Find the coordinates of a point in 3D space

Question

I have a sphere with radius $d$ centered at the origin. The $z$-axis is vertical. I take any point $P$ on that sphere, $P(d \cos\phi \sin\theta,\ d \sin\phi \sin\theta,\ d \cos\theta)$. I take a plane that is tangent to the sphere, and its point of tangency is at point $P$. Therefore, the equation of the plane would be $x \cos\phi \sin\theta + y \sin\phi \sin\theta + z \cos\theta = d$. Point $P_1$ lies on that same plane. Point $P$ is $r$ units away from point $P_1$. Point $P$ has the same $z$-coordinate as point $P_1$, and I am trying to find the $x$-coordinate and $y$-coordinate of point $P_1$, while still preserving the variables. Is this enough information to find these coordinates? If not, what other information do I need to provide?

Thank you in advance.

Answer 1

In the plane tangent at point $P$, point $P_1$ can be on a circle with radius $r$, centered on point $P$. You have $z=\cos\theta$ as a plane. The intersection of two planes is usually a line that belongs to both planes. Since you know that $P$ is on the line, you get a line passing through the center of the circle. Then the intersection will consist of two points, diametrically opposed. But this is the case when $\theta$ is $0$ or $\pi$. Then the tangent plane is at constant $z=\pm1$, so the solution is any point with coordinates $(r\cos\alpha,r\sin\alpha, \pm1$) is fine.

Answer 2

Let $O$ be the origin. Then $\vec{OP} = (\,\cos(\phi) \sin(\theta),\,\, d\sin(\phi) \sin(\theta),\,\, d \cos(\theta) \,)$. The vector $\vec{k} = (0, \, 0, \, 1)$ is the unit vector collinear with the $z-$axis. Then according to the definition of point $P_1$:

1. $P_1$ has the same $z-$coordinate as point $P$, i.e. $P_1$ lies on the plane $p_z$ that passes through point $P$ and is parallel to the $x, y-$coordinate plane. Equivalently, $P_1$ lies on the plane $p_z$, where $p_z$ can also be defined as the plane that passes through point $P$ and is perpendicular to vector $\vec{k}$.

2. $P_1$ lies on the plane $t_P$ tangent to the sphere at point $P$, i.e. $P_1$ lies on the plane $t_P$, which can also be defined as the plane that passes through point $P$ and is perpendicular to vector $\vec{OP}$.

Consequently, point $P_1 \, \in \, p_Z \,\cap\, t_P = l_P$, where $l_P$ is the intersection line of the planes $p_Z$ from point 1 and $t_P$ from point 2. Thus, $l_P$ is perpendicular to both vectors $\vec{k}$ and $\vec{OP}$. Therefore, their normalized cross-product vector $$\frac{\vec{k} \times \vec{OP}}{\|\vec{k} \times \vec{OP}\|}$$ is aligned with the line $l_P$. Hence, the points you are looking for are $$\vec{OP_1}_{+} = \vec{OP} \, + \, \left(\frac{r}{\|\vec{k} \times \vec{OP}\|} \right)\, \vec{k} \times \vec{OP}$$
$$\vec{OP_1}_{-} = \vec{OP} \, - \, \left(\frac{r}{\|\vec{k} \times \vec{OP}\|} \right)\, \vec{k} \times \vec{OP}$$