Let M be the surface defined by z=x(adsbygoogle = window.adsbygoogle || []).push({}); ^{2}+3xy-5y^{2}. Find a unit normal vector field U defined on a neighborhood of p on M.

First, I reparameterized the equation for the surface to get x(u,v)=(u,v,u^{2}+3xy-5y^{2}). Next I found two tangent vectors x_{u}(u,v)=(1,0,2u+3v) and x_{v}=(0,1,3u-10v). The next step is where I'm unsure. In the text it gives an equation for the unit normal function, U=(x_{u}X x_{v})/||x_{u}X x_{v}||. When I use this equation, I come up with

U=(-2u-3v,10v-3u,1)/(sqrt(13u^{2}-54uv+109v^{2}+1)

This just seemed messy to me, not sure if I'm missing something or if that is actually what the answer should look like. Any verification/help would be appreciated

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# Homework Help: Differential Geometry: Unit Normal Field

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