Find Unit Vectors for f(x,y) w/ D_uf=0

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Cpt Qwark
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Homework Statement


For [tex]f(x,y)=x^2-xy+y^2[/tex] and the vector [tex]u=i+j[/tex].
ii)Find two unit vectors such [tex]D_vf=0[/tex]

Homework Equations


N/A.

The Attempt at a Solution


Not sure if relevant but the previous questions were asking for the unit vector u - which I got [tex]\hat{u}=\frac{1}{\sqrt{2}}(i+j)[/tex] for the maximum value of [tex]D_uf[/tex] which was [tex]\sqrt{2}[/tex].
 
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The directional derivative is defined as
[tex]D_{v}f = \nabla f \cdot \mathbf{v}[/tex]
Your task is to find two vectors [itex]\mathbf{v}[/itex] such that [itex]D_{v}f = 0[/itex].
 
Cpt Qwark said:

Homework Statement


For [tex]f(x,y)=x^2-xy+y^2[/tex] and the vector [tex]u=i+j[/tex].
ii)Find two unit vectors such [tex]D_vf=0[/tex]
This problem doesn't have anything to do with the vector "u". Why is that given? [itex]D_vf[/itex] is the dot product of the gradient, [itex]\nabla f[/itex], and a unit vector in the same direction as vector v. Since you want that to be 0, you are looking for two unit vectors perpendicular to [itex]\nabla f[/itex].

2. Homework Equations
N/A.

The Attempt at a Solution


Not sure if relevant but the previous questions were asking for the unit vector u - which I got [tex]\hat{u}=\frac{1}{\sqrt{2}}(i+j)[/tex] for the maximum value of [tex]D_uf[/tex] which was [tex]\sqrt{2}[/tex].
So "u" was used in previous questions?