Question

(a) Describe and find an equation for the surface generated by all points \((x, y, z)\) that are four units from the point (3,-2,5) (b) Describe and find an equation for the surface generated by all points \((x, y, z)\) that are four units from the plane \(4 x-3 y+z=10\)

Short answer

The equations of the surfaces are \( (x-3)^2+(y+2)^2+(z-5)^2=16 \) and \(4x - 3y + z = 10 \pm 4\sqrt{20}\)

Step-by-step solution

Find the Sphere Equation

For a sphere centered at a point (3,-2,5) with radius 4, the equation is \( (x-3)^2+(y+2)^2+(z-5)^2=4^2 \). Thus, the equation becomes \( (x-3)^2+(y+2)^2+(z-5)^2=16 \).

Describe the Sphere

The sphere is the locus of all points in 3-dimensional space that are four units away from the point (3, -2, 5). Every point on this sphere is equidistant from the center point.

Find the Two Planes Equations

For a set of points four units from the plane \(4x - 3y + z = 10\), two parallel planes equidistant from the given plane are obtained. Write the plane equation in normal form, \(4x - 3y + z - 10 = 0\), so the distance is \( \frac{4x - 3y + z - 10}{\sqrt{4^2 + (-3)^2 + 1^2}} = \pm 4\). Solve these equations for the planes, producing \(4x - 3y + z = 10 \pm 4\sqrt{4^2 + (-3)^2 + 1^2}\).

Describe the Two Planes

These two planes are the set of points in 3-dimensional space that are four units distant from the plane \(4x - 3y + z = 10\). Each point on these planes maintains a constant distance of four units from the given plane, and the planes are parallel to the given plane.

Key concepts

3-Dimensional Geometry

Understanding three-dimensional geometry is fundamental when engaging with objects such as spheres, cubes, and planes. Unlike the two dimensions of a flat surface, composed of length and width, 3-dimensional geometry introduces height or depth, allowing us to visualize shapes in the space that surrounds us.

When studying a sphere within this context, as with the exercise involving the sphere centered at a point (3, -2, 5), we are referring to a perfectly round geometrical object in three-dimensional space. Every point on the surface of a sphere is exactly the same distance (called the radius) from its center. In mathematical terms, a sphere can be represented by the equation \( (x-h)^2+(y-k)^2+(z-l)^2=r^2 \), where \( (h, k, l) \) is the center and \( r \) is the radius. This simple yet powerful formula encapsulates the essence of 3-dimensional geometry by providing a relation that holds for every single point on the sphere's surface.

To visualize this in the context of the given problem, every point \( (x, y, z) \) on this sphere's surface follows the equation \( (x-3)^2+(y+2)^2+(z-5)^2=16 \) and is exactly four units from the center point at (3, -2, 5).

Locus of Points in Space

A locus is the set of points that satisfy a particular condition. In the sphere's case, the condition is that all points on the sphere's surface are equidistant from the center. For the first part of the exercise, where we aim to find the sphere's equation, we are essentially identifying the locus of points in space that are all four units from the point (3, -2, 5).

The concept of locus extends to other shapes as well, such as lines, curves, and, as seen in the second part of the exercise, planes. In the case of the two parallel planes, their locus is defined by all the points that maintain a constant distance of four units from the plane defined by the equation \( 4x - 3y + z = 10 \). It illustrates how the locus concept is not confined to solid figures but also applies to flat surfaces. This condition of constant distance creates two distinct planes parallel to the initial plane, which are identified by their respective equations.

Distance from a Plane

The concept of distance from a plane is essential to solve the problem's second part. Typically, to find the distance from a point to a plane, one would use the formula \( d = \frac{|Ax + By + Cz + D|}{\sqrt{A^2 + B^2 + C^2}} \), where \( Ax + By + Cz + D = 0 \) is the equation of the plane, and \( (x, y, z) \) is the point. However, this exercise requires us to find a set of points that are a specific distance away from the plane.

The given equation \( 4x - 3y + z = 10 \) represents a plane in space. To find the locus of points that are four units away from this plane, the distance formula is rearranged to include the distance as a constant value, resulting in two new plane equations parallel to the original. In the terms of our exercise, this gives us \( 4x - 3y + z = 10 \pm 4 \sqrt{16 + 9 + 1} \), which defines two parallel planes, equally spaced and coexisting in 3-dimensional space with the initial plane -- highlighting how 3-dimensional geometry, the notion of a locus of points, and the concept of distance from a plane converge to form comprehensive solutions to complex spatial problems.