Answer to Question #195579 in Analytic Geometry for Dhruv bartwal

Question

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

Answer 1

Let $S$ be the sphere in $R^3$ with center $O(x_0,y_0,z_0)$ and radius $r,$ and let $P$ be the plane with equation $Ax+By+Cz+D=0,$ so that $\vec n=(A,B,C)$is a normal vector of $P.$

If $P_0$ is an arbitrary point on $P,$ the signed distance from the center of the sphere $O$ to the plane $P$ is

$$\rho=\dfrac{(O-P_0)\vec n}{\|\vec n\|}=\dfrac{Ax_0+By_0+Cz_0-D}{\sqrt{A^2+B^2+C^2}}.$$

The intersection $S\cap P$ is a circle if and only if $r<\rho<r,$ and in that case, the circle has radius $r_c=\sqrt{r^2-\rho^2}$ and center

$$C=O+\rho \cdot\dfrac{\vec n}{\|\vec n\|}$$

$$=(x_0, y_0, z_0)+\rho\cdot\dfrac{(A,B,C)}{\sqrt{A^2+B^2+C^2}}$$

In our case:

$O(0,0,0), |r|=4, 2x-y+4z=3$

$S=\{(x,y,z):x^2+y^2+z^2=16\},$

$P=\{(x,y,z):2x-y+4z=3\}\}.$

$$\rho=\dfrac{Ax_0+By_0+Cz_0-D}{\sqrt{A^2+B^2+C^2}}$$

$$=\dfrac{2\cdot0-1\cdot0+4\cdot0-3}{\sqrt{(2)^2+(-1)^2+(4)^2}}=-\dfrac{3}{\sqrt{21}}$$

$$r_c=\sqrt{r^2-\rho^2}=\sqrt{(4)^2-(-\dfrac{3}{\sqrt{21}})^2}=\sqrt{\dfrac{109}{7}}$$

$$\approx3.946$$

$$C=O+\rho \cdot\dfrac{\vec n}{\|\vec n\|}$$

$$=(0, 0, 0)+\big(-\dfrac{3}{\sqrt{21}}\big)\cdot\dfrac{(2,-1,4)}{\sqrt{(2)^2+(-1)^2+(4)^2}}$$

$$=(-\dfrac{2}{7}, \dfrac{1}{7}, -\dfrac{4}{7})$$

$$r_c=\sqrt{\dfrac{109}{7}}\approx3.946, C=(-\dfrac{2}{7}, \dfrac{1}{7}, -\dfrac{4}{7})$$