# Cross product between two vector quantities [closed]

I wish to understand the "why, what, and how" of the vector cross product. Particularly speaking, I'd like to know its interpretation and how it's connected to the area.

Also, what would be the quantity we have when we cross-product two force vectors?

• Looking back at this what exactly do u mean by cross product of two force vectors? Apr 17, 2020 at 21:16
• imagine a drone is loosely tied to a pole and it's trying to move forward but instead, it ends up revolving around the pole this is a normal torque concept if we consider the distance vector and the force vector. but what if we consider the tension vector and force vector.they are also perpendicular. what happens when we cross-product them? what is the resultant vector be? I know the question seems dumb cause we know that we add up those forces to know the tangential force. but then again why did we cross product when it was distance and force and why are we adding when its force and force Apr 18, 2020 at 2:42
• Ah sorry didn't see this yesterday. Well if a force is acting tangentially then the drone will start move around the circle faster and faster. If we cross product the tangential force and tension ... we get a force vector which would be perpendicular to both force vectors. Not sure how exactly that would be useful but that sure is a 'thing'. We did the cross product with distance vector and force vector because the vector operation of cross product completely captures the the propotionalities and physical observations we make Apr 19, 2020 at 13:45
• thank you i was thinking the exact samething. however why did we get another force vector after crossing two force vectors any reason behind it? Apr 20, 2020 at 16:39

Let's leave aside the mathematics and focus simply on physics. Say you have a door and you apply force on some point of the door, would you agree that the amount of 'rotational oomf' given to the door is only dependent on the perpendicular distance from the axis of rotation to the point where you applied the force? Think about it, would the rotation change if you pushed the door at the upper edge or lower edge?

Right, If you really thought of that, we are motivated to talk about the cross-product ( it would help us model the situation)

$$\vec{r} \times \vec{F} = ∥\vec{r}∥ ∥\vec{F}∥ \sin(\theta) \hat{n}$$

Take a piece of paper and draw two vectors say $$\vec{a}$$ and $$\vec{b}$$, and tilt them at an angle. Now to find the component of 'a' perpendicular of to 'b' you just multiply the magnitude of 'a' by $$b\sin(\phi)$$ where $$\phi$$ is the angle between the vectors.

However, you may note that this is exactly equivalent to finding the area of a parallelogram with side lengths as the magnitude of 'a' and 'b' ( coincidence??)

The mathematical side of things: Now we have defined torque using a magnitude but what if we wanted to define it as a vector that is encode the axis of spinning as well?

We will define the cross product as an operation that takes in two vectors and spits out a perpendicular vector, hence these follow

$$\vec{i}\times \vec{j}=\vec{k}$$ $$\vec{j}\times \vec{k}=\vec{i}$$ $$\vec{k}\times \vec{i}=\vec{j}$$

Now, to include the orientation of the cross product ( sort of like how you want to encode if the spinning is clockwise or anti-clockwise), we bring in the signs You just define that

$$\vec{j} \times \vec{i} = - \vec{k}$$ $$\vec{k}\times \vec{j}=-\vec{i}$$ $$\vec{i}\times \vec{k}=-\vec{i}$$

Essentially bring in anti commutativity

Now, let us define

$$\vec{a}= 3 \vec{i} + 2 \vec{j}$$

and, $$\vec{b}=2\vec{i}$$

Now let's try to cross them ... uh oh there are multiple components :( well just define it to be distributive as well then : $$\vec{a} \times \vec{b} = (3\vec{i} + 2\vec{j} ) \times ( 2\vec{i})= 3 * 2 (\vec{i} \times \vec{i}) + 2*2 (\vec{j} \times \vec{i})$$

And here's the next challenge what is the cross product of a vector with it self... hmmm a vector which is perpendicular to 'i' and 'i' ... Let's go back to the magnitude definition $$|\vec{i} \times \vec{i} |= ∥\vec{i}∥ ∥\vec{i}∥ \sin(0) = 0$$

Hence the $$\vec{i} \times \vec{i}$$ term goes to 0 !!

Therefore, $$\vec{a}\times \vec{b} = 0 + 2*2 (\vec{j} \times \vec{i})$$

Now, remember $$\vec{j} \times \vec{i} = - \vec{i} \times \vec{j} = -\vec{k}$$

Finally, we write $$\vec{a}\times \vec{b} = -4\vec{k}$$

So, how does this relate to that original magnitude relation? well it turns out that the magnitude of this perpendicular vector $$\vec{a} \times \vec{b}$$ is the area of a parallelogram with side length $$'∥a∥'$$ and $$'∥b∥'$$

Connection between area and rotation

Refer

• As an exercise try to define the equation of a plane given a point on it ;) Heres the sketch 1. construct a vector on the plane with that and some arbitary point (x,y,z) 2. dot that vector with the normal vector ( remember what the dot product of two perpendicular vectors are) Apr 17, 2020 at 17:37

The existence of a cross-product is a feature that only some vector spaces have. Examples are $$\mathbb R^3$$ and $$\mathbb R^7$$. Under the identification of the dual basis of a Euclidean vector space with the standard basis it is hard to get a grasp of the actual geometric meaning of the cross-product. In $$\mathbb R^3$$, a skew-symmetric form, that is, an anti-symmetric matrix, can be identified with a vector in $$\mathbb R^3$$ because it only has three independent variables. Indeed, the cross-product map $$\omega\times:\mathbb R^3\to\mathbb R^3$$ that sends $$v$$ to $$\omega\times v$$ can be represented with a skew-symmetric matrix $$\Omega\in\operatorname{Mat}_{3\times 3}$$ so that

$$\omega\times v = \Omega v.$$

This displays a link between the cross product and rotations in $$\mathbb R^3$$. That's because skew-symmetric matrices generate rotations under the exponential map.

There is also a distinction between polar and axial vectors. Any ordinary vector is sometimes called a polar vector; on the other hand, any vector that arises as a cross-product of two polar vectors is an axial vector (or pseudovector). Under a reflection of all the axes of $$\mathbb R^3$$, every polar vector flips its sign, whereas axial vectors don't. As to why they are called axial, I couldn't find any references, but my guess is that, from a geometrical point of view, they lie on the axis perpendicular to the plane generated by the two vectors used to compute the cross-product. The magnitude of this vector turns out to be equal to the area of the parallelogram.

Finally, when one considers the exterior algebra over the Euclidean space $$\mathbb R^3$$, the cross-product is the Hodge-dual of the exterior product between two 1-forms. That's because the exterior product of two 1-forms is a 2-form, and the Hodge-dual of a 1-form in $$\mathbb R^3$$ under the Euclidean norm is again a 1-form. Hence

$$\xi\times\eta := \star(\xi\wedge\eta)$$

In $$\mathbb R^3$$, a 1-form is sometimes also called a line form because of its linearity. Therefore, a 2-form is called a surface form (and a 3-form a volume form). A typical example of a 2-form comes from the $$\hat{\mathbb n}\ \text dS$$ that appears in surface integrals. Indeed, $$\hat{\mathbb n}$$ is the vector perpendicular to the integration surface and $$\text dS$$ is its infinitesimal magnitude. Under Hodge-duality, $$\star\hat{\mathbb n}\ \text dS$$ is a surface form.

• thank you for the answer, although I didn't get the entirety of it I got the big picture. Any suggestions where can i find out more about cross product in-depth? Apr 17, 2020 at 13:42
• A good read for me is Differential Forms with Applications to the Physical Science by Flanders. Apr 17, 2020 at 13:54
• I've deleted a number of obsolete comments and/or responses to them. Apr 17, 2020 at 21:37

Here's an attempt of a minimal-math explanation of the cross-product.

1) Why do we use the cross product?

2) what does the resultant vector from the cross product represent?

5) Is such a thing really in existence and if so where do we use it?

A cross-product is nothing but a mathematical invention. Just like the dot-product. They happen to be invented in such a way that:

• a cross-product $$\times$$ (its magnitude) multiplies the perpendicular components of two vectors, whereas
• a dot-product $$\cdot$$ multiplies the parallel components of two vectors.

Now, these two features turn out to appear very often in the real world (in physics). For example:

• Pull a wrench sideways, and you turn the bolt, but pull along with it and nothing turns. The formula for the torque magnitude is $$\tau=F_\perp r$$ (or equivalently) $$\tau=F r_\perp$$).
• Push along with the train cart, and you move it and do work on it, but push sideways and it won't move anywhere. The work formula is $$W=F_\parallel \Delta r$$ (or equivalently $$W=F \Delta r_\parallel$$). Source: A lesson at All That Matters Academy

Instead of having to write them as parallel or perpendicular components, and instead of jumping between two equivalent versions of the formulas, why don't we simplify the formulas by using those mathematical tools that happen to match what we need? So, let's simply write:

$$\vec \tau=\vec r \times \vec F\qquad \text{ and }\qquad W=\vec F \cdot \vec r$$

Also, the cross-product is mathematically invented to not just give a scalar value like the dot-product, but to actually give a vector (whose magnitude is the perpendicular components multiplied). Since we in physics like to think of rotations as vectors, why don't we then utilize this vector-feature of the cross-product as well? And so, it became. Just use the right-hand rule on the torque vector that comes out of a cross-product to figure out the rotation direction.

3) Does the resultant vector mathematically perpendicular to the given input vectors or did we just used the third dimension to represent the area of the parallelogram between the vectors for our convenience?

It turns out that multiplying the perpendicular components of two vectors just happens to give the area they span! We can call this a coincidence, if we want to. Area is after all width-times-length, so whenever you multiply two numbers that belong to non-parallel vectors, you can indeed think of the result as an area.

It is rarely relevant in physics, and I like to think of it as just a mathematical fluke; a coincidence. (It might now and then be a smart trick to remember, which can ease future calculations - the cross-product "recipe" is after all not the simplest or easiest.)

4) what is the cross product of two force vectors called?

Nothing I know of. As explained above, we only grab mathematical tools if they happen to make sense for something we are doing in physics. In real life, the perpendicular components of force and position are obviously involved when turning the bolt with the wrench. And because of that we pick the mathematical tool that fits.

But when are two perpendicular force components involved? If you find a scenario where this makes sense, then feel free to use the cross-product and interpret the result. I don't know of a situation like that of-hand.