Jump to content

Unit vectors


1123581321

Recommended Posts

I was wondering why we use unit vectors, what's happens if we don't and in the definition of a unit vector, does a magnitude of 1 mean a scalar of 1 (for a that vector) or that the vector itself only represents one/single arrow ?

Link to comment
Share on other sites

We can make all of our equations without unit vectors (but not without vectors), just divide and multiply by the magnitude.

[math]\frac{r}{r}\hat{\mathbf{r}}=\frac{1}{r}\mathbf{r}[/math]

In terms of the magnitude of a unit vector (or any vector) it relates to size.

 

A vector is something that carries two pieces of information, direction and size. So 3km east is a vector, or 1 Newton down. We often use unit vectors when we are supplying the size information elsewhere so we'll have

[math]\frac{1}{r^2}[/math] The maginutde (a scalar or number, it carries the units, too) [math]\times\hat{\mathbf{r}}[/math] the direction (in english this means 'in the direction that you measured r') It still carries a magnitude (1), but this is replaced by the other magnitude when they are multiplied (1x something=something).

 

Another unit vector would be 1 unit east, or 'in the direction I drove'

so you could have 5 km (magnitude) 'in the direction I drove'(unit vector), together they make a normal vector.

Link to comment
Share on other sites

It means vector's length is 1. If you know the dot product, it means that [math]\sqrt{\vec{v} \cdot \vec{v}} = 1[/math]. If you don't, it means that if you take a ruler and measure the vector, it'll be 1 unit long.

 

It's convenient because you can do math with unit vectors without changing the length of your original vector. For example, in physics we might do:

 

[math]\vec{F} = G \frac{m_1 m_2}{r^2} \hat{r}[/math]

 

where [math]\hat{r}[/math] is a unit vector pointing in the same direction as [math]\vec{r}[/math]. In this equation, [math]\vec{r}[/math] would be a vector pointing from one object of mass [math]m_1[/math] to another object of mass [math]m_2[/math], and [math]\vec{F}[/math] would be the gravitational force between them.

 

If we don't use a unit vector in the equation, we end up multiplying by a vector of some random length, and we make the gravitational force vector too strong or too weak.

Link to comment
Share on other sites

Across the board? What do you mean by that?

 

Those certainly are not the only notations used, even for three space. You will see [math]\hat x, \hat y, \hat z[/math] to denote the unit vectors along the x, y, and z axes, [math]\hat r, \hat {\theta}, \hat{\phi}[/math] for the spherical unit vectors, and more generically [math]\hat u_1, \hat u_2, \hat u_3[/math] to denote an arbitrary set of unit vectors. That's just a starter.

 

And of course i, j, and k only pertain to three space.

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.