In Pressley's Elementary Differential geometry, Q6.4.7 p148 stomps me.

Question: Let $\sigma(u,v)$ be a surface patch with standard unit normal $N$. Show that $$N \times \sigma_u = \frac{E\sigma_v - F\sigma_u}{(EG-F^2)^{1/2}}$$ with $E = \langle\sigma_u , \sigma_u \rangle , F = \langle \sigma_u , \sigma_v \rangle$ and $ G = \langle \sigma_v , \sigma_v \rangle$.

Answer: $N \times \sigma_u = \alpha \sigma_u+\beta \sigma_v$ and $(N \times \sigma_u) \cdot\sigma_u = 0 , (N \times \sigma_u) \cdot \sigma_v = (EG-F^2)^{1/2}$ (makes sense)
This implies that $\alpha E+\beta F = 0 , \alpha F + \beta G = (EG-F^2)^{1/2}$ (does not make sense) How are the equalities deduced? any hint is appreciated!


Taking the dot product of $N \times \sigma_u$ with $\sigma_u$ we get

$$ 0 = (N \times \sigma_u) \cdot \sigma_u = (\alpha \sigma_u + \beta \sigma_v) \cdot \sigma_u = \alpha (\sigma_u \cdot \sigma_u) + \beta ( \sigma_v \cdot \sigma_u) = \alpha E + \beta F. $$

Taking the dot product of $N \times \sigma_u$ with $\sigma_v$ we get

$$ (EG - F^2)^{1/2} = (N \times \sigma_u) \cdot \sigma_v = (\alpha \sigma_u + \beta \sigma_v) \cdot \sigma_v = \alpha (\sigma_u \cdot \sigma_v) + \beta ( \sigma_v \cdot \sigma_v) = \alpha F + \beta G. $$

  • $\begingroup$ thank you! Silly of me not to have thought of that! $\endgroup$ – rannoudanames Mar 25 '17 at 21:55

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