Question

Find the radius and the center of the circular section of the sphere |r| = 4 cut off by the
plane r·(2i−j+ 4k) = 3

Answer 1

Answer on Question #62051 - Math - Analytic Geometry

Question

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

Solution

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 $Ax + By + Cz = D$ , 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

$$\rho = \frac {(O - P _ {0}) \vec {n}}{| | \vec {n} | |} = \frac {A x _ {0} + B y _ {0} + C z _ {0} - D}{\sqrt {A ^ {2} + B ^ {2} + C ^ {2}}}.$$

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

$$c = O + \rho \cdot \frac {\vec {n}}{| | \vec {n} | |} = (x _ {0}, y _ {0}, z _ {0}) + \rho \cdot \frac {(A , B , C)}{\sqrt {A ^ {2} + B ^ {2} + C ^ {2}}}.$$

In our case:

$$\begin{array}{l} O (0, 0, 0); | r | = 4; 2 x - y + 4 z = 3 \\ S = \{(x, y, z): x ^ {2} + y ^ {2} + z ^ {2} = 1 6 \}, \qquad P = \{(x, y, z): 2 x - y + 4 z = 3 \}. \\ \end{array}$$ $$\begin{array}{l} \rho = \frac {A x _ {0} + B y _ {0} + C z _ {0} - D}{\sqrt {A ^ {2} + B ^ {2} + C ^ {2}}} = \frac {2 \cdot 0 - 1 \cdot 0 + 4 \cdot 0 - 3}{\sqrt {2 ^ {2} + (- 1) ^ {2} + 4 ^ {2}}} = - \frac {3}{\sqrt {2 1}}. \\ r _ {c} = \sqrt {r ^ {2} - \rho^ {2}} = \sqrt {4 ^ {2} - (- \frac {3}{\sqrt {2 1}}) ^ {2}} = \sqrt {\frac {1 0 9}{7}} \approx 3. 9 4 6. \\ \end{array}$$ $$c = O + \rho \cdot \frac {\vec {n}}{| | \vec {n} | |} = (0, 0, 0) + \left(- \frac {3}{\sqrt {2 1}}\right) \cdot \frac {(2 , - 1 , 4)}{\sqrt {2 ^ {2} + (- 1) ^ {2} + 4 ^ {2}}} = \left(- \frac {2}{7}, \frac {1}{7}, - \frac {4}{7}\right).$$

Answer: $\sqrt{\frac{109}{7}}, \left(-\frac{2}{7}, \frac{1}{7}, -\frac{4}{7}\right)$ .

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