- #1
marutkpadhy
- 9
- 0
Find the radius of the circular section of the sphere of the sphere x^2 + y^2 + z^2 = 49 by the plane 2x+3y-z-5 \sqrt{14}= 0
MHB Calculating Radius of Circular Section from Sphere and Plane Intersection
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SUMMARY
The discussion focuses on calculating the radius of the circular section formed by the intersection of the sphere defined by the equation x² + y² + z² = 49 and the plane given by 2x + 3y - z - 5√14 = 0. The method proposed involves determining the distance from the center of the sphere to the plane using the formula d = |Ax₀ + By₀ + Cz₀ + D| / √(A² + B² + C²). Once the distance is established, the radius of the circular section can be derived using the right triangle formed by the radius of the sphere and the distance to the plane.
PREREQUISITES- Understanding of sphere equations in three-dimensional geometry
- Familiarity with plane equations and their geometric implications
- Knowledge of distance formulas in analytical geometry
- Basic trigonometry to comprehend right triangle relationships
- Study the derivation of the distance formula from a point to a plane
- Explore the geometric interpretation of sphere and plane intersections
- Learn about right triangle properties in three-dimensional space
- Investigate applications of circular sections in physics and engineering
Students and professionals in mathematics, physics, and engineering fields who are involved in geometric calculations, particularly those focusing on three-dimensional shapes and their intersections.
- #2
MarkFL
Gold Member
MHB
- 13,284
- 12
Suppose you solve the equation of the plane for $z$, and then substitute for $z$ into the equation of the sphere...what do you get?
- #3
Evgeny.Makarov
Gold Member
MHB
- 2,434
- 4
View attachment 2708
I think it is easier to find the distance $d$ from the center of the sphere to the plane (recall the the distance from $(x_0,y_0,z_0)$ to $Ax+By+Cz+D=0$ is $\dfrac{Ax_0+By_y+Cz_0+D}{\sqrt{A^2+B^2+C^2}}$) and then find the radius $r$ of the required circular section from the right triangle.
I think it is easier to find the distance $d$ from the center of the sphere to the plane (recall the the distance from $(x_0,y_0,z_0)$ to $Ax+By+Cz+D=0$ is $\dfrac{Ax_0+By_y+Cz_0+D}{\sqrt{A^2+B^2+C^2}}$) and then find the radius $r$ of the required circular section from the right triangle.
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