# Why do we need both dot product and cross product?

I was looking for an intuitive definition for dot product and cross product. I have found two similar quesitions in SO, but I am not satisfied with the answers. Finally I found a possible answer here. It says dot product actually gives us a way to depict mathematically how parallel two lines are and on the other side cross products tells us how two lines are perpendicular to each other. So my question is why do we want both. Why cant we just have dot product?

• For a start one gives a vector and the other a scalar. – Quantum spaghettification May 26 '15 at 5:42
• Possible duplicates: physics.stackexchange.com/q/14082/2451 , physics.stackexchange.com/q/115107/2451 and links therein. – Qmechanic May 26 '15 at 5:55
• Because you are living in three dimensions. The other possibility would be seven dimensions, but I think we can kind of rule that out: arxiv.org/pdf/1212.3515. Please note, though, that similar multilinear forms with different physical relevance exist in Euclidean spaces of all dimensions and they do lead to very different "classical physics", if we try to extend physics under the usual assumptions to these higher dimensional spaces. – CuriousOne May 26 '15 at 5:57
• Qmechanic's first link, the top answer probably will explain it best. – Xeren Narcy May 26 '15 at 6:07
• Can you give links to the two similar questions on SO? – ankit Mar 19 '17 at 6:12

There is a sense where you can start with just a scalar product.

If you assume that $v^2=vv=v\cdot v=|v|^2$ holds for any vector then that is a scalar product for a vector with itself.

From that you can get the scalar product of any two vectors $v$ and $w$, $v \cdot w= \frac{1}{2}\left((v+w)^2-v^2-w^2\right)=\frac{1}{2}\left(vw+wv\right)$ where for the last equality we assume the same distributive laws as for matrices.

So the simplest assumption of having a scalar product of any vector with itself can easily give us a scalar product of any two vectors (we can use the scalar products of $v+w, v$ and $w$ with themselves).

But we can also have a general product and allow it to have the regular rules of matrix algebras (associative, distributive, multiplication, addition, scaling by scalars, etcetera). Then we note that for vectors $v$ and $w$ we have $vw=\frac{1}{2}\left(vw+wv\right)+\frac{1}{2}\left(vw-wv\right)$ or $vw=v\cdot w+\frac{1}{2}\left(vw-wv\right)$ and we can notice that the last term $\frac{1}{2}\left(vw-wv\right)$ is different than a scalar or a vector, specifically its square is negative. For vectors $v$ and $w$ we can denote $\frac{1}{2}\left(vw-wv\right)$ by $v \wedge w$.

These new objects, like $v\wedge w$ naturally represent the plane spanned by the vectors $v$ and $w$. They are useful in physics, though for historical reasons we often use the complete accident that in 3d there are vectors orthogonal to planes to represent $v \wedge w$ by the vector orthogonal to it and then call it the cross product.

When you have both products $v \cdot w$ and $v\wedge w$ you have the full information to get the product $vw=v \cdot w+v\wedge w$, and from that you can solve for $v$ or $w$ if you have the other one. For instance $v=vw\frac{w}{w^2}=\left(v \cdot w+v\wedge w\right)\frac{w}{w^2}$. So therefore if you know the scalar product and the wedge product and you know $w$ then you can find out what $v$ was. Neither of the products by themselves allow you that. Many vectors can give the same inner product, and many vectors can give the same wedge product, but knowing both can allow you know the full relative relationship between the two.

It all follows from just the scalar product, but you get the whole package if you want it.

tl;dr

The dot product (symmetric part of the one product) tells you how much they have in common. The other product (antisymmetric part of the one product) tells you how much they orthogonal, specially how much you have to rotate one line to get align it with the other, and if you don't live in just a plane it also tells you the plane in which you need to rotate to send one into the other. That directionality is something you don't get from just a scalar.

But you don't need two products. One product is enough as long as you do it the invertible way. And that way naturally creates numbers, lines, planes and even higher dimensional objects as they come up. Further details above.

• Thanks, but a one liner explanation is what I am looking for. Is that really this complex? You can also answer to my comments for steevan's answer. – ibsenv May 28 '15 at 5:08
• @ibsenv You asked why can't we just use the dot product. But i told you that you can in a sense use just the dot product. My answer is longer because there is a sense where you don't need two products. One product, that gives the dot product when you product a vector with itself, is actually enough. And unlike the cross product accidentalness it works in any dimension, so it works in the actual 4d spacetime in which we live. I'll add a tl;dr version but it does take longer to actually tell you how to use one product yo do the work of two. Others might claim the opposite in less space. – Timaeus May 28 '15 at 5:35
• @ibsenv I updated. But I think you still actively refuse to accept that the symmetric product (dot) and the antisymmetric product (cross) are complimentary. When one is small the other is big. One tells you how things are similar, the other tells you how they are different, and the how different needs a direction to tell you how to turn one to the other. – Timaeus May 28 '15 at 5:47
• Thanks a lot for your patience in explaing this. I am sure this will be helpful for a lot of young students like me. I have one last question, just to make sure I understand you correctly. Is it 'POSSIBLE' for me( even though it may not be approprite ) to stick with only the asymmetric product since it gives me all the information I want to rotate a vector in any plane to another vector in any plane? [ Can I consider using dot products as a 'shortcut' for the same process If I live in a single plane as the direction change is obvious.] – ibsenv May 28 '15 at 8:53
• @ibsenv Firstly, the word asymmetric and antisymmetric are different words, the full product is asymmetric, but giving the same input on both sides gives you the symmetric part and that is enough (plus associativity and distributivity) to give you the rest. If you tried to start with an antisymmetric product that won't work in 1d for sure, you'll never get the symmetric part back. There are some people that start out making the product totally antisymmetric and they have to add a duality operation to get the rest – Timaeus May 29 '15 at 16:40

The dot product or scalar product is an algebraic operation that takes two equal-length sequences of numbers and returns a single number. This operation can be defined either algebraically or geometrically.

The cross product or vector product is a binary operation on two vectors in three-dimensional space and is denoted by the symbol ×. The cross product a × b of two linearly independent vectors a and b is a vector that is perpendicular to both and therefore normal to the plane containing them. It has many applications in mathematics, physics, engineering, and computer programming.

As both of them has got different aspects , we have to use both of them.

In short:

• The dot product gives you the multiplication of the parallel components.
• Example: The work expression $W=\vec F \cdot \vec r$, where only the force component parallel to the direction (or likewise, the position component parallel to the force) is wanted.
• The (magnetude of the) cross product gives you the multiplication of the perpendicular components.
• Example: The torque expression $\vec \tau=\vec F \times \vec r$, where only the force component perpendicular to the "arm" (or likewise vise versa) is wanted.

Why cant we just have dot product?

Simply because we sometimes want the perpendicular variation and not only the parallel one. At some point the cross product was "invented" to describe this in a neat and concise expression.

Just keep in mind that while the dot product (or scalar product) gives you a number only (a scalar), the cross product (or vector product) gives you a number with a direction (that is, a vector). And this direction is perpendicular to both.

• Hey, I understood the first part. I have this question. Why Vector product have direction and not for Dot product? – ibsenv May 26 '15 at 9:43
• @ibsenv The scalar product represents length and angles, so it must have scalar value to extract the information. It is not obvious from elementary definitions, but if you "arrive at" the cross product from some higher mathematics approach (exterior algebra, Lie-groups), it either represents oriented area elements or infinitesimal rotations. In 3-space both have exactly 3 degrees of freedom, thus the set of oriented area elements or infinitesimal rotations can be identified with $\mathbb{R}^3$ itself, hence the vectorial value. – Bence Racskó May 26 '15 at 9:48
• @ibsenv You can either see Uldreth's comment, or you can consider it as "just a convention". In the first example above (the dot product) there is no need for the orientation, as the result is along both vectors. In the second example (with torque) you are "turning" around an axis. It makes sense that this axis is perpendicular to both original vectors, and by defining the cross product in this way, this axis direction can simply be calculated directly as a vector. – Steeven May 26 '15 at 13:50
• @Uldreth, Steevan, so If I understand correctly,In simple words Vector product is just another version of dot product, where your output is also a vector quantity. Both are used for comparing two 2 vectors. There is no other difference. Am I right? – ibsenv May 28 '15 at 4:52
• @Steeven, Uldreth, Can I even say dot products are simply a subset of vector products with the direction practically meaningless or same relative to the direction of interest. (So we ignore the direction and call it scalar)? – ibsenv May 28 '15 at 4:59

The dot and cross products are both recovered as components of the tensor product, which takes two vectors and gives you a rank-2 tensor $v^\mu w^\nu$. The dot product is the trace of this tensor, obtained by a contraction $v^\mu w_\mu$ -- this is useful because it is a scalar, which is invariant under nice transformations (rotations and skews, specifically). The cross product is the anticommutativity of this, evaluated as $v^\mu w^\nu-v^\nu w^\mu$ (for two vectors -- for more vectors you need to use a Le Cevita symbol to sum with signs over the permutations of $\mu,\nu...$).