I don't quite 'get' vectors.
I've plotted the vector field
$$F(x,y)=(y^2-2xy)i+(3xy-6x^2)j$$
http://kevinmehall.net/p/equationexplorer/vectorfield

Now I have to find the set of points (x,y) such that F(x,y)=0

Question

I don't quite 'get' vectors. 
I've plotted the vector field 
$$F(x,y)=(y^2-2xy)i+(3xy-6x^2)j$$
http://kevinmehall.net/p/equationexplorer/vectorfield

Now I have to find the set of points (x,y) such that F(x,y)=0

anonymous · Answer

such that the vector is zero?

anonymous · Answer

Please guide me to wisdom......

anonymous · Answer

haha! okay... in order for F(x,y) to be 0, y^2 - 2xy must equal to xero and 3xy - 6x^2 = 0 
and then, solve for y to get the set of  points (x,y) such that F(x,y)=0!

anonymous · Answer

zero*

anonymous · Answer

Ohhhhhh...it's that simple? 
$$y^2-2xy=0$$
$$y=\sqrt{2xy}$$
and
$$3x\sqrt{2xy}- 6x^2 = 0 $$ and then just solve for x and solve for y
Perfect!

anonymous · Answer

THank you!

anonymous · Answer

humm...not quite what i was thinking! great try though! 
do u want to see my method?

anonymous · Answer

I reread what you just wrote....uhm, why are we solving for only Y and not x?

anonymous · Answer

u actually solve for both! but mysteriously u get y = something for both! 
i will show...

anonymous · Answer

no I solved for y in the first equation and subbed it into the second equation

anonymous · Answer

okay here is what you get when u solve for the first one.. |dw:1344482176416:dw| and then, to solve for the second one, |dw:1344482387900:dw|