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klimenkov
 2 years ago
Best ResponseYou've already chosen the best response.2\(\vec{F}(x,y,z)=y\cdot\vec{k}\)

klimenkov
 2 years ago
Best ResponseYou've already chosen the best response.2dw:1359238654831:dw

Wislar
 2 years ago
Best ResponseYou've already chosen the best response.0I'm not sure what you did there?

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.0I'm not sure what she did there either.

klimenkov
 2 years ago
Best ResponseYou've already chosen the best response.2This 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.

Wislar
 2 years ago
Best ResponseYou've already chosen the best response.0So, 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?

klimenkov
 2 years ago
Best ResponseYou've already chosen the best response.2A 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 ycoordinate = 0, it will be zero vector. y=0 is a plane. Every vector in this plane will be zerovector. 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.

Wislar
 2 years ago
Best ResponseYou've already chosen the best response.0I 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?

klimenkov
 2 years ago
Best ResponseYou've already chosen the best response.2Yes. Sorry, my fault. I have lost the "" sign. I'm happy that you've got it. You are welcome.
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