Question

Find the radius and the center of the circular section of the sphere |r| = 17 cut off by
the plane r . (i + 2j + 2k) = 24.

Answer 1

Answer on Question #68960 – Math – Linear Algebra

Question

Find the radius and the center of the circular section of the sphere $|r| = 17$ cut off by the plane $r \cdot (i + 2j + 2k) = 24$.

Solution

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

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

$$\rho = \frac {\overline {P 0} \cdot \vec {n}}{| | \vec {n} | |} = \frac {D}{\sqrt {A ^ {2} + B ^ {2} + C ^ {2}}}.$$

The intersection $S \cap \Pi$ 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

$$C = O + \rho \cdot \frac {\vec {n}}{| | \vec {n} | |} = \rho \cdot \frac {(A , B , C)}{\sqrt {A ^ {2} + B ^ {2} + C ^ {2}}}.$$

In our case:

$$\begin{array}{l} | r | = 17; x + 2 y + 2 z = 24 \\ S = \{(x, y, z): x ^ {2} + y ^ {2} + z ^ {2} = 289 \}, P = \{(x, y, z): x + 2 y + 2 z = 24 \}. \\ \rho = \frac {24}{\sqrt {1 ^ {2} + 2 ^ {2} + 2 ^ {2}}} = \frac {24}{3}. \\ r _ {c} = \sqrt {R ^ {2} - \rho^ {2}} = \sqrt {17 ^ {2} - \left(\frac {24}{3}\right) ^ {2}} = 15. \\ C = \frac {24}{3} \cdot \frac {(1 , 2 , 2)}{\sqrt {1 ^ {2} + 2 ^ {2} + 2 ^ {2}}} = \left(\frac {24}{9}, \frac {48}{9}, \frac {48}{9}\right). \\ \end{array}$$

Answer: 15; $\left(\frac{24}{9}, \frac{48}{9}, \frac{48}{9}\right)$.

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