If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## Multivariable calculus

### Course: Multivariable calculus>Unit 1

Lesson 4: Visualizing vector-valued functions

# 3d vector fields, introduction

Vector fields can also be three-dimensional, though this can be a bit trickier to visualize.  Created by Grant Sanderson.

## Want to join the conversation?

• I'm sorry if I seem ignorant for asking this here, but if you can, may you please show me where to find the 3-d visualization program that you use?
• It is an application that comes with Mac OS X called Grapher. If you are on a Mac, you should be able to find it in the `/Applications/Utilities/` directory.
• so now it matters where we start drawing the vector, whereas earlier we could move it around?
• That's right, because in this case the place where you start the vector communicates the "input", while the vector itself is the "output". As you say, when you are thinking of a vector off on its own, not as the output of some vector valued function, you can place it wherever you want.

Also, when vector fields come up in physics, there's usually a physical meaning to the location of a vector. For example, it might tell you the velocity of a fluid at that point, so putting it on that point matters.
• i am tripping on the last example.
if output is the identity, then why isn't it a positional vector?i am having hard time wrapping my head around why it is from the input point.
• It's not really an identity. An identity usually means 1, unless identity is used in a different way in math somewhere I am not aware of. Which is totally possible.

I am not sure if there is a word for it, but the idea of this one is you draw a vector to any point, and the vector field at that point will be the vector from the origin to there, but now with the starting point at the point (x,y,z) So it almost doubles the length, or in other words makes a vector <2x, 2y, 2z>, though the first half of those vectors would be erased.

Let me know if that doesn't make sense.
(1 vote)
• I know you can graph these graphs via Grapher (a Mac app), but is there any standalone software (preferably for windows) that can be used to graph these complex functions?
• I think you might be able to do it with MatLab!
(1 vote)
• When he shows f(x, y, z) = [1, 0, 0], why are there so many of the same vector starting at different points? Is it saying for any x, y, z (point) we have [1, 0, 0] coming out from that point?
• Just realised, that compared to the previous video of 2-dimensional space vector field (Vector Field Introduction at ).. this one maps to itself. A point f(2, 2, 2) would map to (2, 2, 2) and there would be no particles flowing outwards, isn't it?
If, otherwise, the output space is supposed to be added to the input space (as it appears to be in the video), the current representation does not indicate that it's a vector field transformation.
• So when we are plotting f(1, 1, 1 ) = [1, 1, 1], we take (1, 1, 1) as starting point, add [1, 1, 1] from it and connect them. Am I guessing correct?