Spheres and Planes Intersection

In summary, the scenario involves a 3D cuboid with planes as front, back, left, right, top and bottom, and three spheres named s1, s2, and s3. The coordinates and radius of each sphere are given. The goal is to find the intersection points of the spheres with the planes, as well as the center and radii of the circles they create on the planes. The equations of the planes and spheres are used to find these intersection points and circles. However, the projection of a sphere onto a plane is different from the intersection of a sphere with a plane. The projections are circles while the intersections can be circles or empty sets.
  • #1
sijad
7
0
Please bear with my ignorance. I will try to explain the complete scenario . I have a 3D cuboid (with planes as front, back, left, right, top and bottom) and three spheres called s1, s2 and s3. s1's center is at (2,4,5) and radius is 2. s2's center is at (-2,3,2) and radius is 1. s3's center is at (2,4,-2) and radius is 3.

Now I want to find where these spheres will intersect the planes and also find the center and radii of circles they will make on the planes. The coordinates of the cuboid walls are (bottom left, bottom right, top right, top left) as follows:

//Back
(-2, 0, -4),(2, 0, -4),(2, 3, -4),(-2, 3, -4)

//Left
(-2, 0, 4),(-2, 0, -4),(-2, 3, -4),(-2, 3, 4)

//Right
(2, 0, -4),(2, 0, 4),(2, 3, 4),(2, 3, -4)

//Front
(2, 0, 4),(-2, 0, 4),(-2, 3, 4),(2, 3, 4)

//Top
(-2, 3, -4),(2, 3, -4),(2, 3, 4),(-2, 3, 4)

//Floor
(-2, 0, 4),(2, 0, 4),(2, 0, -4),(-2, 0, -4)

I hope the question is clear enough to understand. Thank you.
 
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  • #2
sijad said:
Please bear with my ignorance. I will try to explain the complete scenario . I have a 3D cuboid (with planes as front, back, left, right, top and bottom) and three spheres called s1, s2 and s3. s1's center is at (2,4,5) and radius is 2. s2's center is at (-2,3,2) and radius is 1. s3's center is at (2,4,-2) and radius is 3.

Now I want to find where these spheres will intersect the planes and also find the center and radii of circles they will make on the planes. The coordinates of the cuboid walls are (bottom left, bottom right, top right, top left) as follows:

//Back
(-2, 0, -4),(2, 0, -4),(2, 3, -4),(-2, 3, -4)
This plane has equation z= -4. The first sphere has equation [itex](x-2)^2+ (y-4)^2+ (z- 5)^2= 4[/itex]. Setting z= 4, [itex](x- 2)^2+ (y- 4)^2+ 81= 4[/itex]

//Left
(-2, 0, 4),(-2, 0, -4),(-2, 3, -4),(-2, 3, 4)

//Right
(2, 0, -4),(2, 0, 4),(2, 3, 4),(2, 3, -4)

//Front
(2, 0, 4),(-2, 0, 4),(-2, 3, 4),(2, 3, 4)

//Top
(-2, 3, -4),(2, 3, -4),(2, 3, 4),(-2, 3, 4)

//Floor
(-2, 0, 4),(2, 0, 4),(2, 0, -4),(-2, 0, -4)

I hope the question is clear enough to understand. Thank you.
 
  • #3
Thanks for the help. Now how should I find the projected circle from the below equation:

(x−2)^ 2 +(y−4)^ 2 +81=4

I need the circles center and also its radius on the plane. Thanks.
 
  • #4
sijad said:
Thanks for the help. Now how should I find the projected circle from the below equation:

(x−2)^ 2 +(y−4)^ 2 +81=4

I need the circles center and also its radius on the plane. Thanks.
There are no real solutions to that equation.

What do you think that means?
 
  • #5
I know the planes' coordinates, so I should be able to find the projection of the sphere on these planes (the circles that the sphere creates on the planes on intersection).
 
  • #6
"The circles that the sphere creates on the planes on intersection" are NOT the same as "the projection of the sphere on these planes". They are, as I said before, given by putting the equation of the plane into the equation of the sphere and finding which point are on the plane and the sphere- that is, the equation of the circle of intereserction. And SammyS has told you that the first equation, using the first plane, does NOT have any real solution. What does that tell you about the circle of intersection?

Do you not know the general equation of a circle? It is [itex](x- a)^2+ (y- b)^2= R^2[/itex] where (a, b) is the center and R is the radius.
 
  • #7
Thank you all for all the help so far. To be honest I am now more confused than I was before, as you are now saying that the projections of the spheres are not the same as the circles created by these spheres on the planes. If you could be kind enough to point me to some online tutorial or material that could explain this to me in detail. I would really appreciate that as I don't want to ask any further questions being totally lost (the questions would only annoy you all and make me look more foolish :-) Thank you.
 
  • #8
sijad said:
Thank you all for all the help so far. To be honest I am now more confused than I was before, as you are now saying that the projections of the spheres are not the same as the circles created by these spheres on the planes. If you could be kind enough to point me to some online tutorial or material that could explain this to me in detail. I would really appreciate that as I don't want to ask any further questions being totally lost (the questions would only annoy you all and make me look more foolish :-) Thank you.
We are here to help. What tends to annoy us, and also makes those who as the questions look foolish, is when the questioner ignores the questions and advice of those who are trying to help -- merely gives responses without much thought. You don't appear to fit that description.

An example of the difference between, 1) the projection of a sphere onto a plane, and 2) the intersection of a sphere with a plane :

Consider your 3rd sphere, ( radius of 3 units, centered at (2,4,-2) ) and the coordinate planes. BTW: This sphere has the equation, (x-2)2 + (y-4)2 + (z+2)2 = 9

This sphere does not intersect the xz-plane at all, because its center is at y=4, and its radius is 3. However, the projection of this sphere onto the xz-plane is a circle with a radius of 3 units, and a center at (2,0,-2). The equation of this circle is (x-2)2 + (z+2)2 = 9, y=0 .

This sphere does intersect the other two coordinate planes. Its intersection with the xy-plane is given by (x-2)2 + (y-4)2 + (0+2)2 = 9 , which is equivalent to (x-2)2 + (y-4)2 = 5 and of course z = 0 . This circle has radius, √5, and is centered at (2,4,0) and lies in the xy-plane. The projection of the sphere onto the xy-plane is also a circle. This circle has the same center but its radius is 3.

BTW: The projection of the sphere onto any plane is a circle of radius 3.

I hope this helps.
 

FAQ: Spheres and Planes Intersection

What is the concept of Spheres and Planes Intersection?

Spheres and Planes Intersection is a mathematical concept that involves the intersection of a three-dimensional sphere and a two-dimensional plane. It is used to determine the points where these two geometric shapes intersect.

What is the formula for finding the intersection of a sphere and a plane?

The formula for finding the intersection of a sphere and a plane is (x - a)^2 + (y - b)^2 + (z - c)^2 = r^2, where (a, b, c) are the coordinates of the center of the sphere and r is its radius. The plane is represented by the equation Ax + By + Cz + D = 0, where A, B, and C are the coefficients of the plane's normal vector and D is the distance of the plane from the origin.

How many points of intersection can a sphere and plane have?

A sphere and a plane can have either 0, 1, or 2 points of intersection. If the plane is tangent to the sphere, there will be only one point of intersection. If the plane does not intersect the sphere, there will be no points of intersection. And if the plane intersects the sphere at two different points, there will be two points of intersection.

What is the relationship between the center of the sphere and the point of intersection with the plane?

The point of intersection between a sphere and a plane will always lie on the line that connects the center of the sphere and the closest point on the plane to the center of the sphere. This means that the point of intersection will be equidistant from the center of the sphere and the plane.

How is the intersection of a sphere and a plane used in real-life applications?

The concept of Spheres and Planes Intersection has various applications in fields such as engineering, computer graphics, and physics. It is used to determine the path of a projectile, the position of a satellite, and the reflection of light off a spherical object. It is also used in computer algorithms for rendering 3D objects and for designing architectural structures.

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