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

klimenkovBest ResponseYou've already chosen the best response.2
dw:1359238654831:dw
 one year ago

WislarBest ResponseYou've already chosen the best response.0
I'm not sure what you did there?
 one year ago

abb0tBest ResponseYou've already chosen the best response.0
I'm not sure what she did there either.
 one year ago

klimenkovBest ResponseYou've already chosen the best response.2
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.
 one year ago

WislarBest ResponseYou've already chosen the best response.0
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?
 one year ago

klimenkovBest ResponseYou've already chosen the best response.2
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 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.
 one year ago

WislarBest ResponseYou've already chosen the best response.0
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?
 one year ago

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