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Calculate the spherical intersection location
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Having the values (x1 y1 z1), Based on the figure below, I need to find the (x2 y2 z2)
O1 =0 0 0
r1 = R1
O2 = 0 0 L
r2 = R2
analytic-geometry
edited Jul 31 at 8:08
amd
25.7k2943
asked Jul 31 at 7:36
alex
1062
add a comment |Â
up vote
1
down vote
favorite
Having the values (x1 y1 z1), Based on the figure below, I need to find the (x2 y2 z2)
O1 =0 0 0
r1 = R1
O2 = 0 0 L
r2 = R2
analytic-geometry
edited Jul 31 at 8:08
amd
25.7k2943
asked Jul 31 at 7:36
alex
1062
You could start by considering an easier, 2D case. The $z$-direction is not taken into consideration. From the equation of the line, you know that $$ fracx_2y_2 = fracx_1y_2 $$ And then, from the definition of the sphere (with radius $r_2$), $$ x_2^2 + (y_2 - r_2)^2 = r^2 $$ From the two equations you can solve $x_2$ and $y_2$. Then expand the logic to three dimensions.
– Matti P.
Jul 31 at 8:10
@MattiP. thanks, but i don't understand second equation. would you please elaborate more?
– alex
Jul 31 at 8:28
It's the equation of the circle. What is the difficulty you're facing?
– Matti P.
Jul 31 at 8:32
@alex What do you really want? The intersections between the two spheres and the line (light blue)? For any line passing by the origin?
– Cesareo
Jul 31 at 9:13
@Cesareo , yes ,(x1,y1,z1) is intersection between big sphere and blue line that i know the values , and looking for (x2,y2,z2) coordinate that is the intersection between small sphere and blue line
– alex
Jul 31 at 9:58
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Having the values (x1 y1 z1), Based on the figure below, I need to find the (x2 y2 z2)
O1 =0 0 0
r1 = R1
O2 = 0 0 L
r2 = R2
analytic-geometry
edited Jul 31 at 8:08
amd
25.7k2943
asked Jul 31 at 7:36
alex
1062
Having the values (x1 y1 z1), Based on the figure below, I need to find the (x2 y2 z2)
O1 =0 0 0
r1 = R1
O2 = 0 0 L
r2 = R2
analytic-geometry
edited Jul 31 at 8:08
amd
25.7k2943
asked Jul 31 at 7:36
alex
1062
edited Jul 31 at 8:08
amd
25.7k2943
edited Jul 31 at 8:08
amd
25.7k2943
edited Jul 31 at 8:08
amd
25.7k2943
25.7k2943
asked Jul 31 at 7:36
alex
1062
asked Jul 31 at 7:36
alex
1062
asked Jul 31 at 7:36
alex
1062
1062
You could start by considering an easier, 2D case. The $z$-direction is not taken into consideration. From the equation of the line, you know that $$ fracx_2y_2 = fracx_1y_2 $$ And then, from the definition of the sphere (with radius $r_2$), $$ x_2^2 + (y_2 - r_2)^2 = r^2 $$ From the two equations you can solve $x_2$ and $y_2$. Then expand the logic to three dimensions.
– Matti P.
Jul 31 at 8:10
@MattiP. thanks, but i don't understand second equation. would you please elaborate more?
– alex
Jul 31 at 8:28
It's the equation of the circle. What is the difficulty you're facing?
– Matti P.
Jul 31 at 8:32
@alex What do you really want? The intersections between the two spheres and the line (light blue)? For any line passing by the origin?
– Cesareo
Jul 31 at 9:13
@Cesareo , yes ,(x1,y1,z1) is intersection between big sphere and blue line that i know the values , and looking for (x2,y2,z2) coordinate that is the intersection between small sphere and blue line
– alex
Jul 31 at 9:58
add a comment |Â
You could start by considering an easier, 2D case. The $z$-direction is not taken into consideration. From the equation of the line, you know that $$ fracx_2y_2 = fracx_1y_2 $$ And then, from the definition of the sphere (with radius $r_2$), $$ x_2^2 + (y_2 - r_2)^2 = r^2 $$ From the two equations you can solve $x_2$ and $y_2$. Then expand the logic to three dimensions.
– Matti P.
Jul 31 at 8:10
@MattiP. thanks, but i don't understand second equation. would you please elaborate more?
– alex
Jul 31 at 8:28
It's the equation of the circle. What is the difficulty you're facing?
– Matti P.
Jul 31 at 8:32
@alex What do you really want? The intersections between the two spheres and the line (light blue)? For any line passing by the origin?
– Cesareo
Jul 31 at 9:13
@Cesareo , yes ,(x1,y1,z1) is intersection between big sphere and blue line that i know the values , and looking for (x2,y2,z2) coordinate that is the intersection between small sphere and blue line
– alex
Jul 31 at 9:58
You could start by considering an easier, 2D case. The $z$-direction is not taken into consideration. From the equation of the line, you know that $$ fracx_2y_2 = fracx_1y_2 $$ And then, from the definition of the sphere (with radius $r_2$), $$ x_2^2 + (y_2 - r_2)^2 = r^2 $$ From the two equations you can solve $x_2$ and $y_2$. Then expand the logic to three dimensions.
– Matti P.
Jul 31 at 8:10
You could start by considering an easier, 2D case. The $z$-direction is not taken into consideration. From the equation of the line, you know that $$ fracx_2y_2 = fracx_1y_2 $$ And then, from the definition of the sphere (with radius $r_2$), $$ x_2^2 + (y_2 - r_2)^2 = r^2 $$ From the two equations you can solve $x_2$ and $y_2$. Then expand the logic to three dimensions.
– Matti P.
Jul 31 at 8:10
@MattiP. thanks, but i don't understand second equation. would you please elaborate more?
– alex
Jul 31 at 8:28
@MattiP. thanks, but i don't understand second equation. would you please elaborate more?
– alex
Jul 31 at 8:28
It's the equation of the circle. What is the difficulty you're facing?
– Matti P.
Jul 31 at 8:32
It's the equation of the circle. What is the difficulty you're facing?
– Matti P.
Jul 31 at 8:32
@alex What do you really want? The intersections between the two spheres and the line (light blue)? For any line passing by the origin?
– Cesareo
Jul 31 at 9:13
@alex What do you really want? The intersections between the two spheres and the line (light blue)? For any line passing by the origin?
– Cesareo
Jul 31 at 9:13
@Cesareo , yes ,(x1,y1,z1) is intersection between big sphere and blue line that i know the values , and looking for (x2,y2,z2) coordinate that is the intersection between small sphere and blue line
– alex
Jul 31 at 9:58
@Cesareo , yes ,(x1,y1,z1) is intersection between big sphere and blue line that i know the values , and looking for (x2,y2,z2) coordinate that is the intersection between small sphere and blue line
– alex
Jul 31 at 9:58
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
Given
$$
C_1to ||P-O_1|| = R_1\
C_2to ||P-O_2|| = R_2\
P_1 = (x_1,y_1,z_1) in C_1
$$
with
$$
P = (x,y,z)\
O_1 = (0,0,0)\
O_2 = (0,0,L)
$$
how to obtain $P_2 = (x_2,y_2,z_2) in C_2$ such that $O_1, P_1$ and $P_2$ are aligned ?
Choosing the line
$$
Lto P = lambda vec v
$$
with $ vec v = P_1/R_1$ it's intersections with $C_2$ are obtained by solving for $lambda$
$$
||lambdavec v-O_2|| = R_2
$$
or squaring
$$
lambda^2||vec v||^2-2lambda <vec v, O_2 > + ||O_2||^2 = R_2^2
$$
or
$$
lambda = frac< vec v, O_2 > pmsqrt < vec v, O_2 > ^2-
$$
but here $vec v = P_1/R_1$ is an unit vector so
$$
lambda = frac1R_1 < P_1, O_2 >pm sqrtO_2
$$
or
$$
lambda = fracz_1 LR_1pm sqrtfracz_1^2 L^2R_1^2-L^2+R_2^2
$$
and thus we can have solutions for $P_2$ as
$$
P_2 = left(fracz_1 LR_1pm sqrtfracz_1^2 L^2R_1^2-L^2+R_2^2right)fracP_1R_1
$$
NOTE
Here $< cdot, cdot > $ represents the scalar product of two vectors.
answered Aug 1 at 12:19
Cesareo
5,5812412
thanks,i'll test it out soon
– alex
Aug 1 at 15:36
add a comment |Â
up vote
1
down vote
Let $P_2 = (x_1, y_1, z_1)$ be the intersection of the given ray with the sphere centered at $O_2.$
Construct a triangle $triangle O_1 O_2 P_2.$
Let angle $theta = angle O_2 O_1 P_2$; this is the angle between the ray and the positive $z$ axis.
Consider the triangle formed by the points $O_1 = (0,0,0),$
$(x_1,y_1,z_1),$ and $(0,0,z_1).$
This is a right triangle with the same angle $theta$ at $O_1.$
From this right triangle, by examining the ratio of the hypotenuse to the
leg along the $z$ axis we find that
$costheta = z_1/R_1.$
Applying the cosine law to triangle $triangle O_1 O_2 P_2,$
we have
$$ R_2^2 = L^2 + r^2 - 2costheta Lr. $$
Substitute $z_1/R_1$ for $costheta$ and rearrange the terms to make this a quadratic equation over $r$:
$$ r^2 - 2fracz_1R_1 Lr + L^2 - R_2^2 = 0. $$
This can be solved by the usual formula for roots of quadratic equations:
beginalign
r &= frac12 left(2fracz_1R_1 L
pm sqrtleft(2fracz_1R_1 Lright)^2 - 4(L^2 - R_2^2)right)\
&= fracz_1R_1 L
pm sqrtleft(fracz_1R_1 Lright)^2 - L^2 + R_2^2.
endalign
Now we want a point at distance $r$ from the origin in the same direction as
$(x_1,y_1,z_1).$
That is, we want a point $r/R_1$ times as far from the origin as $(x_1,y_1,z_1).$
Based on the diagram, we want to take the positive value of the $pm$ part in the formula above;
the negative value will give us a "backward" projection to the point where
the line from $(x_1,y_1,z_1)$ to $O_1$ intersects the smaller sphere on the other side of $O_1.$
So the final result is $(kx_1,ky_1,kz_1),$
where
$$k = frac1R_1 left(fracz_1R_1 L
+ sqrtleft(fracz_1R_1 Lright)^2 - L^2 + R_2^2right). $$
answered Aug 1 at 16:29
David K
48k340107
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Given
$$
C_1to ||P-O_1|| = R_1\
C_2to ||P-O_2|| = R_2\
P_1 = (x_1,y_1,z_1) in C_1
$$
with
$$
P = (x,y,z)\
O_1 = (0,0,0)\
O_2 = (0,0,L)
$$
how to obtain $P_2 = (x_2,y_2,z_2) in C_2$ such that $O_1, P_1$ and $P_2$ are aligned ?
Choosing the line
$$
Lto P = lambda vec v
$$
with $ vec v = P_1/R_1$ it's intersections with $C_2$ are obtained by solving for $lambda$
$$
||lambdavec v-O_2|| = R_2
$$
or squaring
$$
lambda^2||vec v||^2-2lambda <vec v, O_2 > + ||O_2||^2 = R_2^2
$$
or
$$
lambda = frac< vec v, O_2 > pmsqrt < vec v, O_2 > ^2-
$$
but here $vec v = P_1/R_1$ is an unit vector so
$$
lambda = frac1R_1 < P_1, O_2 >pm sqrtO_2
$$
or
$$
lambda = fracz_1 LR_1pm sqrtfracz_1^2 L^2R_1^2-L^2+R_2^2
$$
and thus we can have solutions for $P_2$ as
$$
P_2 = left(fracz_1 LR_1pm sqrtfracz_1^2 L^2R_1^2-L^2+R_2^2right)fracP_1R_1
$$
NOTE
Here $< cdot, cdot > $ represents the scalar product of two vectors.
answered Aug 1 at 12:19
Cesareo
5,5812412
thanks,i'll test it out soon
– alex
Aug 1 at 15:36
add a comment |Â
up vote
1
down vote
Given
$$
C_1to ||P-O_1|| = R_1\
C_2to ||P-O_2|| = R_2\
P_1 = (x_1,y_1,z_1) in C_1
$$
with
$$
P = (x,y,z)\
O_1 = (0,0,0)\
O_2 = (0,0,L)
$$
how to obtain $P_2 = (x_2,y_2,z_2) in C_2$ such that $O_1, P_1$ and $P_2$ are aligned ?
Choosing the line
$$
Lto P = lambda vec v
$$
with $ vec v = P_1/R_1$ it's intersections with $C_2$ are obtained by solving for $lambda$
$$
||lambdavec v-O_2|| = R_2
$$
or squaring
$$
lambda^2||vec v||^2-2lambda <vec v, O_2 > + ||O_2||^2 = R_2^2
$$
or
$$
lambda = frac< vec v, O_2 > pmsqrt < vec v, O_2 > ^2-
$$
but here $vec v = P_1/R_1$ is an unit vector so
$$
lambda = frac1R_1 < P_1, O_2 >pm sqrtO_2
$$
or
$$
lambda = fracz_1 LR_1pm sqrtfracz_1^2 L^2R_1^2-L^2+R_2^2
$$
and thus we can have solutions for $P_2$ as
$$
P_2 = left(fracz_1 LR_1pm sqrtfracz_1^2 L^2R_1^2-L^2+R_2^2right)fracP_1R_1
$$
NOTE
Here $< cdot, cdot > $ represents the scalar product of two vectors.
answered Aug 1 at 12:19
Cesareo
5,5812412
thanks,i'll test it out soon
– alex
Aug 1 at 15:36
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Given
$$
C_1to ||P-O_1|| = R_1\
C_2to ||P-O_2|| = R_2\
P_1 = (x_1,y_1,z_1) in C_1
$$
with
$$
P = (x,y,z)\
O_1 = (0,0,0)\
O_2 = (0,0,L)
$$
how to obtain $P_2 = (x_2,y_2,z_2) in C_2$ such that $O_1, P_1$ and $P_2$ are aligned ?
Choosing the line
$$
Lto P = lambda vec v
$$
with $ vec v = P_1/R_1$ it's intersections with $C_2$ are obtained by solving for $lambda$
$$
||lambdavec v-O_2|| = R_2
$$
or squaring
$$
lambda^2||vec v||^2-2lambda <vec v, O_2 > + ||O_2||^2 = R_2^2
$$
or
$$
lambda = frac< vec v, O_2 > pmsqrt < vec v, O_2 > ^2-
$$
but here $vec v = P_1/R_1$ is an unit vector so
$$
lambda = frac1R_1 < P_1, O_2 >pm sqrtO_2
$$
or
$$
lambda = fracz_1 LR_1pm sqrtfracz_1^2 L^2R_1^2-L^2+R_2^2
$$
and thus we can have solutions for $P_2$ as
$$
P_2 = left(fracz_1 LR_1pm sqrtfracz_1^2 L^2R_1^2-L^2+R_2^2right)fracP_1R_1
$$
NOTE
Here $< cdot, cdot > $ represents the scalar product of two vectors.
answered Aug 1 at 12:19
Cesareo
5,5812412
Given
$$
C_1to ||P-O_1|| = R_1\
C_2to ||P-O_2|| = R_2\
P_1 = (x_1,y_1,z_1) in C_1
$$
with
$$
P = (x,y,z)\
O_1 = (0,0,0)\
O_2 = (0,0,L)
$$
how to obtain $P_2 = (x_2,y_2,z_2) in C_2$ such that $O_1, P_1$ and $P_2$ are aligned ?
Choosing the line
$$
Lto P = lambda vec v
$$
with $ vec v = P_1/R_1$ it's intersections with $C_2$ are obtained by solving for $lambda$
$$
||lambdavec v-O_2|| = R_2
$$
or squaring
$$
lambda^2||vec v||^2-2lambda <vec v, O_2 > + ||O_2||^2 = R_2^2
$$
or
$$
lambda = frac< vec v, O_2 > pmsqrt < vec v, O_2 > ^2-
$$
but here $vec v = P_1/R_1$ is an unit vector so
$$
lambda = frac1R_1 < P_1, O_2 >pm sqrtO_2
$$
or
$$
lambda = fracz_1 LR_1pm sqrtfracz_1^2 L^2R_1^2-L^2+R_2^2
$$
and thus we can have solutions for $P_2$ as
$$
P_2 = left(fracz_1 LR_1pm sqrtfracz_1^2 L^2R_1^2-L^2+R_2^2right)fracP_1R_1
$$
NOTE
Here $< cdot, cdot > $ represents the scalar product of two vectors.
answered Aug 1 at 12:19
Cesareo
5,5812412
edited Aug 1 at 16:07
answered Aug 1 at 12:19
Cesareo
5,5812412
answered Aug 1 at 12:19
Cesareo
5,5812412
answered Aug 1 at 12:19
Cesareo
5,5812412
5,5812412
thanks,i'll test it out soon
– alex
Aug 1 at 15:36
add a comment |Â
thanks,i'll test it out soon
– alex
Aug 1 at 15:36
thanks,i'll test it out soon
– alex
Aug 1 at 15:36
thanks,i'll test it out soon
– alex
Aug 1 at 15:36
add a comment |Â
up vote
1
down vote
Let $P_2 = (x_1, y_1, z_1)$ be the intersection of the given ray with the sphere centered at $O_2.$
Construct a triangle $triangle O_1 O_2 P_2.$
Let angle $theta = angle O_2 O_1 P_2$; this is the angle between the ray and the positive $z$ axis.
Consider the triangle formed by the points $O_1 = (0,0,0),$
$(x_1,y_1,z_1),$ and $(0,0,z_1).$
This is a right triangle with the same angle $theta$ at $O_1.$
From this right triangle, by examining the ratio of the hypotenuse to the
leg along the $z$ axis we find that
$costheta = z_1/R_1.$
Applying the cosine law to triangle $triangle O_1 O_2 P_2,$
we have
$$ R_2^2 = L^2 + r^2 - 2costheta Lr. $$
Substitute $z_1/R_1$ for $costheta$ and rearrange the terms to make this a quadratic equation over $r$:
$$ r^2 - 2fracz_1R_1 Lr + L^2 - R_2^2 = 0. $$
This can be solved by the usual formula for roots of quadratic equations:
beginalign
r &= frac12 left(2fracz_1R_1 L
pm sqrtleft(2fracz_1R_1 Lright)^2 - 4(L^2 - R_2^2)right)\
&= fracz_1R_1 L
pm sqrtleft(fracz_1R_1 Lright)^2 - L^2 + R_2^2.
endalign
Now we want a point at distance $r$ from the origin in the same direction as
$(x_1,y_1,z_1).$
That is, we want a point $r/R_1$ times as far from the origin as $(x_1,y_1,z_1).$
Based on the diagram, we want to take the positive value of the $pm$ part in the formula above;
the negative value will give us a "backward" projection to the point where
the line from $(x_1,y_1,z_1)$ to $O_1$ intersects the smaller sphere on the other side of $O_1.$
So the final result is $(kx_1,ky_1,kz_1),$
where
$$k = frac1R_1 left(fracz_1R_1 L
+ sqrtleft(fracz_1R_1 Lright)^2 - L^2 + R_2^2right). $$
answered Aug 1 at 16:29
David K
48k340107
add a comment |Â
up vote
1
down vote
Let $P_2 = (x_1, y_1, z_1)$ be the intersection of the given ray with the sphere centered at $O_2.$
Construct a triangle $triangle O_1 O_2 P_2.$
Let angle $theta = angle O_2 O_1 P_2$; this is the angle between the ray and the positive $z$ axis.
Consider the triangle formed by the points $O_1 = (0,0,0),$
$(x_1,y_1,z_1),$ and $(0,0,z_1).$
This is a right triangle with the same angle $theta$ at $O_1.$
From this right triangle, by examining the ratio of the hypotenuse to the
leg along the $z$ axis we find that
$costheta = z_1/R_1.$
Applying the cosine law to triangle $triangle O_1 O_2 P_2,$
we have
$$ R_2^2 = L^2 + r^2 - 2costheta Lr. $$
Substitute $z_1/R_1$ for $costheta$ and rearrange the terms to make this a quadratic equation over $r$:
$$ r^2 - 2fracz_1R_1 Lr + L^2 - R_2^2 = 0. $$
This can be solved by the usual formula for roots of quadratic equations:
beginalign
r &= frac12 left(2fracz_1R_1 L
pm sqrtleft(2fracz_1R_1 Lright)^2 - 4(L^2 - R_2^2)right)\
&= fracz_1R_1 L
pm sqrtleft(fracz_1R_1 Lright)^2 - L^2 + R_2^2.
endalign
Now we want a point at distance $r$ from the origin in the same direction as
$(x_1,y_1,z_1).$
That is, we want a point $r/R_1$ times as far from the origin as $(x_1,y_1,z_1).$
Based on the diagram, we want to take the positive value of the $pm$ part in the formula above;
the negative value will give us a "backward" projection to the point where
the line from $(x_1,y_1,z_1)$ to $O_1$ intersects the smaller sphere on the other side of $O_1.$
So the final result is $(kx_1,ky_1,kz_1),$
where
$$k = frac1R_1 left(fracz_1R_1 L
+ sqrtleft(fracz_1R_1 Lright)^2 - L^2 + R_2^2right). $$
answered Aug 1 at 16:29
David K
48k340107
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let $P_2 = (x_1, y_1, z_1)$ be the intersection of the given ray with the sphere centered at $O_2.$
Construct a triangle $triangle O_1 O_2 P_2.$
Let angle $theta = angle O_2 O_1 P_2$; this is the angle between the ray and the positive $z$ axis.
Consider the triangle formed by the points $O_1 = (0,0,0),$
$(x_1,y_1,z_1),$ and $(0,0,z_1).$
This is a right triangle with the same angle $theta$ at $O_1.$
From this right triangle, by examining the ratio of the hypotenuse to the
leg along the $z$ axis we find that
$costheta = z_1/R_1.$
Applying the cosine law to triangle $triangle O_1 O_2 P_2,$
we have
$$ R_2^2 = L^2 + r^2 - 2costheta Lr. $$
Substitute $z_1/R_1$ for $costheta$ and rearrange the terms to make this a quadratic equation over $r$:
$$ r^2 - 2fracz_1R_1 Lr + L^2 - R_2^2 = 0. $$
This can be solved by the usual formula for roots of quadratic equations:
beginalign
r &= frac12 left(2fracz_1R_1 L
pm sqrtleft(2fracz_1R_1 Lright)^2 - 4(L^2 - R_2^2)right)\
&= fracz_1R_1 L
pm sqrtleft(fracz_1R_1 Lright)^2 - L^2 + R_2^2.
endalign
Now we want a point at distance $r$ from the origin in the same direction as
$(x_1,y_1,z_1).$
That is, we want a point $r/R_1$ times as far from the origin as $(x_1,y_1,z_1).$
Based on the diagram, we want to take the positive value of the $pm$ part in the formula above;
the negative value will give us a "backward" projection to the point where
the line from $(x_1,y_1,z_1)$ to $O_1$ intersects the smaller sphere on the other side of $O_1.$
So the final result is $(kx_1,ky_1,kz_1),$
where
$$k = frac1R_1 left(fracz_1R_1 L
+ sqrtleft(fracz_1R_1 Lright)^2 - L^2 + R_2^2right). $$
answered Aug 1 at 16:29
David K
48k340107
Let $P_2 = (x_1, y_1, z_1)$ be the intersection of the given ray with the sphere centered at $O_2.$
Construct a triangle $triangle O_1 O_2 P_2.$
Let angle $theta = angle O_2 O_1 P_2$; this is the angle between the ray and the positive $z$ axis.
Consider the triangle formed by the points $O_1 = (0,0,0),$
$(x_1,y_1,z_1),$ and $(0,0,z_1).$
This is a right triangle with the same angle $theta$ at $O_1.$
From this right triangle, by examining the ratio of the hypotenuse to the
leg along the $z$ axis we find that
$costheta = z_1/R_1.$
Applying the cosine law to triangle $triangle O_1 O_2 P_2,$
we have
$$ R_2^2 = L^2 + r^2 - 2costheta Lr. $$
Substitute $z_1/R_1$ for $costheta$ and rearrange the terms to make this a quadratic equation over $r$:
$$ r^2 - 2fracz_1R_1 Lr + L^2 - R_2^2 = 0. $$
This can be solved by the usual formula for roots of quadratic equations:
beginalign
r &= frac12 left(2fracz_1R_1 L
pm sqrtleft(2fracz_1R_1 Lright)^2 - 4(L^2 - R_2^2)right)\
&= fracz_1R_1 L
pm sqrtleft(fracz_1R_1 Lright)^2 - L^2 + R_2^2.
endalign
Now we want a point at distance $r$ from the origin in the same direction as
$(x_1,y_1,z_1).$
That is, we want a point $r/R_1$ times as far from the origin as $(x_1,y_1,z_1).$
Based on the diagram, we want to take the positive value of the $pm$ part in the formula above;
the negative value will give us a "backward" projection to the point where
the line from $(x_1,y_1,z_1)$ to $O_1$ intersects the smaller sphere on the other side of $O_1.$
So the final result is $(kx_1,ky_1,kz_1),$
where
$$k = frac1R_1 left(fracz_1R_1 L
+ sqrtleft(fracz_1R_1 Lright)^2 - L^2 + R_2^2right). $$
answered Aug 1 at 16:29
David K
48k340107
answered Aug 1 at 16:29
David K
48k340107
answered Aug 1 at 16:29
David K
48k340107
answered Aug 1 at 16:29
David K
48k340107
48k340107
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You could start by considering an easier, 2D case. The $z$-direction is not taken into consideration. From the equation of the line, you know that $$ fracx_2y_2 = fracx_1y_2 $$ And then, from the definition of the sphere (with radius $r_2$), $$ x_2^2 + (y_2 - r_2)^2 = r^2 $$ From the two equations you can solve $x_2$ and $y_2$. Then expand the logic to three dimensions.
– Matti P.
Jul 31 at 8:10
@MattiP. thanks, but i don't understand second equation. would you please elaborate more?
– alex
Jul 31 at 8:28
It's the equation of the circle. What is the difficulty you're facing?
– Matti P.
Jul 31 at 8:32
@alex What do you really want? The intersections between the two spheres and the line (light blue)? For any line passing by the origin?
– Cesareo
Jul 31 at 9:13
@Cesareo , yes ,(x1,y1,z1) is intersection between big sphere and blue line that i know the values , and looking for (x2,y2,z2) coordinate that is the intersection between small sphere and blue line
– alex
Jul 31 at 9:58