^2 + \|\overrightarrow{v_1}\|^2 -2\|\overrightarrow{v_2}\| \, \|\overrightarrow{v_1}\| \cos \theta &= \|\overrightarrow{v_2} - \overrightarrow{v_1}\|^2 \\ 2x_1 x_2 + 2y_1 y_2 + 2z_1 z_2 &= 2\|\overrightarrow{v_2}\|\, \|\overrightarrow{v_1}\| \cos \theta \\ \cos \theta &= \dfrac{x_1 x_2 + y_1 y_2 + z_1 z_2} {\|\overrightarrow{v_2}\| \, \|\overrightarrow{v_1}\|} \\ \cos \theta &= \dfrac{\overrightarrow{v_2} \circ \overrightarrow{v_1}} {\|\overrightarrow{v_2}\| \, \|\overrightarrow{v_1}\|} \end{align}

And so, $\overrightarrow{v_1} \circ \overrightarrow{v_2} = \|\overrightarrow{v_2}\| \, \|\overrightarrow{v_1}\| \cos \theta$

Note that $\sin \theta$ is non negative for all $0 \le \theta \le \pi$.


\begin{align} \sin^2 \theta &= 1 - \cos^2 \theta \\ &= 1 - \dfrac{(x_1 x_2 + y_1 y_2 + z_1 z_2)^2} {\|\overrightarrow{v_2}\|^2 \, \|\overrightarrow{v_1}\|^2} \\ &= \dfrac{\|\overrightarrow{v_2}\|^2 \, \|\overrightarrow{v_1}\|^2 - (x_1 x_2 + y_1 y_2 + z_1 z_2)^2} {\|\overrightarrow{v_2}\|^2 \, \|\overrightarrow{v_1}\|^2} \\ &= \dfrac{x_2^2 y_1^2 - 2x_1 x_2 y_2 y_1 + x_1^2 y_2^2 + x_2^2 z_1^2 +x_1^2 z_2^2 - 2 x_1 x_2 z_1 z_2 +y_1^2 z_2^2-2 y_2 y_1 z_1 z_2+y_2^2 z_1^2} {\|\overrightarrow{v_2}\|^2 \, \|\overrightarrow{v_1}\|^2} \\ &= \dfrac{(-y_2 z_1+y_1 z_2)^2+(x_2 z_1-x_1 z_2)^2+(-x_2 y_1+x_1 y_2)^2} {\|\overrightarrow{v_2}\|^2 \, \|\overrightarrow{v_1}\|^2} \\ &= \dfrac {\|\overrightarrow{v_1} \times \overrightarrow{v_2} \|^2} {\|\overrightarrow{v_2}\|^2 \, \|\overrightarrow{v_1}\|^2} \end{align}

And we conclude, for purely formal reasons, $\|\overrightarrow{v_1} \times \overrightarrow{v_2} \| = \|\overrightarrow{v_2}\| \, \|\overrightarrow{v_1}\| \sin \theta$


The cross product doesn't actually follow any such rule. It is defined arithmetically as a determinant:

$$\mathbf{u\times v} = \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ u_1&u_2&u_3\\ v_1&v_2&v_3\\ \end{vmatrix}$$

There is no hand rule there. The right hand rule comes from the conventions about how we visualize the result in 3-D vector space.

Firstly, the resulting scalar value is assigned as the length of a vector which is orthogonal to $\mathbf{u}$ and $\mathbf{v}$. That does not give us a right hand rule yet.

If we identify the first vector component with the $x$ axis, the second with the $y$ axis and the third with the $z$ axis, such that the positive ray of the $x$ axis shoots left, $y$ shoots upward, and $z$ points out of the page toward us, then we end up with the right hand rule.

Why? Because the cross product of a unit vector lying in the positive $x$ axis direction, with a unit vector similarly lying in positive $y$ produces a unit vector lying in positive $z$.

Based on how we laid out the axes, this gives us the right hand rule: if we extend our thumb in the direction of $z$, then the fingers curl from $x$ to $y$.

Your question is really about why we lay out the axes this way, and that is merely a convention, like driving on the right side of the road (in the majority of the world's countries, not all, obviously).

The important thing about conventions is only a) to have them and b) have them be consistent. The choice between two equivalent conventions, like whether to choose one system or its exact mirror image, is unimportant; there is no external reason to go one way or the other.

  • $\begingroup$ While this answer has its own issues (see Marc van Leeuwen's answer in a related thread), it is the only answer that correctly points out that, contrary to what the OP claims, there simply isn't any notion of handedness in the (modern) definition of cross product. Almost all other answers here, if not wrong, are unclear or misleading in this regard. $\endgroup$
    – user1551
    Mar 6, 2017 at 15:07

As a complement to Bye_World's illuminating answer, which defines $b \times\ c$ as the vector whose dot product with $a$ equals $\det(a, b, c)$, $\|b\|\|c\|\sin(\theta)$ is the surface of the parallelogram formed by vectors $b$ and $c$. If $a$ is a unit vector orthogonal to the $b$-and-$c$ plane, then $\det(a, b, c)$, which is the (signed) volume of the parallelepiped formed by the 3 vectors, equals the surface of the $b$-and-$c$ parallelogram. In this case, i.e.$\|a\|=1$ and angle from $a$ to $b$ (and to $c$) equal to positive $\pi/2$, one can picture $b \times c$ as $a$ times the surface of the parallelogram.


I have answers for your question. If it was defined relative to the left-hand rule, you'd currently be asking why not the right-hand rule. It's how we deal with ambiguity: by establishing conventions. Depending on how you choose the definition of cross product, the result could be one of two different vectors. This led to the establishment of a convention. If there were no conventions with cross products, things would get very confusing because some people would be coming up with vectors in the opposite direction from other people, and their results would be different, which would lead to confusion. It's also likely that some theorems wouldn't work out because they're based on one definition and not the other, etc; one would keep having to convert between definitions all the time; and it would get very annoying.

The answer to your second questions? Upon establishing the definition of cross product, it just happened that the geometry worked out with sine and not cosine. It's just the behaviors of geometry in light of the constraints placed on it as a result having decided on a definition.

The same answer goes for the scalar product. Once the definition was made, that's how the geometry worked out.

However, it could be, though that the definitions were made so that they'd coincide with sin and cosine the way they did to unite a variety of approaches to the same set of problems, some people using trig, others using algebra to attack the same set of problems until some brilliant person united all the approaches with one definition. Such a thing often happens in mathematics, as anyone who's familiar with the concept of research might tell you.

Hope that helps.


The real reason is because it is convention to use a right-handed coordinate system. If you switched to a left-handed coordinate system for some reason (I'm not sure why), then you would need to do your cross-products using a left-hand rule to get the same answers. That's really as simple as it gets. It's a result of our conventions in defining our coordinates the way we do.


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