Question

Find the radius and the center of the circular section of the sphere jrj = 26 cut off
by the plane r
(2i+6j+3k) = 70.

Answer 1

Answer on Question #79920 – Math – Analytic Geometry

Question

Find the radius and the center of the circular section of the sphere $|r| = 26$ cut off by the plane

$$r \cdot (2i + 6j + 3k) = 70$$

Solution

Let

$$\vec{n} = (2, 6, 3)$$ $$r_0 = 26$$ $$c_0 = 70$$

Then

$$\langle r, \vec{n} \rangle = \langle \lambda \vec{n}, \vec{n} \rangle = c_0$$ $$\lambda = \frac{c_0}{|\vec{n}|^2}$$ $$|\vec{n}|^2 = 4 + 36 + 9 = 49$$ $$\lambda = \frac{70}{49} = \frac{10}{7}$$

The center of the circular section

$$p_c = \frac{c_0}{|\vec{n}|^2} \vec{n} = \frac{10}{7} \cdot (2, 6, 3) = \left(\frac{20}{7}, \frac{60}{7}, \frac{30}{7}\right)$$

The radius of the circular section

$$r_c = \sqrt{|p_c| (r_0 - |p_c|)}$$ $$|p_c| = \frac{c_0}{|\vec{n}|} = \frac{70}{7} = 10$$ $$r_c = \sqrt{10 \cdot (26 - 10)} = \sqrt{160} = 4\sqrt{10}$$

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