Eelipsoid and line intersection

  • Thread starter Thread starter ppmko
  • Start date Start date
  • Tags Tags
    Intersection Line
Click For Summary
SUMMARY

The discussion focuses on determining the intersection point P(x,y,z) of a line extending from the center of an ellipsoid to a point A(1,2,3) located inside the ellipsoid. The ellipsoid is defined by the equation x²/a² + y²/b² + z²/c² = 1, with the user initially using a=1, b=2, and c=3, which incorrectly places point A outside the ellipsoid. The correct approach requires selecting values for a, b, and c that are greater than the coordinates of point A to ensure it lies within the ellipsoid. The user is advised to adjust the axes of the ellipsoid accordingly to achieve the desired configuration.

PREREQUISITES
  • Understanding of ellipsoid equations and their parameters (a, b, c).
  • Familiarity with parametric equations of lines in 3D space.
  • Knowledge of distance formulas in three-dimensional geometry.
  • Ability to solve quadratic equations and interpret their solutions.
NEXT STEPS
  • Learn how to derive the parametric equations of a line in 3D space.
  • Study the properties of ellipsoids and how to manipulate their equations.
  • Explore methods for determining points of intersection between lines and surfaces.
  • Investigate the implications of changing the semi-axis lengths on the position of points relative to the ellipsoid.
USEFUL FOR

Students studying geometry, mathematicians working with ellipsoids, and anyone involved in computational geometry or 3D modeling.

ppmko
Messages
9
Reaction score
0
Homework Statement

I have an ellipsoid with center (000). There is a point A inside the ellipsoid with known coordinates(1,2,3) I draw a line from center to point A and extend it to cut the ellipsoid on on point p(x,y,z).




Homework Equations



I want to find the coordinates of point P(x,y,z)


The Attempt at a Solution




The equation of ellipsoid for p is x^2/a^2+y^2/b^2+z^2/c^2=1
i have the values of a,b and c
i want to know if the ellipsoid equation is applicable to coordinates of A and coordinates of P

and how can i create equation using the coordinates of p with the coordinates of A
by equation of line method as both the points lie on a straight line with one end on (000) as the third point.
 
Physics news on Phys.org
Express the line through (0,0,0) and (1,2,3) in parametric form, i.e. x=t, y=2t, z=3t. Put that into the ellipsoid equation. Solve for t.
 
Thank you very much . I tried that method but the problem I am facing is say my point A (1,2,3) is inside the ellipsoid and center(000) and the point on the ellipsoid i solve using
x^2/a2+y^2/b2+Z^2/c^2=1 by substituting x=t,y=2t,z=3t and solving for t. But the distance between center and point (x,y,z) should be equal to the sum of the dist between center and A and A and point(x,y,z) . that is not matching .i am using the formula for dist between two pints say(x1,y1,z1) and (x2,y2,z2) as sqrt(x2-x1)^2+(y2-y1)^2+(z2-z1)^2.
can you tell me where i am going wrong
 
ppmko said:
Thank you very much . I tried that method but the problem I am facing is say my point A (1,2,3) is inside the ellipsoid and center(000) and the point on the ellipsoid i solve using
x^2/a2+y^2/b2+Z^2/c^2=1 by substituting x=t,y=2t,z=3t and solving for t. But the distance between center and point (x,y,z) should be equal to the sum of the dist between center and A and A and point(x,y,z) . that is not matching .i am using the formula for dist between two pints say(x1,y1,z1) and (x2,y2,z2) as sqrt(x2-x1)^2+(y2-y1)^2+(z2-z1)^2.
can you tell me where i am going wrong

What did you use for a, b and c, what did you get for t and hence for (x,y,z)? Of course you should find that the distance from O to A plus the distance from A to (x,y,z) should equal the distance from O to (x,y,z). But it's impossible to say what you are doing wrong until you tell us what you did.
 
Thank you for your responce.
I used a=1
b=2
c=3
for coordinates of A(1,2,3) and center(0,0,0)

now my equaation becomes for point p(x,y,z) on ellipsoid
x=0+(1-0)t=t
y=0+(2-0)t=2t
z=0+(3-0)t=3t

put in ellipsoid equation
t^2/1+4t^2/4+9t^2/9=1

3t^2=1
t=sqrt(1/3)=.57

x=.57
y=1.15
z=1.71


now pa=sqrt[(x-1)^2+(y-2)^2+(z-3)^2]=1.603
point A and center(000)=3.74
whereas my p to center is 3.30
i cannoot understand where i am going wrong
 
I don't think your distance from p to the center is right. But more importantly, if you choose a=1, b=2, and c=3 then your selected point A=(1,2,3) is OUTSIDE of the ellipsoid. I thought you were going to put it inside.
 
i need the point A to be inside . I guess I should have given the arbitraty values for the axis as a=3 b=2 and c=1 . I hope a,b,c are the semimajor axis,semiminor axis and the z axis repectively in the ellipsoid equation . and the coordinates of A(1,2,3) are separate.
the point on the ellipsoid will have x^2/a^2+y^2/b^2+z^2/3^3=1.where the x=t,y=2t,z=3t as the line touching the point on ellipsoid pwill pass through the center(000) and A(1,2,3) located inside the ellipsoid.
I tried that too but still it is not matching.
 
Sure, a, b and c are the axes of the ellipse. But if you want A(1,2,3) to be inside the ellipse, you need to make a, b and c larger than 1, 2 and 3.
 

Similar threads

Replies
4
Views
3K
Replies
23
Views
7K
Replies
4
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
4
Views
2K