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Diameter \( \mathbb{P Q} \) where \( \mathbb{P}(-1,5,7) \) and \( Q(-2,7,3) \) Equation citte Schtere: \( (x-\text { Number })^{2}+(y-\text { Numter })^{2}+\left(z--\frac{\text { Number }}{}\right)^{2}= \) Numter

Ask by Clarke Chan. in the United States
Jan 17,2025

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Answer

The equation of the sphere is \( (x + 1.5)^2 + (y - 6)^2 + (z - 5)^2 = \frac{21}{4} \).

Solution

To find the equation of the sphere where points \( \mathbb{P}(-1, 5, 7) \) and \( \mathbb{Q}(-2, 7, 3) \) are the endpoints of its diameter, follow these steps: ### 1. Find the Center of the Sphere The center of the sphere \( \mathbb{C}(h, k, l) \) is the midpoint of the line segment joining points \( \mathbb{P} \) and \( \mathbb{Q} \). \[ h = \frac{x_P + x_Q}{2} = \frac{-1 + (-2)}{2} = \frac{-3}{2} = -1.5 \] \[ k = \frac{y_P + y_Q}{2} = \frac{5 + 7}{2} = \frac{12}{2} = 6 \] \[ l = \frac{z_P + z_Q}{2} = \frac{7 + 3}{2} = \frac{10}{2} = 5 \] So, the center of the sphere is \( \mathbb{C}(-1.5, 6, 5) \). ### 2. Calculate the Radius of the Sphere Since \( \mathbb{P} \) and \( \mathbb{Q} \) are endpoints of the diameter, the radius \( r \) is half the distance between these two points. First, find the distance \( d \) between \( \mathbb{P} \) and \( \mathbb{Q} \): \[ d = \sqrt{(x_Q - x_P)^2 + (y_Q - y_P)^2 + (z_Q - z_P)^2} \] \[ = \sqrt{(-2 - (-1))^2 + (7 - 5)^2 + (3 - 7)^2} \] \[ = \sqrt{(-1)^2 + 2^2 + (-4)^2} = \sqrt{1 + 4 + 16} = \sqrt{21} \] Thus, the radius \( r \) is: \[ r = \frac{d}{2} = \frac{\sqrt{21}}{2} \] ### 3. Formulate the Equation of the Sphere The standard equation of a sphere with center \( (h, k, l) \) and radius \( r \) is: \[ (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2 \] Substituting the values we found: \[ (x - (-1.5))^2 + (y - 6)^2 + (z - 5)^2 = \left(\frac{\sqrt{21}}{2}\right)^2 \] Simplify the equation: \[ (x + 1.5)^2 + (y - 6)^2 + (z - 5)^2 = \frac{21}{4} \] ### Final Equation \[ \boxed{(x + 1.5)^2 + (y - 6)^2 + (z - 5)^2 = \frac{21}{4}} \] Alternatively, to express the center using fractions instead of decimals: \[ \boxed{\left(x + \frac{3}{2}\right)^2 + (y - 6)^2 + (z - 5)^2 = \frac{21}{4}} \] This is the equation of the sphere with diameter endpoints at \( \mathbb{P}(-1, 5, 7) \) and \( \mathbb{Q}(-2, 7, 3) \).

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Bonus Knowledge

To find the diameter \( \mathbb{PQ} \), we first need to calculate the distance between the two points \( \mathbb{P}(-1,5,7) \) and \( \mathbb{Q}(-2,7,3) \). The distance formula in 3D is given by: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \] Plugging in our points, we get: \[ d = \sqrt{((-2) - (-1))^2 + (7 - 5)^2 + (3 - 7)^2} = \sqrt{(-1)^2 + (2)^2 + (-4)^2} = \sqrt{1 + 4 + 16} = \sqrt{21} \] Thus, the diameter \( \mathbb{PQ} \) is \( \sqrt{21} \). Now, to put it into the standard sphere equation format, we find the center of the diameter, which is the midpoint between point \( P \) and point \( Q \): \[ \text{Midpoint} = \left(\frac{-1 - 2}{2}, \frac{5 + 7}{2}, \frac{7 + 3}{2}\right) = \left(-\frac{3}{2}, 6, 5\right) \] The radius \( r \) will be half of the distance \( d \): \[ r = \frac{\sqrt{21}}{2} \] Now, inserting these into the standard sphere equation, we have: \[ \left(x + \frac{3}{2}\right)^2 + (y - 6)^2 + (z - 5)^2 = \left(\frac{\sqrt{21}}{2}\right)^2 = \frac{21}{4} \] To express it out completely: \[ \left(x + \frac{3}{2}\right)^2 + (y - 6)^2 + (z - 5)^2 = \frac{21}{4} \] --- Understanding spheres does not just belong in math class! The concept of a sphere has fascinating historical roots tracing back to ancient civilizations, with the Greeks pondering the very definition of space and shape. Think about the philosophers and mathematicians like Euclid and Archimedes; they not only studied spheres but also the universe's geometry, laying the groundwork for both mathematics and astronomy. In practical applications, spheres are everywhere! You might not think about it, but from sports balls to bubbles and even planets, they represent a perfect symmetry and efficiency in volume. Engineers often use spherical designs in various fields, like aerodynamics and structural integrity, allowing for optimal performance in everything from airplane wings to pressure vessels. So next time you kick a soccer ball, remember you’re engaging with a shape rich in history and engineering prowess!

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