## Wislar Group Title How do I draw a vector field by hand? F(x,y,z)=-y k one year ago one year ago

1. Wislar Group Title

@phi @Zarkon @inkyvoyd

2. klimenkov Group Title

$$\vec{F}(x,y,z)=-y\cdot\vec{k}$$

3. Wislar Group Title

Yes

4. klimenkov Group Title

|dw:1359238654831:dw|

5. Wislar Group Title

I'm not sure what you did there?

6. abb0t Group Title

I'm not sure what she did there either.

7. klimenkov Group Title

This field doesn't depent on $$x$$ and $$z$$. That means that there will be the same vectors in the planes $$y=\text{const}$$. $$\vec{k}$$ is a vector that has length = 1 and is collinear to $$z$$-axis. I drew 3 planes $$y=\text{const}$$ for different values of this constant.

8. Wislar Group Title

So, say (1,0,0).. it would be the zero vector, so wouldn't it just be a point and not a line of length 1?

9. klimenkov Group Title

A vector field is a vector function. Its argument is a vector (point) and its value is also a vector. For example if you take a point (1,0,0), thats mean that $$x=1,y=z=0.$$. So, $$\vec{F}(1,0,0)=0\cdot\vec{k}=\vec{0}$$. For every vector that has y-coordinate = 0, it will be zero vector. y=0 is a plane. Every vector in this plane will be zero-vector. If you take (0,1,0), then $$\vec{F}(0,1,0)=1\cdot\vec{k}=\vec{k}$$. Every vector in the plane y=1 will be $$\vec{k}$$. And so on.

10. Wislar Group Title

I get it!! Thank you so much! So everything with a negative y value would be going up and everything with a positive y value would be going down?

11. klimenkov Group Title

Yes. Sorry, my fault. I have lost the "-" sign. I'm happy that you've got it. You are welcome.

12. Wislar Group Title

Thank you!