Equation of a Sphere (Vector Form)

The equation of a sphere in vector form is an elegant way to represent a sphere in three-dimensional space using vectors. This form is particularly useful in physics and engineering where vector calculus is commonly used.

Standard Equation of a Sphere

Before we delve into the vector form, let's recall the standard equation of a sphere in Cartesian coordinates. A sphere with center at point ( (h, k, l) ) and radius ( r ) is described by the equation:

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

Vector Form of a Sphere's Equation

In vector form, the equation of a sphere is given by:

[ \lVert \mathbf{r} - \mathbf{c} \rVert^2 = r^2 ]

where:

• ( \mathbf{r} ) is the position vector of a generic point ( P(x, y, z) ) on the sphere,
• ( \mathbf{c} ) is the position vector of the center of the sphere ( C(h, k, l) ),
• ( r ) is the radius of the sphere,
• ( \lVert \cdot \rVert ) denotes the Euclidean norm (or magnitude) of a vector.

Breaking Down the Vector Form

The vector form of the equation essentially states that for any point on the sphere, the distance from that point to the center of the sphere is constant and equal to the radius ( r ). The distance is calculated using the Euclidean norm of the vector difference ( \mathbf{r} - \mathbf{c} ).

Important Points and Differences

Aspect	Cartesian Form	Vector Form
Center of Sphere	Given by coordinates ( (h, k, l) )	Given by vector ( \mathbf{c} )
Radius	Denoted by scalar ( r )	Denoted by scalar ( r )
Position of Point on Sphere	Given by coordinates ( (x, y, z) )	Given by vector ( \mathbf{r} )
Equation	( (x - h)^2 + (y - k)^2 + (z - l)^2 = r^2 )	( \lVert \mathbf{r} - \mathbf{c} \rVert^2 = r^2 )
Representation of Distance	Squared differences of coordinates	Norm of the vector difference
Applicability	Easy for algebraic manipulations	Convenient for vector calculus operations
Visualization	Intuitive in Cartesian grid	Requires understanding of vectors

Examples

Example 1: Finding the Vector Equation

Given a sphere with center at ( (2, -3, 4) ) and radius ( 5 ), find the vector equation.

Solution:

The center of the sphere in vector form is ( \mathbf{c} = \begin{bmatrix} 2 \ -3 \ 4 \end{bmatrix} ), and the radius is ( r = 5 ).

The vector equation of the sphere is:

[ \lVert \mathbf{r} - \begin{bmatrix} 2 \ -3 \ 4 \end{bmatrix} \rVert^2 = 5^2 ]

Example 2: Verifying a Point on the Sphere

Verify if the point ( P(3, -2, 7) ) lies on the sphere with the vector equation ( \lVert \mathbf{r} - \begin{bmatrix} 2 \ -3 \ 4 \end{bmatrix} \rVert^2 = 5^2 ).

Solution:

The position vector of point ( P ) is ( \mathbf{r} = \begin{bmatrix} 3 \ -2 \ 7 \end{bmatrix} ).

Substitute ( \mathbf{r} ) into the sphere's equation:

[ \lVert \begin{bmatrix} 3 \ -2 \ 7 \end{bmatrix} - \begin{bmatrix} 2 \ -3 \ 4 \end{bmatrix} \rVert^2 = \lVert \begin{bmatrix} 1 \ 1 \ 3 \end{bmatrix} \rVert^2 = 1^2 + 1^2 + 3^2 = 1 + 1 + 9 = 11 ]

Since ( 11 \neq 5^2 ), the point ( P(3, -2, 7) ) does not lie on the sphere.

Understanding the vector form of the equation of a sphere is crucial for solving problems in 3D geometry, especially when dealing with vector fields and operations in physics and engineering. It provides a more general and powerful way to describe spheres in three-dimensional space.