Answer
F(x,y)=0 for those (x,y) that lie on the line y=2x.
Work Step by Step
The function is \({\bf{F}}(x,y) = \left( {{y^2} - 2xy} \right){\bf{i}} + \left( {3xy - 6{x^2}} \right){\bf{j}}\).
Let \(D\) be a set in \({\mathbb{R}^2}\) (a plane region). A vector field on \({\mathbb{R}^2}\) is a function \({\bf{F}}\) that assigns to each point \((x,y)\) in \(D\) a two-dimensional vector \({\bf{F}}(x,y)\).
Plot the vector field as follows:
\({\bf{F}}(x,y) = \left( {{y^2} - 2xy} \right){\bf{i}} + \left( {3xy - 6{x^2}} \right){\bf{j}}\)
Length of a vector in field \({\bf{F}}\) increases the farther it is from \((0,0)\).
Notice that vectors with length of almost \(0\) lie around a line.
Find \((x,y)\) such that \({\bf{F}}(x,y) = 0\) as follows:
\(\begin{aligned}{\bf{F}}(x,y) & = 0\\\left( {{y^2} - 2xy} \right){\bf{i}} + \left( {3xy - 6{x^2}} \right){\bf{j}} & = 0\\y(y - 2x){\bf{i}} + 3x(y - 2x){\bf{j}} & = 0\\(y - 2x)(y{\bf{i}} + 3x{\bf{j}}) & = 0\end{aligned}\)
Simplify further as follows:
\(\begin{aligned}y - 2x & = 0\\y &= 2x\end{aligned}\)
Therefore, \({\bf{F}}(x,y) = 0\) for those \((x,y)\) that lie on the line \(y = 2x\).
