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Wislar

  • one year ago

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

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  1. Wislar
    • one year ago
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    @phi @Zarkon @inkyvoyd

  2. klimenkov
    • one year ago
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    \(\vec{F}(x,y,z)=-y\cdot\vec{k}\)

  3. Wislar
    • one year ago
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    Yes

  4. klimenkov
    • one year ago
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    |dw:1359238654831:dw|

  5. Wislar
    • one year ago
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    I'm not sure what you did there?

  6. abb0t
    • one year ago
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    I'm not sure what she did there either.

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

  12. Wislar
    • one year ago
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    Thank you!

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