Question

Find the angle between the surfaces x^{2}+y^{2}+z^{2}=9 and z+3=x^{2}+y^{2} at the point (-2,1,2)

Answer 1

$Suppose \space \phi_1=x^{2}+y^{2}+z^{2}-9 \\ \phi_2= x^{2}+y^{2}-z-3\\ So, \Delta \phi_1 = 2x \hat{i}+2y \hat{j}+2z \hat{k}\\ \Delta \phi_2 = 2x \hat{i}+2y \hat{j}\\ (\Delta \phi_1)_{(-2,1,2)} = -4 \hat{i}+2 \hat{j}+4 \hat{k}\\ (\Delta \phi_2)_{(-2,1,2)} = -4\hat{i}+2 \hat{j}\\ \mathrm{Computing\:the\:angle\:between\:the\:vectors}:\quad \cos \left(\theta \right)\:=\frac{\vec{a\:}\cdot \vec{b\:}}{\left|\vec{a\:}\right|\cdot \left|\vec{b\:}\right|}\\ \theta = \arccos [\frac{\Delta \phi_1*\Delta \phi_2}{|\Delta \phi_1||\Delta \phi_2|}]\\ θ=\arccos \left(\cos \left(θ\right)\right)=\arccos \left(\frac{\sqrt{5}}{3}\right)\\ θ=41.81031^{\circ \:}$