Find the coordinates of the point on the ellipsoid where the major axis meet

[ppmko]

Homework Statement

I have a point p(xp,yp,zp) inside an ellipsoid and i want to find the angle of that point from the center of the ellipsoid(xc,yc,zc) .

I also have
the major axis length 'a' ,with length ax,ay and az components

I calculated the unit vector of axis a with formula length of axis 'a"/sqrt(ax^2+ay^2+az^2).

Homework Equations

how do i find the angle at which p makes with the center of ellipsoid from the from axis a

The Attempt at a Solution

I calculated the coordinates of the point that intersect the major axis on the ellipsoid using the unit vector by calculating the xcoordinate as ax/sqrt(ax^2+ay^2+az^2) similarly for y and z coordinates
this give me 2 set of coordinates .now i have new coordinates of the point that intersects the major axis on the ellipsoid and the coordinates of p . Now i use distance formula and then use the cosine law for triangles to find the angle between the point p and the center of the ellipsoid.pls let me know if this is correct.

 

[nvn]

The coordinates of points C, A, and P are (xc,yc,zc), (xa,ya,za), and (xp,yp,zp), respectively. The length of vector CA is La = sqrt[(xa-xc)^2 + (ya-yc)^2 + (za-zc)^2]. The length of vector CP is Lp = sqrt[(xp-xc)^2 + (yp-yc)^2 + (zp-zc)^2]. Therefore, the angle between vectors CA and CP is as follows.

theta = acos{[(xa-xc)(xp-xc) + (ya-yc)(yp-yc) + (za-zc)(zp-zc)]/(La*Lp)}

 

[ppmko]

but " a" is not a point. see the coordinates of p and center are given. length of axis and components are given. I am not sure if I can consider components of vector A to be coordinates.Then I can use the cosine formula for the triangle.

 

[nvn]

Good catch. So the coordinates of points C and P are (xc,yc,zc) and (xp,yp,zp). And the components of the length of the ellipsoid semi-major axis are ax, ay, and az. We can call the semi-major axis vector CA. The coordinates of point A are therefore ((xc+ax),(yc+ay),(zc+az)). The length of vector CA is a. The length of vector CP is Lp = sqrt[(xp-xc)^2 + (yp-yc)^2 + (zp-zc)^2]. Therefore, the angle between vectors CA and CP is as follows.

theta = acos{[ax*(xp-xc) + ay*(yp-yc) + az*(zp-zc)]/(a*Lp)}