Equation of a Sphere with the Extremities of Diameter Being Given (Vector Form)

Understanding the equation of a sphere in vector form is an important concept in 3D geometry, especially when dealing with the extremities of a diameter. The general equation of a sphere in Cartesian coordinates is given by:

$$(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2$$

where $(h, k, l)$ is the center of the sphere and $r$ is the radius.

Vector Form of the Equation of a Sphere

In vector form, the equation of a sphere with center $\vec{C}$ and radius $r$ is given by:

$$|\vec{r} - \vec{C}|^2 = r^2$$

where $\vec{r}$ is the position vector of any point on the sphere, and $\vec{C}$ is the position vector of the center of the sphere.

Extremities of a Diameter

If we are given the extremities of a diameter of the sphere, say $\vec{A}$ and $\vec{B}$, we can find the center $\vec{C}$ of the sphere by taking the midpoint of the line segment joining $\vec{A}$ and $\vec{B}$:

$$\vec{C} = \frac{\vec{A} + \vec{B}}{2}$$

The radius $r$ of the sphere can be found by taking the distance between one of the extremities and the center:

$$r = \frac{|\vec{A} - \vec{B}|}{2}$$

Equation with Given Extremities

Using the above information, the equation of the sphere with extremities $\vec{A}$ and $\vec{B}$ is:

$$|\vec{r} - \frac{\vec{A} + \vec{B}}{2}|^2 = \left(\frac{|\vec{A} - \vec{B}|}{2}\right)^2$$

Important Points and Differences

Aspect	Description
Center of Sphere	The midpoint of the diameter's extremities.
Radius of Sphere	Half the distance between the extremities of the diameter.
Position Vector	A vector that represents the position of a point in space relative to the origin.
Equation in Cartesian	Uses coordinates $(x, y, z)$ to define the sphere.
Equation in Vector Form	Uses position vectors and is independent of the coordinate system.

Example

Let's consider an example to illustrate these concepts.

Suppose we have two points $A$ and $B$ with position vectors $\vec{A} = \begin{pmatrix} 2 \ 4 \ 6 \end{pmatrix}$ and $\vec{B} = \begin{pmatrix} 8 \ 10 \ 12 \end{pmatrix}$, which are the extremities of a diameter of a sphere.

1. Find the center $\vec{C}$:

$$\vec{C} = \frac{\vec{A} + \vec{B}}{2} = \frac{1}{2}\begin{pmatrix} 2 + 8 \ 4 + 10 \ 6 + 12 \end{pmatrix} = \begin{pmatrix} 5 \ 7 \ 9 \end{pmatrix}$$

1. Find the radius $r$:

$$r = \frac{|\vec{A} - \vec{B}|}{2} = \frac{1}{2}\sqrt{(2 - 8)^2 + (4 - 10)^2 + (6 - 12)^2} = \frac{1}{2}\sqrt{36 + 36 + 36} = \frac{1}{2}\sqrt{108} = 3\sqrt{3}$$

1. Write the equation of the sphere:

$$|\vec{r} - \vec{C}|^2 = r^2$$

$$|\vec{r} - \begin{pmatrix} 5 \ 7 \ 9 \end{pmatrix}|^2 = (3\sqrt{3})^2$$

$$|\vec{r} - \begin{pmatrix} 5 \ 7 \ 9 \end{pmatrix}|^2 = 27$$

So, the equation of the sphere in vector form with the given extremities of the diameter is:

$$|\vec{r} - \begin{pmatrix} 5 \ 7 \ 9 \end{pmatrix}|^2 = 27$$

This equation represents a sphere centered at the point $(5, 7, 9)$ with a radius of $3\sqrt{3}$.