Eelipsoid and line intersection

  • Thread starter Thread starter ppmko
  • Start date Start date
  • Tags Tags
    Intersection Line
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.
 
Back
Top