Three-dimensional Geometry

Contents

**Recall Three-dimensional Geometry** 2

**Describe the distance formulation in 3D** 3

**Describe Direction Ratios and Direction Cosines of a Line** 5

**Describe the Equation of a Line through a given point and parallel to the given Vector** 7

**Describe the Equation of a Line passing through two given Points** 8

**Recall the Angle between two Lines** 10

**Recall the distance between two Skew Lines and two Parallel Lines** 12

**Describe the Coplanarity of two lines** 15

**Recall the distance formulae of a Point from a Plane** 15

**Describe the Angle between two Planes** 17

**Describe the Angle between a Line and a Plane** 19

**Describe the equation of a Sphere in Centre Radius Form** 24

**Describe the general equation of a Sphere** 26

**Determine the equation of a Sphere in Centre Radius Form** 28

**Determine the Centre and the Radius of a Sphere** 30

**Recall Three-dimensional Geometry**

Three-dimensional geometry is the study of geometric shapes and figures in three dimensions, i.e., space. In three-dimensional space, we have three perpendicular axes – the x-axis, y-axis, and z-axis. These axes divide the space into eight octants. The positive directions of the axes are indicated by arrows.

Points, lines, planes, and other geometric shapes can be represented in three-dimensional space using coordinates. The coordinates of a point in three-dimensional space are given by a triplet (x, y, z) where x represents the displacement of the point along the x-axis, y represents the displacement of the point along the y-axis, and z represents the displacement of the point along the z-axis.

Lines in three-dimensional space can be described using a variety of methods, including parametric equations, symmetric equations, and vector equations. For example, the parametric equations of a line passing through the point (x1, y1, z1) and parallel to the vector a = <a1, a2, a3> can be given by:

x = x1 + a1t

y = y1 + a2t

z = z1 + a3t

where t is a parameter that can take any real value.

Planes in three-dimensional space can also be described using a variety of methods, including point-normal form, two-point form, and general form. For example, the point-normal form of a plane passing through the point (x1, y1, z1) and normal to the vector n = <a, b, c> can be given by:

a(x – x1) + b(y – y1) + c(z – z1) = 0

where a, b, and c are the direction cosines of the normal vector.

Three-dimensional geometry also involves studying the properties of geometric solids such as spheres, cylinders, cones, and polyhedra. These solids can be analyzed using techniques such as surface area, volume, and moment of inertia. The study of three-dimensional geometry has applications in many fields, including physics, engineering, architecture, and computer graphics.

**Describe the distance formulation in 3D**

Distance formulation is an essential concept in 3D geometry. It is used to calculate the distance between two points in a three-dimensional space. The distance between two points can be defined as the length of the straight line segment connecting the two points.

The distance formula between two points (x_{1}, y_{1}, z_{1}) and (x_{2}, y_{2}, z_{2}) in three-dimensional space is given by:

d = √[(x2 – x1)^{2} + (y2 – y1)^{2} + (z2 – z1)^{2}]

where d is the distance between the two points.

For example, suppose we want to find the distance between two points P and Q with coordinates (2, 4, 5) and (-1, 3, 7), respectively. We can use the distance formula as follows:

d = √[(−1−2)^{2} + (3−4)^{2 }+ (7−5)^{2}]

= √[(-3)^{2} + (-1)^{2} + 2^{2}]

= √14

Therefore, the distance between P and Q is √14.

The distance formula can also be used to find the distance between a point and a plane or between two parallel or skew lines. In these cases, the distance is calculated by finding the perpendicular distance between the objects in question.

In summary, the distance formula in 3D is a fundamental concept in three-dimensional geometry that is used to calculate the distance between two points in a three-dimensional space. The distance formula can also be used to find the distance between a point and a plane or between two parallel or skew lines.

**Recall the Section Formula**

The section formula is a concept in 3D geometry that allows us to find the coordinates of a point that divides a line segment joining two given points in a particular ratio. This formula is also known as the internal division formula or the partition formula.

Suppose we have two points A and B in three-dimensional space with coordinates (x1, y1, z1) and (x2, y2, z2), respectively. Let P be a point on the line segment AB that divides the segment in the ratio m:n, where m and n are positive integers. Then the coordinates of point P are given by:

Px = (mx2 + nx1)/(m + n)

Py = (my2 + ny1)/(m + n)

Pz = (mz2 + nz1)/(m + n)

Here, Px, Py, and Pz are the coordinates of point P.

For example, suppose we have two points A and B with coordinates (1, 2, 3) and (4, 5, 6), respectively. We want to find the coordinates of a point P that divides the line segment AB in the ratio 2:3. Using the section formula, we can find the coordinates of P as follows:

Px = (3*4 + 2*1)/(2+3) = 13/5

Py = (3*5 + 2*2)/(2+3) = 19/5

Pz = (3*6 + 2*3)/(2+3) = 27/5

Therefore, the coordinates of P are (13/5, 19/5, 27/5).

The section formula is a useful tool in 3D geometry, and it has many applications in engineering, physics, and other fields. It allows us to find the coordinates of a point on a line segment without having to know the equation of the line. The section formula can also be used to find the centroid of a triangle or the center of mass of a solid object, among other things.

**Describe Direction Ratios and Direction Cosines of a Line**

In 3D geometry, a line can be represented by a set of coordinates of two points lying on the line. However, there are other ways to represent a line, including using direction ratios and direction cosines.

Direction ratios are the ratios of the differences in the x, y, and z coordinates of two points on the line. Suppose we have two points A and B with coordinates (x1, y1, z1) and (x2, y2, z2), respectively. Then the direction ratios of the line AB are given by:

rx = x2 – x1

ry = y2 – y1

rz = z2 – z1

Direction cosines are the cosines of the angles that a line makes with the positive x, y, and z-axes. The direction cosines of a line can be obtained by dividing the direction ratios of the line by the magnitude of the line, which is the square root of the sum of the squares of the direction ratios. Thus, the direction cosines are given by:

lx = rx / sqrt(rx^{2} + ry^{2} + rz^{2})

ly = ry / sqrt(rx^{2} + ry^{2} + rz^{2})

lz = rz / sqrt(rx^{2} + ry^{2} + rz^{2})

Direction ratios and direction cosines provide useful information about the orientation of a line in 3D space. For example, if the direction ratios or direction cosines of two lines are equal, then the lines are parallel. Moreover, if the dot product of the direction cosines of two lines is zero, then the lines are perpendicular.

Let’s take an example to understand the concept of direction ratios and direction cosines. Consider a line AB with coordinates (2, 3, 4) and (5, 6, 7). The direction ratios of the line are:

rx = 5 – 2 = 3

ry = 6 – 3 = 3

rz = 7 – 4 = 3

Therefore, the direction ratios of the line AB are (3, 3, 3).

The magnitude of the line AB is given by:

sqrt(rx^{2} + ry^{2} + rz^{2}) = sqrt(3^{2} + 3^{2} + 3^{2}) = 3*sqrt(3)

Therefore, the direction cosines of the line AB are:

lx = 3/sqrt(27) = 1/sqrt(3)

ly = 3/sqrt(27) = 1/sqrt(3)

lz = 3/sqrt(27) = 1/sqrt(3)

Thus, the direction cosines of the line AB are (1/sqrt(3), 1/sqrt(3), 1/sqrt(3)).

In summary, direction ratios and direction cosines are important tools for describing the orientation of a line in 3D space. These concepts are essential in many fields, including physics, engineering, and computer graphics.

**Describe the Equation of a Line through a given point and parallel to the given Vector**

In three-dimensional space, a line can be represented by a point and a vector. Given a point and a direction vector, we can find the equation of a line in vector form. Let’s say we have a point P(x1, y1, z1) and a direction vector A(a, b, c). Then the equation of the line through the point P and parallel to A is given by:

where r is the position vector of any point on the line, and t is a scalar parameter. The vector equation can be written as three component equations:

Here, x, y, and z are the coordinates of any point on the line.

Example:

Let’s find the equation of a line through the point (1, 2, 3) and parallel to the vector A = (2, -1, 4).

The point P = (1, 2, 3)

The direction vector A = (2, -1, 4)

The equation of the line can be written as:

r = P + tA

r = (1, 2, 3) + t(2, -1, 4)

r = (1 + 2t, 2 – t, 3 + 4t)

This is the vector equation of the line. We can also write it in component form as:

x = 1 + 2t

y = 2 – t

z = 3 + 4t

Note that there are infinitely many points on the line for different values of t.

**Describe the Equation of a Line passing through two given Points**

Given two points P(x1, y1, z1) and Q(x2, y2, z2) in three-dimensional space, we can find the equation of the line passing through these points. The direction vector of the line is obtained by subtracting the coordinates of one point from the coordinates of the other point. Let’s call this vector PQ.

So, the direction vector of the line is given by:

PQ = Q – P = (x2 – x1, y2 – y1, z2 – z1)

Now, we can use the equation of a line through a point and parallel to a vector to get the equation of the line passing through P and Q.

Let’s say we want to find the equation of the line passing through P and Q. Then the equation of the line can be written as:

r = P + tPQ

where r is the position vector of any point on the line, and t is a scalar parameter. The vector equation can be written as three component equations:

x = x1 + t(x2 – x1)

y = y1 + t(y2 – y1)

z = z1 + t(z2 – z1)

Here, x, y, and z are the coordinates of any point on the line.

Example:

Let’s find the equation of a line passing through the points P(1, 2, 3) and Q(4, 5, 6).

The point P = (1, 2, 3)

The point Q = (4, 5, 6)

The direction vector PQ = Q – P = (4-1, 5-2, 6-3) = (3, 3, 3)

The equation of the line passing through P and Q can be written as:

r = P + tPQ

r = (1, 2, 3) + t(3, 3, 3)

r = (1 + 3t, 2 + 3t, 3 + 3t)

This is the vector equation of the line. We can also write it in component form as:

x = 1 + 3t

y = 2 + 3t

z = 3 + 3t

Note that there are infinitely many points on the line for different values of t.

**Recall the Angle between two Lines**

When two lines are intersecting in 3D space, they create an angle between them. The angle between two lines can be found using the dot product of the direction vectors of the lines. The dot product of two vectors is defined as:

A · B = ||A|| ||B|| cos(θ)

where A and B are the two vectors, ||A|| and ||B|| are the magnitudes of A and B respectively, and θ is the angle between them.

If two lines have direction vectors A and B, the angle θ between them can be found as:

θ = cos⁻¹((A · B) / (||A|| ||B||))

If the two lines are not intersecting, we can find the angle between them by considering the angle between the normal vectors of the two lines. The normal vector to a line is given by the cross product of the direction vector and any other vector on the line.

If two lines have direction vectors A and B, and normal vectors n1 and n2 respectively, the angle θ between them can be found as:

θ = cos⁻¹((n1 · n2) / (||n1|| ||n2||))

For example, let’s consider two lines in 3D space with direction vectors A = 2i + 3j – k and B = 4i – j + 2k. To find the angle between these lines, we first calculate the dot product of A and B as:

A · B = (2 * 4) + (3 * -1) + (-1 * 2) = 5

We also calculate the magnitudes of A and B as:

||A|| = sqrt(2^{2} + 3^{2} + (-1)^{2}) = sqrt(14)

||B|| = sqrt(4^{2} + (-1)^{2} + 2^{2}) = sqrt(21)

Using these values, we can find the angle between the two lines as:

θ = cos⁻¹((A · B) / (||A|| ||B||)) = cos⁻¹(5 / (sqrt(14) * sqrt(21))) ≈ 57.3 degrees

Therefore, the angle between the two lines is approximately 57.3 degrees.

**Recall the distance between two Skew Lines and two Parallel Lines**

Distance between Two Skew Lines:

Two lines in three-dimensional space are said to be skew lines if they are not parallel and do not intersect. The distance between two skew lines is defined as the shortest distance between any two points lying on the two lines.

Let L1 and L2 be two skew lines, and P1 and P2 be two points on L1 and L2, respectively. Let d be the distance between the lines L1 and L2. Let N be the unit vector perpendicular to both L1 and L2. Then, the distance between the lines can be calculated using the formula:

d = |P1P2 x N| / |N|

where x denotes the cross product of two vectors and | | represents the magnitude of a vector.

Distance between Two Parallel Lines:

Two lines in three-dimensional space are said to be parallel if they have the same direction vector or are coincident. The distance between two parallel lines is defined as the distance between any two parallel planes perpendicular to the lines.

Let L1 and L2 be two parallel lines with direction vectors D1 and D2, respectively. Let P1 and P2 be two points on L1 and L2, respectively. Then, the distance between the lines can be calculated using the formula:

d = |P1P2 x D1| / |D1|

where x denotes the cross product of two vectors and | | represents the magnitude of a vector.

If the direction vectors D1 and D2 of the parallel lines are given in terms of their direction cosines, then the distance between the lines can be calculated as:

d = |a_{1}x + b_{1}y + c_{1}z – a_{2}x – b_{2}y – c_{2}z| / sqrt(a1^{2} + b1^{2} + c1^{2})

where (a1, b1, c1) and (a2, b2, c2) are the direction cosines of D1 and D2, respectively, and (x, y, z) are the coordinates of any point on the line.

**Describe the equations of a Plane: In Normal Form, Perpendicular to a given Vector and passing through a Point, and In Intercept Form**

A plane is a two-dimensional flat surface that extends infinitely in all directions. In geometry, a plane is represented by a mathematical equation that describes its position in three-dimensional space. There are two common forms of the equation of a plane: the normal form and the intercept form.

- Normal Form:

The normal form of the equation of a plane is given by:

ax + by + cz = d,

where a, b, and c are the direction ratios of a vector that is perpendicular to the plane, and x, y, and z are the coordinates of a point on the plane. The constant d is the distance of the plane from the origin along the direction of the normal vector.

For example, consider a plane that passes through the point (1,2,3) and is perpendicular to the vector (2,3,4). The normal form of the equation of this plane is:

2x + 3y + 4z = 2*1 + 3*2 + 4*3 = 2 + 6 + 12 = 20.

2. Perpendicular to a given Vector and passing through a Point:

To find the equation of a plane that is perpendicular to a given vector and passes through a given point, we can use the normal form of the equation. Let the given point be P(x1, y1, z1) and the given vector be a = (a1, a2, a3). Then the normal form of the equation of the plane is:

a1(x – x1) + a2(y – y1) + a3(z – z1) = 0.

For example, consider a plane that passes through the point (1,2,3) and is perpendicular to the vector (2,3,4). Using the formula, we get:

2(x – 1) + 3(y – 2) + 4(z – 3) = 0.

which simplifies to:

2x + 3y + 4z = 20.

This is the same equation we obtained in the first example using the normal form.

3. Intercept Form:

The intercept form of the equation of a plane is given by:

x/a + y/b + z/c = 1,

where a, b, and c are the intercepts of the plane on the x-axis, y-axis, and z-axis respectively.

For example, consider a plane that intercepts the x, y, and z axes at the points (2,0,0), (0,3,0), and (0,0,4) respectively. The intercept form of the equation of this plane is:

x/2 + y/3 + z/4 = 1.

**Describe the Coplanarity of two lines**

Coplanarity is the property of two or more objects lying in the same plane. Two lines are said to be coplanar if they lie in the same plane. In other words, if two lines intersect, they are coplanar. If they do not intersect, they may or may not be coplanar.

Examples:

- Coplanar lines: Two lines lying in the same plane are said to be coplanar.
- Non-coplanar lines: Two lines that do not lie in the same plane are said to be non-coplanar.
- Parallel lines: Parallel lines are coplanar as they lie in the same plane.
- Intersecting lines: Intersecting lines are coplanar as they lie in the same plane.

Conclusion:

Coplanarity is an important concept in geometry that helps to determine the position of lines and points in space. Two lines are said to be coplanar if they lie in the same plane. In real-life situations, coplanarity is used in various fields such as architecture, engineering, and design. For instance, architects use coplanarity to determine the position of walls, doors, and windows in a building. Engineers use coplanarity to design and construct bridges, roads, and other structures.

**Recall the distance formulae of a Point from a Plane**

The distance formula between a point and a plane is a mathematical formula used to determine the perpendicular distance between a point and a plane in a 3-dimensional space. This formula is useful in various applications, such as in engineering, physics, and geometry.

The formula to calculate the distance between a point and a plane is given by:

distance = |ax + by + cz + d| / √(a^{2} + b^{2} + c^{2})

Where,

a, b, and c are the coefficients of the variables x, y, and z in the equation of the plane.

d is the constant term in the equation of the plane.

x, y, and z are the coordinates of the point.

|.| denotes the absolute value of the expression.

√(.) denotes the square root of the expression.

Example:

Let’s consider the point P(2, 3, 4) and the plane 2x + 3y – z = 7.

The coefficients a, b, and c are 2, 3, and -1, respectively, and the constant term d is 7.

To calculate the distance between the point P and the plane, we can substitute the values of a, b, c, d, x, y, and z into the formula:

distance = |2(2) + 3(3) – 1(4) – 7| / √(2^{2} + 3^{2} + (-1)^{2})

= |-4| / √14

= 2√14 / 7

Therefore, the distance between the point P(2, 3, 4) and the plane 2x + 3y – z = 7 is 2√14 / 7.

In conclusion, the distance formula between a point and a plane is an essential mathematical tool used in various fields, including engineering, physics, and geometry. The formula enables us to determine the perpendicular distance between a point and a plane in a 3-dimensional space.

**Describe the Angle between two Planes**

The angle between two planes is the angle between their respective normal vectors. Normal vectors are perpendicular to the planes they define and the angle between them is the same as the angle between the planes.

Method:

To find the angle between two planes, follow these steps:

- Find the normal vectors of both planes.
- Use the dot product formula to find the angle between the normal vectors.
- Take the absolute value of the result to get the angle between the planes.

Example 1:

Find the angle between the planes 2x + y – 3z = 4 and x – 2y + z = 5.

Solution:

To find the normal vectors of the planes, we can look at the coefficients of x, y, and z in the plane equations. The normal vector of the first plane is (2,1,-3) and the normal vector of the second plane is (1,-2,1).

Using the dot product formula, we get:

(2,1,-3) · (1,-2,1) = (2)(1) + (1)(-2) + (-3)(1) = -3

The angle between the normal vectors is given by:

cos θ = (-3) / (√14 * √6) ≈ -0.685

Thus, θ ≈ 131.8 degrees.

Since we want the angle between the planes, we take the absolute value of θ to get:

|θ| ≈ 131.8 degrees

Example 2:

Find the angle between the planes x + y + z = 1 and x – y + 2z = 3.

Solution:

The normal vectors of the planes are (1,1,1) and (1,-1,2).

Using the dot product formula, we get:

(1,1,1) · (1,-1,2) = (1)(1) + (1)(-1) + (1)(2) = 2

The angle between the normal vectors is given by:

cos θ = 2 / (√3 * √6) ≈ 0.515

Thus, θ ≈ 59.4 degrees.

Taking the absolute value of θ, we get:

|θ| ≈ 59.4 degrees

Conclusion:

The angle between two planes is the angle between their respective normal vectors. The normal vectors are found by looking at the coefficients of x, y, and z in the plane equations. The dot product formula is used to find the angle between the normal vectors, which is then taken as the angle between the planes.

**Describe the Angle between a Line and a Plane**

The angle between a line and a plane is the angle between the line and the plane’s normal vector. This angle can be used to determine whether the line is parallel or perpendicular to the plane.

Method:

To find the angle between a line and a plane, follow these steps:

- Find the normal vector of the plane.
- Find the directional vector of the line.
- Use the dot product formula to find the angle between the line and the plane’s normal vector.
- Take the absolute value of the result to get the angle between the line and the plane.

Example 1:

Find the angle between the line x = 1 + t, y = 2 – t, z = 3 + 2t and the plane 2x – y + 3z = 5.

Solution:

The normal vector of the plane is (2,-1,3).

The directional vector of the line is (1,-1,2).

Using the dot product formula, we get:

(1,-1,2) · (2,-1,3) = (1)(2) + (-1)(-1) + (2)(3) = 9

The angle between the line and the plane’s normal vector is given by:

cos θ = 9 / (√6 * √14) ≈ 0.600

Thus, θ ≈ 53.13 degrees.

Taking the absolute value of θ, we get:

|θ| ≈ 53.13 degrees

Since this angle is not 0 or 90 degrees, we can conclude that the line is neither parallel nor perpendicular to the plane.

Example 2:

Find the angle between the line x = 2 – t, y = -1 + 2t, z = 3t and the plane 2x – y + 3z = 4.

Solution:

The normal vector of the plane is (2,-1,3).

The directional vector of the line is (-1,2,3).

Using the dot product formula, we get:

(-1,2,3) · (2,-1,3) = (-1)(2) + (2)(-1) + (3)(3) = 5

The angle between the line and the plane’s normal vector is given by:

cos θ = 5 / (√14 * √14) ≈ 0.357

Thus, θ ≈ 69.41 degrees.

Taking the absolute value of θ, we get:

|θ| ≈ 69.41 degrees

Since this angle is not 0 or 90 degrees, we can conclude that the line is neither parallel nor perpendicular to the plane.

Conclusion:

The angle between a line and a plane is the angle between the line and the plane’s normal vector. This angle can be used to determine whether the line is parallel or perpendicular to the plane. The normal vector of the plane and the directional vector of the line can be found by looking at the coefficients of x, y, and z in their respective equations. The dot product formula is used to find the angle between the line and the plane’s normal vector.

**Recall Sphere**

A sphere is a three-dimensional object that consists of all points in space that are a fixed distance from a given point called the center. The distance between the center and any point on the surface of the sphere is known as the radius. A sphere is a perfectly symmetrical object and has no corners or edges.

Properties:

- Center: The center of the sphere is the fixed point in space from which all points on the surface of the sphere are equidistant.
- Radius: The radius of a sphere is the distance from the center to any point on the surface of the sphere.
- Diameter: The diameter of a sphere is twice the radius and is the distance between any two points on the surface of the sphere that pass through the center.
- Surface area: The surface area of a sphere is given by the formula 4πr
^{2}, where r is the radius. - Volume: The volume of a sphere is given by the formula (4/3)πr
^{3}, where r is the radius.

Example 1:

Consider a sphere with a radius of 5 cm. Find the diameter, surface area, and volume of the sphere.

Solution:

The diameter of the sphere is 2 × radius = 2 × 5 = 10 cm.

The surface area of the sphere is 4πr^{2} = 4π × 5^{2} = 100π cm^{2}.

The volume of the sphere is (4/3)πr^{3} = (4/3)π × 5^{3} = 523.6 cm^{3}.

Example 2:

A sphere has a surface area of 36π cm^{2}. Find its radius and volume.

Solution:

The surface area of the sphere is given by 4πr^{2}, so we have:

4πr^{2} = 36π

Simplifying, we get:

r^{2} = 9

Taking the square root, we get:

r = 3 cm

The volume of the sphere is (4/3)πr^{3} = (4/3)π × 3^{3} = 36π cm^{3}.

Conclusion:

A sphere is a three-dimensional object that consists of all points in space that are a fixed distance from a given point called the center. The distance between the center and any point on the surface of the sphere is known as the radius. A sphere is a perfectly symmetrical object and has no corners or edges. The diameter of a sphere is twice the radius, and the surface area and volume of a sphere can be calculated using formulas involving the radius.

** ****Describe the equation of a Sphere in Centre Radius Form**

The equation of a sphere in center-radius form represents a sphere in three-dimensional space with its center at the point (h, k, l) and radius r. The equation can be written as:

(x – h)^{2} + (y – k)^{2} + (z – l)^{2} = r^{2}

where x, y, and z are the coordinates of any point on the surface of the sphere.

Properties:

- Center: The center of the sphere is represented by the point (h, k, l).
- Radius: The radius of the sphere is represented by r.
- Diameter: The diameter of the sphere is twice the radius and is given by 2r.

Example 1:

Write the equation of a sphere with center (3, 4, -2) and radius 5.

Solution:

The equation of the sphere in center-radius form is given by:

(x – 3)^{2} + (y – 4)^{2} + (z + 2)^{2} = 5^{2}

Example 2:

Write the equation of a sphere with center (-2, 0, 1) and radius 3.

Solution:

The equation of the sphere in center-radius form is given by:

(x + 2)^{2} + y^{2} + (z – 1)^{2} = 3^{2}

Example 3:

Find the center and radius of the sphere with equation (x – 1)^{2} + (y + 3)^{2} + (z – 2)^{2} = 25.

Solution:

Comparing the given equation with the center-radius form equation, we get:

Center = (1, -3, 2)

Radius = √25 = 5

Conclusion:

The equation of a sphere in center-radius form represents a sphere in three-dimensional space with its center at the point (h, k, l) and radius r. The equation is (x – h)^{2} + (y – k)^{2} + (z – l)^{2} = r^{2}, where x, y, and z are the coordinates of any point on the surface of the sphere. The center of the sphere is represented by the point (h, k, l), and the radius of the sphere is represented by r. The diameter of the sphere is twice the radius, and the center and radius of a sphere can be found by comparing the equation of the sphere with the center-radius form equation.

** ****Describe the general equation of a Sphere**

The general equation of a sphere is a second-degree equation in three variables x, y, and z that represents a sphere in three-dimensional space. The equation can be written as:

Ax^{2} + By^{2} + Cz^{2} + Dx + Ey + Fz + G = 0

where A, B, C, D, E, F, and G are constants.

Properties:

- Center: The center of the sphere can be found by the values of the coefficients of the variables Dx, Ey, and Fz, which are (-D/2A, -E/2B, -F/2C).
- Radius: The radius of the sphere can be calculated by the formula r = √(-(D
^{2}+ E^{2}+ F^{2}– 4ACG – 4BCG – 4ABF)/(4ABC))

Example 1:

Find the center and radius of the sphere with equation x^{2} + y^{2} + z^{2} – 2x – 4y – 6z + 13 = 0.

Solution:

The given equation can be written as:

(x – 1)^{2} + (y – 2)^{2 }+ (z – 3)^{2} = 9

Comparing this equation with the general equation of the sphere, we can see that:

A = B = C = 1, D = -2, E = -4, F = -6, and G = 13.

Therefore, the center of the sphere is (-D/2A, -E/2B, -F/2C) = (1, 2, 3), and the radius can be calculated by the formula r = √(-(D^{2} + E^{2} + F^{2} – 4ACG – 4BCG – 4ABF)/(4ABC)) = 3.

Hence, the center of the sphere is (1, 2, 3) and the radius is 3.

Example 2:

Find the center and radius of the sphere with equation 2x^{2} + 2y^{2} + 2z^{2} – 12x + 6y – 6z + 1 = 0.

Solution:

The given equation can be written as:

(x – 3)^{2} + (y – 1)^{2} + (z + 1)^{2} = (3/2)^{2}

Comparing this equation with the general equation of the sphere, we can see that:

A = B = C = 2, D = -12, E = 6, F = -6, and G = 1.

Therefore, the center of the sphere is (-D/2A, -E/2B, -F/2C) = (3, -1, 1), and the radius can be calculated by the formula r = √(-(D^{2} + E^{2} + F^{2} – 4ACG – 4BCG – 4ABF)/(4ABC)) = 3/2.

Hence, the center of the sphere is (3, -1, 1) and the radius is 3/2.

Conclusion:

The general equation of a sphere is a second-degree equation in three variables x, y, and z that represents a sphere in three-dimensional space.

**Determine the equation of a Sphere in Centre Radius Form**

The equation of a sphere in center-radius form is a second-degree equation in three variables x, y, and z that represents a sphere in three-dimensional space. The equation can be written as:

(x – a)^{2} + (y – b)^{2} + (z – c)^{2} = r^{2}

where (a, b, c) is the center of the sphere, and r is its radius.

Properties:

- Center: The center of the sphere is (a, b, c).
- Radius: The radius of the sphere is r.

Example 1:

Find the equation of the sphere with center (-1, 2, 3) and radius 4.

Solution:

The equation of the sphere in center-radius form is given by:

(x – a)^{2} + (y – b)^{2} + (z – c)^{2} = r^{2}

Substituting the given values, we get:

(x – (-1))^{2} + (y – 2)^{2} + (z – 3)^{2} = 4^{2}

Simplifying, we get:

(x + 1)^{2} + (y – 2)^{2} + (z – 3)^{2} = 16

Hence, the equation of the sphere is (x + 1)^{2} + (y – 2)^{2} + (z – 3)^{2} = 16.

Example 2:

Find the equation of the sphere with center (3, -1, 1) and radius 3/2.

Solution:

The equation of the sphere in center-radius form is given by:

(x – a)^{2} + (y – b)^{2} + (z – c)^{2} = r^{2}

Substituting the given values, we get:

(x – 3)^{2} + (y + 1)^{2} + (z – 1)^{2} = (3/2)^{2}

Simplifying, we get:

(x – 3)^{2} + (y + 1)^{2} + (z – 1)^{2} = 9/4

Hence, the equation of the sphere is (x – 3)^{2} + (y + 1)^{2} + (z – 1)^{2} = 9/4.

Conclusion:

The equation of a sphere in center-radius form is a second-degree equation in three variables x, y, and z that represents a sphere in three-dimensional space. The center of the sphere is (a, b, c), and the radius is r. The equation of the sphere can be obtained by substituting the values of the center and radius in the general equation of a sphere in center-radius form.

**Determine the Centre and the Radius of a Sphere**

The center of a sphere is the point that is equidistant from all points on the sphere. The radius is the distance from the center to any point on the sphere.

Properties:

- Center: The center of the sphere is the point equidistant from all points on the sphere.
- Radius: The radius of the sphere is the distance from the center to any point on the sphere.

Example 1:

Determine the center and radius of the sphere with the equation (x – 2)^{2} + (y + 1)^{2} + (z – 3)^{2} = 16.

Solution:

Comparing the given equation with the general equation of a sphere in center-radius form, we get:

Center = (2, -1, 3)

Radius = sqrt(16) = 4

Hence, the center of the sphere is (2, -1, 3), and the radius is 4.

Example 2:

Determine the center and radius of the sphere with the equation x^{2} + y^{2} + z^{2} – 4x – 6y – 8z + 25 = 0.

Solution:

Rewriting the given equation, we get:

(x – 2)^{2} + (y – 3)^{2} + (z – 4)^{2} = 2^{2} + 3^{2} + 4^{2} – 25

= 14

Comparing the above equation with the general equation of a sphere in center-radius form, we get:

Center = (2, 3, 4)

Radius = sqrt(14)

Hence, the center of the sphere is (2, 3, 4), and the radius is sqrt(14).

Conclusion:

The center of a sphere is the point that is equidistant from all points on the sphere, and the radius is the distance from the center to any point on the sphere. The center and radius of a sphere can be determined by comparing the given equation with the general equation of a sphere in center-radius form.