Question

Imagine three unit spheres (radius equal to 1 ) with centers at \(O(0,0,0), P(\sqrt{3},-1,0)\) and \(Q(\sqrt{3}, 1,0) .\) Now place another unit sphere symmetrically on top of these spheres with its center at \(R\) (see figure). a. Find the coordinates of \(R\). (Hint: The distance between the centers of any two spheres is 2.) b. Let \(\mathbf{r}_{i j}\) be the vector from the center of sphere \(i\) to the center of sphere \(j .\) Find \(\mathbf{r}_{O P}, \mathbf{r}_{O Q}, \mathbf{r}_{P Q}, \mathbf{r}_{O R},\) and \(\mathbf{r}_{P R}\).

Short answer

Answer: The coordinates of point R are (0, 0, -2/3), and the position vectors between the centers of the spheres are: \(\mathbf{r}_{OP} = (\sqrt{3}, -1, 0)\) \(\mathbf{r}_{OQ} = (\sqrt{3}, 1, 0)\) \(\mathbf{r}_{PQ} = (0, 2, 0)\) \(\mathbf{r}_{OR} = (0, 0, -\frac{2}{3})\) \(\mathbf{r}_{PR} = (-\sqrt{3}, 1, -\frac{2}{3})\)

Step-by-step solution

a. Finding the coordinates of Point R

First, we need to find the coordinates of R, which is the center of the fourth sphere placed symmetrically on top of the other three spheres. The symmetry of the four spheres forms a tetrahedron with equal side lengths. Since the distance between any two centers of the spheres is given to be 2, the height of this tetrahedron is given by the formula: \(H_{tetrahedron} = \frac{\sqrt{2^2 - \frac{2^2}{3}}}{\sqrt{2}} = \frac{\sqrt{4 - \frac{4}{3}}}{\sqrt{2}} = \frac{\sqrt{\frac{8}{3}}}{\sqrt{2}} = \sqrt{\frac{4}{3}}\) Now we can find the z-coordinate of point R, as it lies at the height of the tetrahedron: \(z_R = -\frac{H_{tetrahedron}}{2} = -\frac{\sqrt{\frac{4}{3}}}{2} = -\frac{\sqrt{4}}{3} = -\frac{2}{3}\) Since R lies on the plane of symmetry between points P and Q, its x and y coordinates are the average of the coordinates of P and Q: \(x_R = \frac{x_P + x_Q}{2} = \frac{\sqrt{3} - \sqrt{3}}{2} = 0\) \(y_R = \frac{y_P + y_Q}{2} = \frac{-1+1}{2} = 0\) Hence, the coordinates of point R are: \(R(0, 0, -\frac{2}{3})\)

b. Finding position vectors between sphere centers

Now we can find the position vectors requested: \(\mathbf{r}_{OP} = P - O = (\sqrt{3}-0, -1-0, 0-0) = (\sqrt{3}, -1, 0)\) \(\mathbf{r}_{OQ} = Q - O = (\sqrt{3}-0, 1-0, 0-0) = (\sqrt{3}, 1, 0)\) \(\mathbf{r}_{PQ} = Q - P = (\sqrt{3}-\sqrt{3}, 1-(-1), 0-0) = (0, 2, 0)\) \(\mathbf{r}_{OR} = R - O = (0-0, 0-0, -\frac{2}{3}-0) = (0, 0, -\frac{2}{3})\) \(\mathbf{r}_{PR} = R - P = (0-\sqrt{3}, 0-(-1), -\frac{2}{3}-0) = (-\sqrt{3}, 1, -\frac{2}{3})\) The position vectors are: \(\mathbf{r}_{OP} = (\sqrt{3}, -1, 0)\) \(\mathbf{r}_{OQ} = (\sqrt{3}, 1, 0)\) \(\mathbf{r}_{PQ} = (0, 2, 0)\) \(\mathbf{r}_{OR} = (0, 0, -\frac{2}{3})\) \(\mathbf{r}_{PR} = (-\sqrt{3}, 1, -\frac{2}{3})\)

Key concepts

Position Vectors

Position vectors are crucial in vector geometry because they help us define the location of a point in space relative to a specific origin. Imagine you're standing at the origin, which is a fixed point of reference in a coordinate system. A position vector will point from this origin to your desired destination in the space. It effectively acts as a roadmap, telling us how to "travel" from the origin to the object's location.

For example, to find the position vector from one sphere's center to another, we subtract the initial point's coordinates from the final point's coordinates. In our exercise, to find the position vector \( \mathbf{r}_{OP} \), we take the coordinates of point \( P(\sqrt{3},-1,0) \) and subtract those of point \( O(0,0,0) \), leading to \( (\sqrt{3}, -1, 0) \).

This subtraction essentially provides us with the directional shift and distance required to move from \( O \) to \( P \). Position vectors simplify these processes, transforming spatial relationships into manageable mathematical expressions that we can further analyze or use for calculation.

Tetrahedron

A tetrahedron might sound complex, but think of it as a pyramid with a triangular base. It's a solid shape that consists of four triangular faces. In this exercise, when we place four spheres such that their centers create a tetrahedron, we are forming a symmetrical three-dimensional shape.

The interesting part about tetrahedrons is their symmetry and equal edge lengths. In our case, each edge length equals 2, the distance between sphere centers. It forms not just any tetrahedron but a regular tetrahedron, which has consistent angles and edges.

For this tetrahedron, the height mentioned is calculated using a detailed formula derived from geometry principles, which ensures all spheres are symmetrically balanced. The height ends up being a magical point where symmetry beautifully maintains equilibrium among all spheres.

Symmetry

Symmetry in geometry is a sense of harmonious and proportional balance. It's like when patterns look the same after certain transformations such as rotations or reflections. In this problem, symmetry helps us determine the position of the fourth sphere.

Due to symmetry, we decided that the fourth sphere's center lies directly above the plane determined by the first three spheres. Thus, its x and y coordinates are midpoints of the x and y coordinates of points \( P \) and \( Q \).

• This leads us to find the coordinates \( x_R = 0 \) and \( y_R = 0 \) for the point \( R \).

Understanding symmetry not only eases complex computations but it also showcases the predictability and order in the geometric arrangements of points and constructs.

Sphere Centers

Sphere centers in this context describe the points from where each sphere extends its radius in all directions. These centers are crucial because they allow for analysis using vectors and, consequently, the determination of spatial relationships and arrangements.

In this exercise, the centers of the three initial spheres are given, and by leveraging the geometric symmetry, we calculated the center of the fourth sphere. The beauty of using 3D coordinates for these centers is it allows us to visualize and mathematically formulate how objects interact in space.

• Setting the sphere centers apart by their distances turns the task into a vector operation, making it easier to solve problems through straightforward equations.

Therefore, understanding sphere centers fosters better comprehension of its structural role and relationships in vector geometry, like finding distances or symmetries.