Question

Find the radius and the centre of the circular section of the sphere I r I = 4 cut off by the plane r . (2i -j + 4k) = 3.

Answer 1

$\displaystyle\textsf{Let S be the sphere in}\, R^3\, \textsf{with center}\\ O(x_0, y_0, z_0)\, \textsf{and radius}\, r, \, \textsf{and let}\\ P\, \textsf{be the plane with equation}\, Lx + My + Nz = Q,\\ \textsf{so that}\, \overrightarrow{n} = (L, M, N)\, \textsf{is an orthogonal}\\ \textsf{vector of}\, P.\\ \textsf{If}\, P_0\, \textsf{is an arbitrary point on}\,P, \, \textsf{the}\\ \textsf{distance from the center of the sphere}\, O \\ \textsf{to the plane}\, P\, \textsf{is}\\ \rho = \frac{(O - P_0)\overrightarrow{n}}{||\overrightarrow{n}||} = \frac{Lx_0 + My_0 + Nz_0 - Q}{\sqrt{L^2 + M^2 + N^2}}\\ \textsf{The intersection of the plane and the sphere}\\ \textsf{is a circle, if and only if}\, |\rho| < r,\, \textsf{and in}\\ \textsf{that case, the circle has radius}\,r_c = \sqrt{r^2 - \rho^2}\\ \textsf{and center}\\ c = O + \frac{\rho\cdot\overrightarrow{n}}{||\overrightarrow{n}||} = (x_0, y_0, z_0) + \frac{\rho(L\textbf{i} + M\textbf{j} + N\textbf{k})}{\sqrt{L^2 + M^2 + N^2}}\\ \textsf{From the given information, we can deduce that,}\\ O(x_0, y_0, z_0) = O(0, 0, 0), |r| = 4; 2x - y + 4z = 3\\ S = \{(x, y, z): x^2 + y^2 + z^2 = 16\}, \,P = \{(x, y, z): 2x - 4y + 4z = 3\\ \rho = \frac{Lx_0 + My_0 + Nz_0 - Q}{\sqrt{L^2 + M^2 + N^2}} = \frac{2(0) - 1(0) + 4(0) - 3}{\sqrt{2^2 + (-1)^2 + 4^2}} = \frac{-3}{\sqrt{21}}\\ r_c = \sqrt{r^2 - \rho^2} = \sqrt{4^2 - \left(\frac{-3}{\sqrt{21}}\right)^2} = \sqrt{\frac{109}{7}}\\ c = O + \frac{\rho\cdot\overrightarrow{n}}{||\overrightarrow{n}||} = (0, 0, 0) + \left(\frac{-3}{\sqrt{21}}\right)\cdot\frac{(2, -1, 4)}{\sqrt{2^2 + (-1)^2 + 4^2}} = \left(\frac{-2}{7}, \frac{1}{7}, \frac{-4}{7}\right)\\ \therefore \textsf{The radius is}\,\sqrt{\frac{109}{7}}\, \textsf{and the centre is}\, \left(\frac{-2}{7}, \frac{1}{7}, \frac{-4}{7}\right)$