## anonymous 3 years ago How do I draw a vector field by hand? F(x,y,z)=-y k

1. anonymous

@phi @Zarkon @inkyvoyd

2. klimenkov

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

3. anonymous

Yes

4. klimenkov

|dw:1359238654831:dw|

5. anonymous

I'm not sure what you did there?

6. abb0t

I'm not sure what she did there either.

7. klimenkov

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. anonymous

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

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. anonymous

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

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

12. anonymous

Thank you!