Answer to Question #323083 in Differential Geometry | Topology for Dan

Question

Find the angle nearest to the whole number between the surfaces x2+y2+z2=9 and z=x2+y2-3 at the point (2, -1, 2).

Answer 1

$x^2+y^2+z^2=9: kn_1=\left( 2x,2y,2z \right) =\left( 4,-2,4 \right) \Rightarrow n_1=\frac{\left( 4,-2,4 \right)}{\sqrt{4^2+2^2+4^2}}=\left( \frac{2}{3},-\frac{1}{3},\frac{2}{3} \right) \\z=x^2+y^2-3:kn_2=\left( 2x,2y,-1 \right) =\left( 4,-2,-1 \right) \Rightarrow n_2=\frac{\left( 4,-2,-1 \right)}{\sqrt{4^2+2^2+1^2}}=\left( \frac{4}{\sqrt{21}},-\frac{2}{\sqrt{21}},-\frac{1}{\sqrt{21}} \right) \\\left( n_1,n_2 \right) =\frac{1}{3\sqrt{21}}\left( 2\cdot 4-1\cdot \left( -2 \right) +2\cdot \left( -1 \right) \right) =\frac{8}{3\sqrt{21}}\\\cos \theta =\frac{8}{3\sqrt{21}}\Rightarrow \theta =\mathrm{a}\cos \frac{8}{3\sqrt{21}}=0.949716\approx 1$