Find a formula F⃗ =⟨ F1(x,y), F2(x,y) ⟩ for the vector field in the plane with the following properties: 1. F⃗ is not defined at the origin 2. At any other point (x,y)≠(0,0) the vector field F⃗ points toward the origin with magnitude inversely proportional to the square of the distance from (x,y) to the origin. 3. At the point (1,0) the vector field F⃗ has magnitude 1.

Consider two vectors a and b. Then, the vector b-a is a vector whose tail is at the tip of vector a and whose tip is at the tip of vector b. That is, it is a vector that points from the tip of vector a to the tip of vector b. To achieve the second restriction, we will take b as the 0 vector and to be the vector that points at point (x,y). Then, we have that at point (x,y) the vector that points to the origin is (-x,-y). Recall that given a vector (a,b), its magnitude is given by [tex]\sqrt[]{a^2+b^2}[/tex]. Then, in this case we want that

[tex]\sqrt[]{F_1(x,y)^2+F_2(x,y)} = \frac{1}{x^2+y^2}[/tex]

Note that given a vector v, if we divided it by the cube of its magnitude, we get that

[tex]||\frac{v}{||v||^3}|| = \frac{||v||}{||v||^3} = \frac{1}{||v||^2}[/tex]. Then, consider dividing the vector (-x,-y) by the cube of its magnitude. That is, consider the vector

[tex](\frac{-x}{(x^2+y^2)^{\frac{3}{2}}},\frac{-y}{(x^2+y^2)^{\frac{3}{2}}})[/tex]

Then,

[tex]\sqrt[]{\frac{x^2+y^2}{(x^2+y^2)^3}} = \sqrt[]{\frac{1}{(x^2+y^2)^2}}= \frac{1}{x^2+y^2}[/tex].

Note that if (x,y) = (1,0) we get the vector (-1,0), which has magnitude one and that if (x,y) = (0,0) the field is not defined