On the direction of the cross product of vectors

One of my math professors always told me:

Understand the concept and not the definition

A lot of times I have fallen into this pitfall where I seem to completely understand how to methodically do something without actually comprehending what it means. And only after several years after I first encountered the notion of cross products did I actually understand what they really meant. When I did, it was purely ecstatic!

 

Why on earth is the direction of cross product orthogonal ? Like seriously…

I mean this is one of the burning questions regarding the cross product and yet for some reason, textbooks don’t get to the bottom of this. One way to think about this is :

It is modeling a real life scenario!!

The scenario being :

When you try to twist a screw (clockwise screws being the convention) inside a block in the clockwise direction like so, the nail moves down and vice versa.

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i.e When you move from the screw from u to v, then the direction of the cross product denotes the direction the screw will move.

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That’s why the direction of the cross product is orthogonal. It’s really that simple!

 

Another perspective

Now that you get a physical feel for the direction of the cross product, there is another way of looking at the direction too:

Displacement is a vector. Velocity is a vector. Acceleration is a vector. As you might expect, angular displacement, angular velocity, and angular acceleration are all vectors, too.

But which way do they point ?

crossproduct

Let’s take a rolling tire. The velocity vector of every point in the tire is pointed in every other direction. BUT every point on a rolling tire has to have the same angular velocity – Magnitude and Direction.

How can we possibly assign a direction to the angular velocity ?

cross-product-simple

Well, the only way to ensure that the direction of the angular velocity is the same for every point is to make the direction of the angular velocity perpendicular to the plane of the tire.
Problem solved!

 

 

 

Basis Vectors are instructions !

Basis vectors are best thought of in the context of roads.

Imagine you are in a city – X which has only roads that are perpendicular to one another.

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You can reach any part of the city but the only constraint is that you need to move along these perpendicular roads to get there.

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Now lets say you go to another city-Y which has a different structure of roads.

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In this case as well you can get from one part of the city to any other, but you have to travel these ‘Sheared cubic’ pathways to get there.

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Just like these roads determine how you move about in the city, Basis Vectors encode information on how you move about on a plane. What do I mean by that ?

The basis vector of City-X is given as:

screenshot-from-2017-02-19-021951

\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

This to be read as – ” If you would like to move in City-X you can only do so by taking 1 step in the x-direction or 1 step in the y-direction ”

The basis vector of City-Y is given as:

screenshot-from-2017-02-19-021644

\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}

This to be read as – ” If you would like to move in City-Y you can only do so by taking 1 step in the x-direction or  1 step along the diagonal OB ”

 

Conclusion:

By having the knowledge about the Basis Vectors of any city, you can travel to any destination by merely scaling these basis vectors.

As an example, lets say need to get to the point (3,2), then in City-X,  you would take 2 steps in the x-direction and 3 steps in the y-direction

\begin{bmatrix} 3 \\  2 \end{bmatrix}  =  3* \begin{bmatrix} 1 \\  0 \end{bmatrix}  +  2 * \begin{bmatrix} 0 \\  1 \end{bmatrix}

And similarly in City-Y, you would take 1 step along the x -direction and 2 steps along the diagonal OB.

\begin{bmatrix} 3 \\  2 \end{bmatrix}  =  1* \begin{bmatrix} 1 \\  0 \end{bmatrix}  +  2 * \begin{bmatrix} 1 \\  1 \end{bmatrix}

Destination Arrived 😀