Pythagorean Theorem and Cross Product of the Vectors

Below post is a back-up of my posts in my Quora blog. I take a copy of them to here in case if something goes wrong in there.

In previous post, I have mentioned that I will explain the motion based relation of dimensions in my next post. I believe that this is the most counter intuitive idea which I came up with long time ago , but since then I couldn’t figure out a way to explain it in a meaningful way. Recently, while I was trying to understand the Pythagorean Theorem in depth, I have noticed that all I needed was to apply Cross product to the vectors in my model.

When somebody talks about the vectors, an average person usually becomes dull. In our daily lives, we usually don’t use vectors and scalar quantities like speed, distance, weight are enough for all of us. Therefore, explaining something using vectors can be meaningless to the most. But what I have noticed while playing with the Pythagorean Theorem, even its geometrical explanation with a different interpretation can be useful to visualize and understand what the cross product means. Let’s see how;

We have been thought in the school that the Pythagorean Theorem is a useful way to find the longest edge of a perpendicular triangle and its geometric explanation is usually given as in the below sketch:

There are even other geometrical convincing proofs like the below one. You can see more inPythagorean theorem Wikipedia page.

Simple, isn’t it? That means, sum of square of the short edges’ lengths on a perpendicular triangle is equal to the square of the long one’s length. With this formula, once you know the length of any 2 sides of a perpendicular triangle, you can find the 3rd one. Very useful for the geometry problems.

Interestingly, this theorem is also applicable in higher dimensions. You can find the derivation of it in 3 dimensions below:

As per our previous proof, it is clear that the red line’s length can be calculated from the x and y coordinates of that triangle. After calculating that length, by using the z coordinate we can also calculate the length of the brown line which is located in a 3 dimensional coordinate system. It is obvious that the same logic can be even taken further into the higher dimensions. What a useful idea.

But now our picture has changed from our starting one in 2 dimensions. In 2 dimensions, we were using it on a triangle but now we have started to use it in a coordinate system. What if, we step back and use the coordinate system logic in our 2D example again. Take a look at the below sketch to understand what I mean:

Now, everything seems a bit different even though our formula still applies. The perpendicular triangle is no longer there. You can only have it by translating the blue “a” line to the location of dashed red line. Then the triangle is visible. But, without having that triangle, how can we apply the first geometrical proof that we have seen? You see 2 attempts for that below:

It is not as nice and neat as we did before, isn’t it. Our green square is messing with the squares of the other sides and it losses the true meaning of our proof. I wish there was an extra space that we could locate our green square.

Then I saw it. There was an available extra space; 3rd spatial axis! If I place our square along that axis, it would not mess with any other squares. That kind of thinking literally gave a new depth to my look at the problem. Check out the below sketch to see what I mean:

Seeing it this way, have changed my look at the Pythagorean Theorem completely. Before, the whole theory was about finding the hypotenuse but now it seems like the real result of it lies on the z axis and only its geometrical proof which is the c² sized area has its one side on the x-y plane and this could be interpreted as an equivalent to the hypotenuse after translating a or b to the dashed locations.

After seeing that, I had the pleasure of discovering a paradigm shifting concept for a while. That was before I find out its relation to the cross product in the Wikipedia page of Pythagorean Theorem. Then I have started to remember our classes in the high school where they were teaching us about the vectors and how to take their products. Then all these ideas started to make sense (by leaving me wondering; why they don’t teach us this stuff like that at school).

I wanted to take this one step further and try to see how it could be applied to 3 dimensions. I have figured out the directions for the squares of the lengths along x, y, z as in below sketch. They are shown with blue, yellow and green shaded squares and not to mess the big sketch, I have given their intersection view next to it separately.

(I know, some of them doesn’t look like a square at all, but this is the best I could draw after a half day of struggle, so please assume that green, blue and yellow shades are squares.)

Now, there is one important thing is missing compare to our previous 2D version. As the theory says, there should be a d² sized square which its area is equal to the sums of a²+b²+c². The problem is, there is no available space to draw that into. In our previous 2D example, we also couldn’t find any available space and had to use the axis in one higher dimension. Likewise, we should also draw our d² square onto the 4th spatial axis but unfortunately we don’t have one.

Time direction could be a good one to draw that onto, since it is also known as 4th dimension but instead of calling it as 4th dimension, physicists today name our universe as space-time, meaning it has 3 spatial dimensions + 1 temporal time dimension. That means, time dimension cannot be treated like a spatial one. Although, in general relativity, it is possible to convert it into a spatial dimension by multiplying it with a constant called “c”. This constant is speed of light. Coincidentally, this is exactly the same thing to do for calculating the distance we have covered on time axis, if we were travelling in that direction with speed of light as in the universe model which I currently try to work out. Things start to fit perfectly and its my main reason of writing this post about the Pythagorean Theorem. I will use it in my future posts for further explanations.

Even if you are not interested in new universe models, I hope this post could give you some new look on a very well known theorem.