Cross product query.

Question

In cross product, we do it like: $\mathbf{a} \times \mathbf{b} = |a||b|\sin(\theta)$

From where does this $\sin(\theta)$ came from? Can someone please derive the cross product and explain it.

Answer 1

In addition to these intuitive answers, I also wanted to show an alternate, non-geometric approach to derive the cross product given two three-dimensional vectors $\vec{u}, \vec{v} \in \Bbb R^3$.

First, we'll show that $\lVert \vec{u} \times \vec{v}\rVert = \sqrt{\lVert \vec{u} \rVert^2 \lVert \vec{v} \rVert^2 - \left( \vec{u} \cdot \vec{v} \right)^2}$, or rather $\sqrt{\lVert \vec{u} \rVert^2 \lVert \vec{v} \rVert^2 - \left( \vec{u} \cdot \vec{v} \right)^2} = \lVert \vec{u} \times \vec{v}\rVert$ since I will work on the left hand side first. Much easier to prove this way.

Now $$\begin{align} \lVert \vec{u} \rVert^2 \lVert \vec{v} \rVert^2 & = \left( u_1^2+u_2^2+u_3^2 \right) \left( v_1^2+v_2^2+v_3^2 \right) \\ & = u_1^2v_1^2+u_1^2v_2^2+u_1^2v_3^2 + u_2^2v_1^2+u_2^2v_2^2+u_2^2v_3^2 + u_3^2v_1^2+u_3^2v_2^2+u_3^2v_3^2 \end{align}$$

And $$\left( \vec{u} \cdot \vec{v} \right)^2 = \left( u_1v_1+u_2v_2+u_3v_3 \right)^2 = u_1^2v_1^2 + u_2^2v_2^2 + u_3^2v_3^2 + 2u_1v_1u_2v_2 + 2u_1v_1u_3v_3 + 2u_2v_2u_3v_3$$

So $$\begin{align} \lVert \vec{u} \rVert^2 \lVert \vec{v} \rVert^2 - \left( \vec{u} \cdot \vec{v} \right)^2 & = u_1^2v_1^2+u_1^2v_2^2+u_1^2v_3^2 + u_2^2v_1^2+u_2^2v_2^2+u_2^2v_3^2 + u_3^2v_1^2+u_3^2v_2^2+u_3^2v_3^2 - u_1^2v_1^2 - u_2^2v_2^2 - u_3^2v_3^2 - 2u_1v_1u_2v_2 - 2u_1v_1u_3v_3 - 2u_2v_2u_3v_3 \\ &= u_1^2v_2^2+u_1^2v_3^2 + u_2^2v_1^2+u_2^2v_3^2 + u_3^2v_1^2+u_3^2v_2^2 - 2u_1v_1u_2v_2 - 2u_1v_1u_3v_3 - 2u_2v_2u_3v_3 \\ \end{align}$$

That massive equation couldn't possibly be helpful, right? Well, let's look at the right hand side of our equation, $\lVert \vec{u} \times \vec{v}\rVert$.

Your question is about deriving the formula the magnitude of the cross product, but the cross product itself returns a vector perpendicular to the two vectors used for the product. Specifically, the vector returned by $\vec{u} \times \vec{v}$ is of the form $ \vec{a} = \left( u_2v_3 - u_3v_2 \right)\vec{i} - \left( u_1v_3 - u_3v_1 \right)\vec{j} + \left( u_1v_2 - u_2v_1 \right)\vec{k} $.

Of course, we're looking for the magnitude of $\vec a$. This would, by definition, be given by: $$\begin{align} \lVert\vec{a}\rVert & = \sqrt{\left( u_2v_3 - u_3v_2 \right)^2 + \left( u_1v_3 - u_3v_1 \right)^2 + \left( u_1v_2 - u_2v_1 \right)} \\ & = \sqrt{u_1^2v_2^2+u_1^2v_3^2 + u_2^2v_1^2+u_2^2v_3^2 + u_3^2v_1^2+u_3^2v_2^2 - 2u_1v_1u_2v_2 - 2u_1v_1u_3v_3 - 2u_2v_2u_3v_3} \\ \end{align}$$

Look familiar? Here, let me square that magnitude for you:

$$ \lVert\vec{a}\rVert^2 = u_1^2v_2^2+u_1^2v_3^2 + u_2^2v_1^2+u_2^2v_3^2 + u_3^2v_1^2+u_3^2v_2^2 - 2u_1v_1u_2v_2 - 2u_1v_1u_3v_3 - 2u_2v_2u_3v_3 $$

Hey, isn't that just $\lVert \vec{u} \rVert^2 \lVert \vec{v} \rVert^2 - \left( \vec{u} \cdot \vec{v} \right)^2$? So if we take the square root of $\lVert \vec{u} \rVert^2 \lVert \vec{v} \rVert^2 - \left( \vec{u} \cdot \vec{v} \right)^2$, we'll end up with $\lVert \vec{u} \times \vec{v}\rVert$! Awesome!

So now we proved that $\lVert \vec{u} \times \vec{v}\rVert = \sqrt{\lVert \vec{u} \rVert^2 \lVert \vec{v} \rVert^2 - \left( \vec{u} \cdot \vec{v} \right)^2}$. This is cool, but it doesn't answer your question. Luckily, there isn't too much to be done to do so.

We can use the definition of the dot product as follows: $\vec{u} \cdot \vec{v} = \lVert\vec{u}\rVert \lVert\vec{v}\rVert \cos \theta$, where $\theta$ is the angle between the two vectors. I'm not going to prove the definition of the dot product, but the proof is always out there on the internet somewhere. For now, just roll with it.

This means that: $$\begin{align} \sqrt{\lVert \vec{u} \rVert^2 \lVert \vec{v} \rVert^2 - \left( \vec{u} \cdot \vec{v} \right)^2} & = \sqrt{\lVert \vec{u} \rVert^2 \lVert \vec{v} \rVert^2 - \left( \lVert\vec{u}\rVert \lVert\vec{v}\rVert \cos \theta \right)^2} \\ & = \sqrt{\lVert \vec{u} \rVert^2 \lVert \vec{v} \rVert^2 - \lVert\vec{u}\rVert^2 \lVert\vec{v}\rVert^2 \cos^2 \theta} \\ & = \sqrt{\left(\lVert \vec{u} \rVert^2 \lVert \vec{v} \rVert^2 \right) \left(1 - \cos^2 \theta \right)} \\ & = \lVert \vec{u} \rVert \lVert \vec{v} \rVert \sqrt{1 - \cos^2 \theta} \\ \end{align}$$

Now, thanks to the almighty, S-tier trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$, we can say that $1 - \cos^2 \theta = \sin^2 \theta$. Bask in the glory of this identity.

Time to finish what we started:

$$\begin{align} \lVert \vec{u} \rVert \lVert \vec{v} \rVert \sqrt{1 - \cos^2 \theta} & = \lVert \vec{u} \rVert \lVert \vec{v} \rVert \sqrt{\sin^2 \theta} \\ & = \lVert \vec{u} \rVert \lVert \vec{v} \rVert \sin \theta \\ \end{align}$$

Thus $\lVert \vec{u} \times \vec{v}\rVert = \sqrt{\lVert \vec{u} \rVert^2 \lVert \vec{v} \rVert^2 - \left( \vec{u} \cdot \vec{v} \right)^2} = \lVert \vec{u} \rVert \lVert \vec{v} \rVert \sin \theta $.

Hope this helps!

Answer 2

I know what you're thinking, where the hell did \sin \thetacome from? You have to realize that the cross product is defined to be an operation on vectors, that by definition gives you a vector which has magnitude and direction perpendicular to the plane which contains a &b with conventional orientation based on order etc. Now, why did we choose the magnitude to be this in particular? And why is it \sin \theta? And why is it \cos \thetain the dot product? First of all, forget about vectors, what matters is the quantities, we later make objects and call them vectors just to model easily some physical phenomena. Well, for the dot product, we found EXPERIMENTALLY that the work done by the force F, say, pushing an object horizontally on a table in the gravitational field of the earth, and the force makes an angle theta with the table, the work is given by F\delta x where delta x is the distance travelled etc. (I choose a very specific situation just for easier illustration.) Now why is that? Doesnt that depend on what work is and how its defined? Yes, we mean by work, at least in this context, the quantity that is conserved which will have the form 1/2 m v^2 when the object moves and so on. So we did not CHOOSE \cos \thetaarbitrarily, we FOUND that that is precisely the factor you need to multiply the magnitude of the force and displacement by to get the quantity which is characterized to be work. So, similarly for the cross product, the factor which comes in the formula for torque, moment etc is experimentally found to be \sin \theta, there may be theoretical derivations I dont know about, but what matters is the experiment. So, now we can define such operations for vectors and obtain the quantities in terms of the coordinates (like a1b1+a2b2+a3b3 for the dot product) so that its easier to work with when using some types of coordinate systems like cartesian coordinates. So I'm not immersed in mechanics now, so there may be slight flaws in my answer. Sorry for the long answer, and I kind of feel like this should be on physics stackexchange but I guess in things like this we can't really seperate them.

Answer 3

Definition

The cross-product $\vec{u}\times \vec{v}$ is defined as the vector normal to the plane spanned by $\vec{u},\vec{v}$ with a magnitude equal to the area of the parallelogram formed by these two vectors (see above image). You may have meant to say $||\vec{a}\times\vec{b}||=ab\sin\theta$ (where $a,b$ are the respective vector magntiudes) as the expression on the RHS is a scalar quantity.

Proof

We know that the area of a parallelogram is $A=bh$ where $b$ is the length of one base and $h$ is the altitude to the other vector endpoint. From the diagram above, we see that $u$ is the length of a base and $v\sin\theta$ is the altitude to $\vec{v}$'s endpoint. We could have done this the other way by swapping $v$ for $u$. Thus, the area of this parallelogram, and therefore the magnitude of $\vec{u}\times\vec{v}$, is $uv\sin\theta$.

Answer 4

The cross product of two vectors u and v yields a vector perpendicular to both u and v, with a magnitude that is the area of the parallelogram formed by the vectors u and v. Let’s look at an example of vectors u and v in R2, just to get a sense of where the sin(theta) comes from. Consider the following depiction of vectors u and v.

Let’s form a parallelogram using u and v, with an area of A. h is the depicted height.

The magnitude of the cross product of the two vectors u and v is the area of the parallelogram shown. To find this area, we must multiply the magnitude of vector v with h, the shown height.

As shown, h is the magnitude of vector u multiplied by the sine of the angle between vectors u and v.

Thus, the area of the parallelogram would be the magnitude of v multiplied by the magnitude of u multiplied by the sin of the angle between the vectors u and v. That is where the sin(theta) comes from — the height of the parallelogram shown involves the sine of the angle between vectors u and v.