Reflection Across a Sphere

This is a higher-dimensional version of the previous problem.

Known parameters:

$(x_0,y_0,z_0) = (0,0,2) \\ (A_x,A_y,A_z) = (5,3,0) \\ (B_x,B_y,B_z) = (3,5,5) \\ R = 2 \\ R_B = \sqrt{B_x^2 + B_y^2 + B_z^2}$

Coordinates of point $C$ :

$C_x = x_0 + R \, \cos \theta \, \sin \phi \\ C_y = y_0 + R \, \sin \theta \, \sin \phi \\ C_z = z_0 + R \, \cos \phi$

Incident vector from $A$ to $C$ :

$v_x = C_x - A_x \\ v_y = C_y - A_y \\ v_z = C_z - A_z$

Normal vector to sphere:

$N_x = \cos \theta \, \sin \phi \\ N_y = \sin \theta \, \sin \phi \\ N_z = \cos \phi$

Also calculate two tangent vectors to the sphere. Find $(T_{1x},T_{1y},T_{1z})$ such that the dot product of $\vec{N}$ and $\vec{T}_1$ is zero. Then find $(T_{2x},T_{2y},T_{2z})$ by taking the cross product of $\vec{N}$ and $\vec{T}_1$ .

Express $\vec{v}$ in terms of the normal and tangent vectors:

$v_x = \alpha \, N_x + \beta \, T_{1x} + \gamma \, T_{2x} \\ v_y = \alpha \, N_y + \beta \, T_{1y} + \gamma \, T_{2y} \\ v_z = \alpha \, N_z + \beta \, T_{1z} + \gamma \, T_{2z}$

Solve the system for $(\alpha,\beta, \gamma)$ and construct the reflected vector $\vec{v}_2$ :

$v_{2x} = -\alpha \, N_x + \beta \, T_{1x} + \gamma \, T_{2x} \\ v_{2y} = -\alpha \, N_y + \beta \, T_{1y} + \gamma \, T_{2y} \\ v_{2z} = -\alpha \, N_z + \beta \, T_{1z} + \gamma \, T_{2z}$

Trace the reflected ray from point $C$ until it intersects the sphere on which Point $B$ lies. Determine the travel distance until intersection.

$(C_x + \sigma \, v_{2x})^2 + (C_y + \sigma \, v_{2y})^2 + (C_z + \sigma \, v_{2z})^2 = R_B^2$

Solve the quadratic for the positive value of $\sigma$ .

Intersection point:

$x_i = C_x + \sigma \, v_{2x} \\ y_i = C_y + \sigma \, v_{2y} \\ z_i = C_z + \sigma \, v_{2z}$

Sweep the parameters $\theta$ and $\phi$ and store the values of $\theta$ and $\phi$ for which the following are true.

$x_i = B_x \\ y_i = B_y \\ z_i = B_z$

This results in:

$\theta \approx 44.376^\circ \\ \phi \approx 86.046^\circ \\ C_x \approx 1.426 \\ C_y \approx 1.395 \\ C_z \approx 2.138$

Using the reflection law, the incident ray, the reflected ray, and the normal to the surface of the sphere, lie in the same plane. Now, the normal to the sphere at any point passes through its center. Therefore, the plane of reflection passes through points $A$ , $B$ , and the center $O = (0, 0, 2 )$ . Hence the problem now is exactly the same as the 2D version featured here .

With some bookkeeping, the results of the 2D version apply here as well. First define the vector

$OA = A - O = (5, 3, 0) - (0, 0, 2) = (5, 3, -2)$ and $OB = B - O = (3, 5, 5) - (0, 0, 2) = (3, 5, 3)$ . In addition define the constants $a = | OA | = \sqrt{ 5^2 + 3^2 + (-2)^2 } = \sqrt{ 38 }$ and $b = | OB | = \sqrt{ 3^2 + 5^2 + 3^2 } = \sqrt{ 43 }$ , while the radius of the sphere is $r = 2$ .

Also, let $\phi$ be the angle between the two vectors $OA$ and $OB$ , which can computed using the dot product, as follows,

$\phi = \cos^{-1} \left( \dfrac{OA \cdot OB }{ |OA| |OB| } \right)$

Now using the results the 2D version, the angle $\theta$ between $OC$ and $OA$ , satisifes the following equation,

$(b r \sin \phi ) \cos \theta + (- a r - b r \cos \phi) \sin \theta + (-a b \sin \phi )\cos 2 \theta + (a b \cos \phi) \sin 2 \theta = 0$

The solution of this equation was studied in this problem , and we select the solution $\theta$ that satisfies $0 \lt \theta \lt \phi$ . Now we have to find the point $C$ which lies on the surface of the sphere. To that end, we'll define a unit vector along $OA$

$u_1 = \dfrac{ OA }{|OA | }$

Next, we'll find the normal to the reflection plane, and normalize it

$n = \dfrac{ OA \times OB } { | OA \times OB | }$

Finally, well find the perpendicular unit vector to $u_1$ and $n$ ,

$u_2 = n \times u_1$

Now, we can express the point $C$ as,

$C = O + r ( \cos \theta u_1 + \sin \theta u_2 )$

And when we do that, we find that $C = (1.425886 , 1.395591, 2.138474 )$ , making the answer, $\approx 4.960$