Cross Product Intuition

Question

I know the cross product between a vector $a$ and a vector $b$ is just a vector whose magnitude is the product of magnitude of $b$ times the magnitude of the perpendicular component of $a$ in relation to $b$.This arises in many applications like in calculating the torque and the magnetic force.

The magnitude of this vectors definies how perpendicular $a$ and $b$ are ... The more perpendicular, the higher the magnitude.This fact implies that, in order for it to be a measure of perpendicularity, we need the magnitude of $a \times b$ to be $\lvert a \rvert \lvert b \rvert \sin(\theta)$ and that should be a no-brainer.

It turns out that measuring the "perpendicularity" is the same as measuring the area between the paralelogram formed by the two vectors, the more perpendicularity, higher the area that is formed.

Concluding, I fully understand the reasoning between why the magnitude of the cross product is the way it is. But i want to keep building my intuition on Cross Product and I'm kinda stuck with two problems:

1- What is the reason (other than the fact that the magnetic force is vector perpendicular to both $v$ and $b$, and that the normal vector to a plane is a vector perpendicular to two vectors in the plane, for example) that this measure of perpendicularity ($\lvert a \rvert \lvert b \rvert \sin(\theta)$) was attributed to the magnitude of a vector? Why wasn't the cross product defined as just this magnitude? Was the orthogonal vector just some convenient form of killing two birds with a stone (getting both the measure of perpendicularity and getting the normal vector to the plane spanned by $a$ and $b$)?

2 - Is there any intuition that the components of cross product $a \times b$ are:
$ \langle(a_y b_z - a_z b_y), (a_z b_x - a_x b_z), (a_x b_y - a_y b_x) \rangle$? I'm saying intuition because I know the proof that the only vector (well one of the two possibles) who holds magnitude $\lvert a \rvert \lvert b \rvert \sin(\theta)$ and is perpendicular to both $a$ and $b$ should have this components.

Answer 1

Just a remark concerning 'problem 2':

The coordinates of the cross product $\bf{a}\times\bf{b}$ are the determinants of the projections of $\bf{a}$ and $\bf{b}$ onto the coordinate planes. So the $x$-coordinate of $\bf{a}\times\bf{b}$ is the area of the parallelogram spanned by the projections of $\bf{a}$ and $\bf{b}$ onto the $yz$-plane. I hope this helps your intuition a bit.

Answer 2

For 1; The reason why the cross product isn't just the magnitude, is that we simply want it to also be a vector with its direction perpendicular to the other 2 vectors in the plane, etc. Now, why? Well because in the example about the magnetic force, we found that (from observations/discoveries/equations etc...) the force's direction is perpendicular to v and B. But in the same time, the magnitude of this force is also VBSin(theta).

So to MODEL this phenomena, we define the cross product operation to be what it is, just like dot product and work, but in the case of work, it is useless to think of work as a vector, because its just a quantity that's conserved and that's all we care about. :) I hope that makes sense.

Oh, and, about the components, I tried to derive the formula for them based on the criteria that the vector must be perp. to the other 2 and has this magnitude etc, I got pretty close and I'm working on it. I'm in fact thinking about posting a question on this.

Answer 3

Matt L.'s answer is the one I would give but it could use some elaboration.

First, get comfortable with the fact that a n-D determinant is the signed area of a n-D parallelepiped. We're concerned with cross products in 3-D. This entails 2-D determinants -- areas of parallelograms. Here's a good video on this.

Now, what's the significance of the projections of the areas formed by two vectors onto coordinate planes?

Before even worrying about the cross product or areas formed by two vectors, let's motivate the idea of projecting area by thinking about tops:

(ignore the shadow in this picture. The light source is not straight vertically down, so it doesn't fit into the analogy)

The reason tops come to mind is because it has everything we need:

1. area. The area of the "circle" part of the top. We don't really care what the area actually is and we certainly don't care that it involves $\pi$. All we care about is that it's some amount of area that lives in its own flat plane.
2. a "vector" orthogonal to that area. The handle part sticking out of the circle is orthogonal to the plane that the circle lies in.

Let's say we start out with the top upright. So the circle lies in the $x,y$ plane, and the handle points along the $z$ axis.

Let's say the area of the top's circle is $A$. If we shine an idealized flashlight straight down onto the top, the area of the shadow on the table underneath the top should be $A$, Right? The light rays are shining down along the $z$ axis, and striking the table (flat, parallel to $x,y$ plane), except for those rays blocked by the top's circle.

Then, what if we tilt the top in an arbitrary direction? Its physical area $A$ will not change, but the area of the shadow will now decrease, right? Let's call the area of its shadow after tilting $S_{tilted}$.

The key insight is that $\frac{S_{tilted}}{A} = cos(\theta)$, where $\theta$ is the angle between the top's surface and the $x, y$ plane, which is also the angle between the top's handle and the $z$ axis.

That is, as we tilt our top, the handle forms a widening angle with the $z$ axis, and since the handle is rigidly orthogonal to the circle, the circle (the circle's plane) is simultaneously forming that same angle with the $x,y$ plane.

The same thing holds for the other coordinate planes. We could rename $S_{tilted}$ to $S_{table}$, and then also measure $S_{western-wall}$, and $S_{southern-wall}$.

$S_{western-wall}$ is the area of the shadow we get by shining the flashlight "down" the $x$ axis and making a shadow on our western wall, the $y, z$ plane. Likewise for $S_{southern-wall}$.

We can use only these shadow areas and no knowledge of the top's handle to produce a vector that points in the same direction as the top's handle. This will be analogous to using the projections of the area formed by two vectors to get a third vector orthogonal to both of them -- i.e., the cross product.

So if we put these shadow measurements into a vector, we could have

$$ \begin{align} \left( \begin{matrix} S_{western-wall} \\ S_{southern-wall} \\ S_{table} \end{matrix} \right) & = A \cdot \left( \begin{matrix} \text{ cosine of angle between top and y, z plane}\\ \text{ cosine of angle between top and z, x plane}\\ \text{ cosine of angle between top and x, y plane}\\ \end{matrix} \right) \\ \\ & = A \cdot \left( \begin{matrix} \text{ cosine of angle between handle and x axis}\\ \text{ cosine of angle between handle and y axis}\\ \text{ cosine of angle between handle and z axis}\\ \end{matrix} \right) \\ \\ & = A \cdot \text{direction of handle} \end{align} $$

specificially, $A$ times a unit vector pointing in the direction of the handle. Because knowing the angle made between the handle and each of the axes tells us the the direction of the handle.

So all we did was use a flashlight to measure the various shadows (projections), and this gave us a vector pointing in the direction of the handle, with length $A$ (roughly the "cross product", the thing we really wanted). Notice we never measured $A$ directly, and we never measured anything about the handle directly. We only used the knowledge that the handle is orthogonal to the top's surface.

Remaining things we need to figure out

1. What exactly is the angle "between two planes"? Why does cosine/trig still apply here like it does with angles between lines?
2. This was all done with the area of a circle. What about the area of a parallelogram formed with our two vectors we want a cross product of?
3. How does the formula for a cross product give us the areas of the shadows (projections)?

Imagine an opened door. The door forms an angle with the wall / doorway. Imagine if the door and wall was painted with thin horizontal stripes. Each stripe "gets its own angle" between the door and the wall when you open the door. If you're facing the wall when the door opens, the line segments on the door will be "visually scaled" by a factor of $cos(\theta)$. When the door is fully open to 90 degrees, you won't see the line segments on the door at all ($cos(90) = 0$).

If you outlined a circle on the door, the area of that circle is made up of lots of thin horizontal stripes. Each stripe gets scaled by the same amount -- $cos(\theta)$. Therefore, the entire area of the circle is scaled by this factor.

But it doesn't really matter that you drew a circle, does it? This would apply to any drawing. It could be a drawing of a star shape, or it could be a drawing of a parallelogram formed with two vectors. All area in the door plane is scaled by the same factor as you watch the door open.

Any two planes are equivalent to a wall and a door. They just might not be aligned with gravity. The line of intersection between the two planes is where the door's hinges and the "crack" of the door are.

You could also think about starting with the top in its original $x, y$ plane position, grabbing onto the tip of the handle, and tilting the top. As you tilt the top, the tip of the handle traces out a section of a circle in the air between its original $z$ axes aligned position, and its new position upon tilting. You could paint the top's surface with lots of thin lines that are parallel to that circle of tilt. Each of these lines moves away from its original position inside the $x, y$ plane with the same angle that the handle is being tilted from the $z$ axis.

That sort of addresses 1 and 2, which leaves 3.

If you have a parallelogram formed with two vectors $v$ and $w$, how do you get its projection on the $x, y$ plane? You delete the $z$ coordinates of $v$ and $w$ (set them to 0). Now you have two 2-D vetors in the $x, y$ plane. If you look closely at the cross product formula, you'll see that you're using the determinant of these 2-D vectors to get the third coordinate of the cross product. That's the area of the shadow on the table, $S_{table}$.

This hasn't covered the matter of orientation at all. That could be covered elsewhere. A small hint would be... notice I've referred to the $z, x$ plane as the "southern wall" one. Why not the $x, z$ plane? In a sense, $x, z$ is out of order.