Cartesian Equation of a Plane

The general Cartesian equation of a plane is:

ax + by + cz = d

where:

• a, b, and c determine the orientation of the plane.

• d determines its position relative to the origin.

• (a, b, c) is a normal vector.

The normal vector is:

n = (a, b, c)

For example:

2x + 3y − z = 6

has normal vector:

n = (2, 3, −1)

Why the Normal Vector Matters

Every direction vector lying within the plane is perpendicular to the normal vector.

Therefore, if v is a direction vector in the plane:

n · v = 0

This relationship is fundamental to three-dimensional geometry.

Finding the Equation of a Plane from a Point and Normal Vector

If a plane passes through:

P = (x₀, y₀, z₀)

and has normal vector:

n = (a, b, c)

then its equation is:

a(x − x₀) + b(y − y₀) + c(z − z₀) = 0

This is called the point-normal form of a plane.

Worked Example

Find the equation of a plane passing through:

P = (1, 2, 3)

with normal vector:

n = (2, −1, 4)

Step 1: Write the point-normal equation

2(x − 1) − (y − 2) + 4(z − 3) = 0

Step 2: Expand

2x − 2 − y + 2 + 4z − 12 = 0

Step 3: Simplify

2x − y + 4z − 12 = 0

Final Answer

2x − y + 4z = 12

Verification

Substitute P = (1, 2, 3):

2(1) − 2 + 4(3)

= 2 − 2 + 12

= 12

The point satisfies the equation.

Vector Equations of Planes

A plane can also be represented using two nonparallel direction vectors.

The general vector equation is:

r = a + su + tv

where:

• a is the position vector of a known point on the plane.

• u and v are nonparallel direction vectors lying in the plane.

• s and t are real-valued parameters.

Unlike a line, which requires one direction vector and one parameter, a plane requires two independent direction vectors and two parameters.

Worked Example

Write a vector equation of a plane passing through:

P = (1, 2, 0)

with direction vectors:

u = (2, 0, 1)

v = (0, 3, 1)

Solution

r = (1, 2, 0) + s(2, 0, 1) + t(0, 3, 1)

The corresponding parametric equations are:

x = 1 + 2s

y = 2 + 3t

z = s + t

Different combinations of s and t generate points throughout the plane.

Finding a Plane Through Three Points

Three non-collinear points determine a unique plane.

The cross product can be used to calculate a normal vector.

General Method

Given three points P, Q, and R:

1. Calculate PQ = Q − P.

2. Calculate PR = R − P.

3. Find the normal vector n = PQ × PR.

4. Use the point-normal equation.

Worked Example

Find the equation of the plane through:

P = (1, 0, 0)

Q = (0, 2, 0)

R = (0, 0, 3)

Step 1: Calculate two direction vectors

PQ = (−1, 2, 0)

PR = (−1, 0, 3)

Step 2: Calculate the cross product

n = PQ × PR

nₓ = (2)(3) − (0)(0) = 6

nᵧ = (0)(−1) − (−1)(3) = 3

n_z = (−1)(0) − (2)(−1) = 2

Therefore:

n = (6, 3, 2)

Step 3: Apply point-normal form

6(x − 1) + 3(y − 0) + 2(z − 0) = 0

Step 4: Simplify

6x − 6 + 3y + 2z = 0

Final Answer

6x + 3y + 2z = 6

Verification

For P:

6(1) + 3(0) + 2(0) = 6

For Q:

6(0) + 3(2) + 2(0) = 6

For R:

6(0) + 3(0) + 2(3) = 6

All three points lie on the plane.

Important Observation

If the three points are collinear, their direction vectors are parallel and their cross product is zero.

In that case, the three points do not determine a unique plane.

Parallel and Perpendicular Lines

Two lines can be parallel, intersecting, or skew.

In three-dimensional space, skew lines are lines that neither intersect nor are parallel.

Parallel Lines

Two lines are parallel when their direction vectors are scalar multiples of one another.

If:

d₁ = kd₂

for some nonzero scalar k, the directions are parallel.

Worked Example

Line 1:

r = (1, 2, 3) + t(2, 4, 6)

Line 2:

r = (3, 1, 0) + s(1, 2, 3)

Compare the direction vectors:

d₁ = (2, 4, 6)

d₂ = (1, 2, 3)

Since:

d₁ = 2d₂

the lines have parallel directions.

The lines are distinct because the point (3, 1, 0) does not lie on Line 1.

Conclusion

The lines are parallel.

Perpendicular Line Directions

Two direction vectors are perpendicular when their dot product is zero.

d₁ · d₂ = 0

However, in three dimensions, perpendicular direction vectors do not necessarily mean that two lines intersect.

The lines must intersect to be perpendicular as geometric lines.

Worked Example

Consider:

d₁ = (2, 1, 0)

d₂ = (1, −2, 3)

Calculate:

d₁ · d₂ = (2)(1) + (1)(−2) + (0)(3)

= 2 − 2

= 0

The direction vectors are perpendicular.

If lines with these directions intersect, the lines are perpendicular.

Skew Lines

Skew lines occur only in three-dimensional geometry.

They are neither parallel nor intersecting.

Worked Example

Consider:

Line 1:

r = (0, 0, 0) + t(1, 0, 0)

Line 2:

r = (0, 1, 1) + s(0, 1, 0)

The direction vectors are:

d₁ = (1, 0, 0)

d₂ = (0, 1, 0)

They are not parallel.

To test for intersection, equate the coordinates:

t = 0

0 = 1 + s

0 = 1

The final equation is impossible.

Therefore, the lines do not intersect.

Conclusion

The lines are skew.

Although their direction vectors are perpendicular, the lines themselves do not intersect.

Finding the Intersection of Two Lines

Two lines intersect when they share a common point.

Their vector equations can be equated to find the parameters corresponding to that point.

Worked Example

Find the intersection of:

Line 1:

r = (1, 2, 0) + t(1, 1, 2)

Line 2:

r = (3, 1, 4) + s(−1, 2, 0)

Step 1: Write the parametric equations

Line 1:

x = 1 + t

y = 2 + t

z = 2t

Line 2:

x = 3 − s

y = 1 + 2s

z = 4

Step 2: Equate the z-coordinates

2t = 4

t = 2

Step 3: Equate the x-coordinates

1 + 2 = 3 − s

3 = 3 − s

s = 0

Step 4: Check the y-coordinates

Line 1:

y = 2 + 2 = 4

Line 2:

y = 1 + 2(0) = 1

The y-coordinates do not agree.

Conclusion

The lines do not intersect.

Since their direction vectors are not parallel, they are skew.

Important Lesson

In three dimensions, solving two coordinate equations is not enough.

All three coordinates must agree.

Worked Example: Intersecting Lines

Consider:

Line 1:

r = (1, 0, 2) + t(2, 1, −1)

Line 2:

r = (3, 1, 1) + s(1, −1, 2)

Step 1: Equate coordinates

1 + 2t = 3 + s

t = 1 − s

2 − t = 1 + 2s

Step 2: Solve

From the second equation:

t = 1 − s

Substitute into the first:

1 + 2(1 − s) = 3 + s

3 − 2s = 3 + s

−3s = 0

s = 0

Therefore:

t = 1

Step 3: Check the z-coordinate

2 − 1 = 1 + 2(0)

1 = 1

Correct.

Step 4: Find the intersection

Using Line 1:

x = 1 + 2(1) = 3

y = 1

z = 2 − 1 = 1

Final Answer

Intersection point = (3, 1, 1)

Parallel and Perpendicular Planes

The relationship between two planes can be determined by comparing their normal vectors.

Parallel Planes

Two planes are parallel when their normal vectors are parallel.

For planes with normals n₁ and n₂:

n₁ = kn₂

for some nonzero scalar k.

Worked Example

Plane 1:

2x + 4y − 6z = 8

Plane 2:

x + 2y − 3z = 10

The normal vectors are:

n₁ = (2, 4, −6)

n₂ = (1, 2, −3)

Since:

n₁ = 2n₂

the planes have parallel orientations.

Dividing Plane 1 by 2 gives:

x + 2y − 3z = 4

Plane 2 is:

x + 2y − 3z = 10

They have different constants.

Conclusion

The planes are distinct and parallel.

Coincident Planes

If one entire plane equation is a nonzero scalar multiple of the other, both equations describe the same plane.

For example:

2x + 4y − 6z = 8

and:

x + 2y − 3z = 4

describe the same plane.

Perpendicular Planes

Two planes are perpendicular when their normal vectors are perpendicular.

Therefore:

n₁ · n₂ = 0

Worked Example

Plane 1:

2x + y − z = 4

Plane 2:

x − y + z = 2

The normal vectors are:

n₁ = (2, 1, −1)

n₂ = (1, −1, 1)

Calculate the dot product:

n₁ · n₂ = 2(1) + 1(−1) + (−1)(1)

= 2 − 1 − 1

= 0

Conclusion

The planes are perpendicular.

Relationships Between Lines and Planes

A line and a plane may:

• Intersect at one point.

• Be parallel without intersecting.

• Have the entire line contained in the plane.

• Be perpendicular.

These relationships can be determined using the line's direction vector and the plane's normal vector.

Let:

d = direction vector of the line

n = normal vector of the plane

Line Parallel to a Plane

A line has a direction parallel to a plane when:

d · n = 0

The line may lie entirely within the plane or may be separate from it.

Line Perpendicular to a Plane

A line is perpendicular to a plane when its direction vector is parallel to the plane's normal vector.

Therefore:

d = kn

for some nonzero scalar k.

Worked Example

Consider a line with direction vector:

d = (2, 4, −2)

and a plane with normal vector:

n = (1, 2, −1)

Since:

d = 2n

the line is perpendicular to the plane.

Worked Example: Parallel Direction

A line has direction vector:

d = (2, −1, 0)

A plane has normal vector:

n = (1, 2, 3)

Calculate:

d · n = (2)(1) + (−1)(2) + (0)(3)

= 2 − 2

= 0

The line's direction is parallel to the plane.

Additional information about the line's position is needed to determine whether it lies within the plane or is separate from it.

Finding the Intersection of a Line and a Plane

To find the intersection of a line and a plane:

1. Write the parametric equations of the line.

2. Substitute them into the plane equation.

3. Solve for the line parameter.

4. Substitute the parameter back into the line equation.

5. Verify the intersection point.

Worked Example

Find the intersection of:

Line:

r = (1, 2, 3) + t(2, −1, 1)

Plane:

x + 2y + z = 10

Step 1: Write the parametric equations

x = 1 + 2t

y = 2 − t

z = 3 + t

Step 2: Substitute into the plane equation

(1 + 2t) + 2(2 − t) + (3 + t) = 10

Step 3: Expand

1 + 2t + 4 − 2t + 3 + t = 10

Step 4: Simplify

8 + t = 10

t = 2

Step 5: Find the coordinates

x = 1 + 2(2) = 5

y = 2 − 2 = 0

z = 3 + 2 = 5

Final Answer

Intersection point = (5, 0, 5)

Verification

Substitute into the plane equation:

5 + 2(0) + 5 = 10

10 = 10

The point lies on the plane.

When a Line Does Not Intersect a Plane at One Point

Sometimes substituting the line equation into the plane produces no unique solution.

Parallel but Separate

Suppose:

Line:

r = (0, 0, 2) + t(1, 1, 0)

Plane:

z = 0

The line always has:

z = 2

The plane requires:

z = 0

Therefore, the line never intersects the plane.

Line Contained in a Plane

Consider:

Line:

r = (1, 2, 0) + t(1, −1, 0)

Plane:

z = 0

Every point on the line has:

z = 0

Therefore, every point on the line lies in the plane.

Important Principle

Substitution into the plane equation may produce:

• One solution: a single intersection point.

• No solution: the line is parallel and separate.

• An identity true for every parameter value: the line lies entirely in the plane.

Finding the Intersection of Two Planes

Two distinct nonparallel planes intersect along a line.

The direction vector of their intersection line is perpendicular to both plane normals.

Therefore:

d = n₁ × n₂

Worked Example

Find the line of intersection of:

Plane 1:

x + y + z = 6

Plane 2:

x − y + z = 2

Step 1: Subtract the equations

(x + y + z) − (x − y + z) = 6 − 2

2y = 4

y = 2

Step 2: Substitute into Plane 1

x + 2 + z = 6

x + z = 4

Step 3: Introduce a parameter

Let:

z = t

Then:

x = 4 − t

Step 4: Write the parametric equations

x = 4 − t

y = 2

z = t

Final Answer

r = (4, 2, 0) + t(−1, 0, 1)

Verification Using Cross Products

The plane normals are:

n₁ = (1, 1, 1)

n₂ = (1, −1, 1)

Their cross product is:

n₁ × n₂ = (2, 0, −2)

This is a scalar multiple of:

(−1, 0, 1)

Therefore, the direction vector is correct.

Distance from a Point to a Plane

The perpendicular distance from a point to a plane can be calculated using the plane equation.

For a plane:

ax + by + cz + d = 0

and a point:

P = (x₀, y₀, z₀)

the distance is:

Distance = |ax₀ + by₀ + cz₀ + d| / √(a² + b² + c²)

The absolute value ensures that distance is non-negative.

Worked Example

Find the distance from:

P = (1, 2, 3)

to the plane:

2x − y + 2z − 5 = 0

Step 1: Identify the coefficients

a = 2

b = −1

c = 2

d = −5

Step 2: Substitute

Distance = |2(1) − 1(2) + 2(3) − 5| / √(2² + (−1)² + 2²)

Step 3: Simplify the numerator

|2 − 2 + 6 − 5| = |1| = 1

Step 4: Simplify the denominator

√(4 + 1 + 4) = √9 = 3

Final Answer

Distance = 1/3 unit

Distance Between Parallel Planes

For parallel planes written with the same normal coefficients:

ax + by + cz = d₁

ax + by + cz = d₂

the perpendicular distance between them is:

Distance = |d₂ − d₁| / √(a² + b² + c²)

Worked Example

Find the distance between:

2x + 2y + z = 4

and:

2x + 2y + z = 13

Step 1: Identify the constants

d₁ = 4

d₂ = 13

Step 2: Calculate the difference

|13 − 4| = 9

Step 3: Calculate the normal magnitude

√(2² + 2² + 1²)

= √9

= 3

Step 4: Calculate the distance

Distance = 9/3

Final Answer

Distance = 3 units

Angles Between Lines and Planes

Dot products can be used to determine angles involving lines and planes.

Angle Between Two Lines

For two nonzero direction vectors d₁ and d₂:

cos θ = (d₁ · d₂)/(|d₁||d₂|)

This calculates the angle between the chosen direction vectors.

For the smaller angle between the geometric lines, use:

cos θ = |d₁ · d₂|/(|d₁||d₂|)

where 0° ≤ θ ≤ 90°.

Angle Between Two Planes

The angle between two planes is determined by the angle between their normal vectors.

For the smaller angle:

cos θ = |n₁ · n₂|/(|n₁||n₂|)

Angle Between a Line and a Plane

The angle between a line and a plane is complementary to the smaller angle between the line's direction vector and the plane's normal.

Therefore:

sin θ = |d · n|/(|d||n|)

where θ is the acute angle between the line and plane, or 0° or 90° in the limiting cases.

Worked Example

A line has direction vector:

d = (1, 0, 1)

A plane has normal vector:

n = (0, 0, 1)

Find the angle between the line and plane.

Step 1: Calculate the dot product

d · n = 1

Step 2: Calculate the magnitudes

|d| = √2

|n| = 1

Step 3: Apply the formula

sin θ = 1/√2

Step 4: Calculate

θ = 45°

The line forms a 45° angle with the plane.

Real-World Application: Architecture and Engineering

Architects and engineers use three-dimensional geometry to model structural components, surfaces, and intersections.

Examples include:

• Roof planes.

• Support beams.

• Walls and floors.

• Structural joints.

• Bridge components.

• Inclined surfaces.

Worked Example: A Support Beam Intersecting a Roof

A beam follows the line:

r = (0, 0, 1) + t(2, 1, 3)

A roof surface is represented by:

z = 7

Determine where the beam meets the roof.

Step 1: Write the beam equations

x = 2t

y = t

z = 1 + 3t

Step 2: Use the roof equation

1 + 3t = 7

3t = 6

t = 2

Step 3: Find the coordinates

x = 2(2) = 4

y = 2

z = 7

Final Answer

The beam intersects the roof at (4, 2, 7).

This type of calculation helps determine the locations of structural connections.

Real-World Application: Computer Graphics

Computer graphics systems use lines and planes to represent:

• Camera rays.

• Walls and floors.

• Object surfaces.

• Lighting directions.

• Collision boundaries.

• Three-dimensional models.

A camera ray can be represented by a line equation:

r = a + td

An object's surface may be represented by a plane.

To determine where the ray strikes the surface, the line equation is substituted into the plane equation.

This process is called ray-plane intersection.

Important Observation

In ray tracing, a ray usually requires:

t ≥ 0

This ensures that the calculated intersection lies in front of the ray's starting point rather than behind it.

A mathematical line extends infinitely in both directions, but a ray extends in only one direction.

Real-World Application: Navigation and Robotics

Robots and autonomous systems use three-dimensional coordinates to determine positions and movement paths.

A movement path can be modelled using:

r = a + td

Surfaces and obstacles can be represented using plane equations.

These models allow systems to calculate:

• Where a movement path intersects a surface.

• Whether a path is parallel to an obstacle.

• The distance from a position to a wall.

• The orientation of a surface.

• The angle between a movement direction and a surface.

Example

A robotic tool follows:

r = (1, 2, 3) + t(0, 0, −1)

A work surface is:

z = 0

Find the point where the tool's path meets the surface.

The z-coordinate is:

z = 3 − t

Set z = 0:

3 − t = 0

t = 3

Therefore:

Intersection = (1, 2, 0)

This represents the position where the tool reaches the work surface.

Real-World Application: Aircraft Navigation

Aircraft movement can be approximated using vector equations over short periods when velocity is nearly constant.

Suppose an aircraft begins at:

P = (10, 20, 5)

and follows a displacement direction:

d = (3, 4, 1)

Its path can be modelled by:

r = (10, 20, 5) + t(3, 4, 1)

A horizontal altitude surface can be represented by:

z = constant

An inclined surface can be represented by a more general plane equation.

These mathematical models can be used to study path intersections, distances, and orientations.

Real aircraft navigation requires more sophisticated models that account for changing velocity, atmospheric conditions, and other factors.

Solving Multi-Step Three-Dimensional Geometry Problems

More advanced problems combine line equations, plane equations, cross products, dot products, and intersection calculations.

Worked Example: Line Through Two Points and Plane Intersection

A line passes through:

A = (1, 2, 0)

B = (3, 5, 4)

A plane has equation:

x + y + z = 15

Find the intersection point.

Step 1: Find the line direction vector

AB = B − A

AB = (2, 3, 4)

Step 2: Write the line equation

r = (1, 2, 0) + t(2, 3, 4)

Step 3: Write the parametric equations

x = 1 + 2t

y = 2 + 3t

z = 4t

Step 4: Substitute into the plane

(1 + 2t) + (2 + 3t) + 4t = 15

Step 5: Simplify

3 + 9t = 15

9t = 12

t = 4/3

Step 6: Calculate the intersection coordinates

x = 1 + 2(4/3)

x = 11/3

y = 2 + 3(4/3)

y = 6

z = 4(4/3)

z = 16/3

Final Answer

Intersection = (11/3, 6, 16/3)

Verification

11/3 + 6 + 16/3

= 27/3 + 6

= 9 + 6

= 15

The point satisfies the plane equation.

Comparing Lines and Planes

Important Distinction

A line has one independent direction.

A plane has two independent directions.

A plane's normal vector is perpendicular to both directions lying within the plane.

A Systematic Method for Solving Lines and Planes Problems

Identify the Geometric Objects

Determine whether the problem involves points, lines, planes, or combinations of these objects.

Identify the Given Information

Look for:

• Known points.

• Direction vectors.

• Normal vectors.

• Cartesian equations.

• Parametric equations.

Choose an Appropriate Representation

Use vector equations for lines and planes when direction information is important.

Use Cartesian plane equations when substituting coordinates or finding intersections.

Determine Direction Relationships

Use scalar multiples to test parallelism.

Use dot products to test perpendicularity.

Use cross products to calculate plane normals or intersection directions.

Solve Algebraically

Equate coordinate expressions or substitute parametric equations into Cartesian equations.

Check All Conditions

In three-dimensional geometry, all three coordinates must satisfy the relevant equations.

Interpret the Result

Explain whether the objects intersect, are parallel, are perpendicular, coincide, or are skew.

Verify

Substitute calculated points into the original equations whenever possible.

Common Mistakes

Confusing position vectors and direction vectors

A position vector identifies a point relative to the origin, while a direction vector describes the orientation of a line.

Using a zero direction vector

A valid line equation requires a nonzero direction vector.

Using parallel direction vectors to define a plane

A plane requires two nonparallel direction vectors.

Confusing a plane's normal vector with a direction vector in the plane

The normal vector is perpendicular to every direction in the plane.

Assuming all nonparallel lines intersect

In three dimensions, nonparallel lines may be skew.

Assuming perpendicular direction vectors guarantee intersecting lines

The lines must actually intersect to be perpendicular as geometric lines.

Failing to check all three coordinates

An apparent line intersection must satisfy x-, y-, and z-coordinate equations simultaneously.

Using different parameter values for coordinates of the same point on one line

One common parameter value must satisfy all coordinates of a point on a given line.

Confusing parallel and coincident planes

Parallel normals do not necessarily mean the planes are identical.

Assuming a line with direction parallel to a plane never intersects it

The line may lie entirely within the plane.

Confusing line-plane perpendicularity with line-plane parallelism

A line perpendicular to a plane has a direction parallel to the plane's normal.

A line with direction parallel to the plane has a direction perpendicular to the plane's normal.

Forgetting to use absolute values in distance formulas

Distance cannot be negative.

Using an incorrect cross product

A sign error in the cross product can produce an incorrect plane normal.

Ignoring special cases in symmetric equations

A zero direction component produces a constant coordinate, not a fraction with zero denominator.

Assuming every pair of planes intersects in a line

Parallel planes may be distinct or coincident.

Failing to verify calculated intersection points

Substitution into the original equations helps detect algebraic errors.

Key Terms

• Three-Dimensional Space: A geometric system described using three independent coordinate directions.

• Cartesian Coordinate System: A system that locates points using coordinates along perpendicular axes.

• Origin: The point (0, 0, 0) where the coordinate axes meet.

• Position Vector: A vector describing a point's location relative to the origin.

• Direction Vector: A nonzero vector describing the direction of a line or movement.

• Line: A one-dimensional geometric object extending infinitely in two directions.

• Plane: A flat two-dimensional surface extending infinitely in three-dimensional space.

• Vector Equation of a Line: An equation expressing points on a line using a known point and direction vector.

• Parametric Equation: An equation describing coordinates in terms of one or more parameters.

• Parameter: A variable used to generate points on a line or plane.

• Symmetric Equation: A line equation obtained by eliminating the parameter from parametric equations.

• Vector Equation of a Plane: An equation using a known point and two nonparallel direction vectors.

• Cartesian Equation of a Plane: An equation of the form ax + by + cz = d.

• Normal Vector: A vector perpendicular to a plane.

• Point-Normal Form: A plane equation constructed from a known point and normal vector.

• Dot Product: An operation producing a scalar from two vectors.

• Cross Product: An operation producing a vector perpendicular to two given vectors.

• Parallel Lines: Distinct lines with parallel direction vectors that do not intersect.

• Perpendicular Lines: Lines that intersect at a right angle.

• Skew Lines: Lines in three dimensions that are neither parallel nor intersecting.

• Parallel Planes: Distinct planes with parallel normal vectors.

• Perpendicular Planes: Planes whose normal vectors are perpendicular.

• Coincident Planes: Different equations describing the same plane.

• Intersection: A point or set of points shared by geometric objects.

• Line-Plane Intersection: A point or line shared by a line and a plane.

• Plane-Plane Intersection: A line shared by two distinct nonparallel planes.

• Scalar Multiple: A vector obtained by multiplying another vector by a number.

• Collinear Points: Points lying on the same straight line.

• Non-Collinear Points: Points that do not all lie on one straight line.

• Orthogonal Vectors: Vectors whose dot product is zero.

• Perpendicular Distance: The shortest distance from a point to a line or plane.

• Ray: A geometric object beginning at a point and extending indefinitely in one direction.

• Ray-Plane Intersection: The point where a ray meets a plane, when such a point exists.

• Geometric Model: A mathematical representation of a geometric or physical situation.

• Verification: Checking that a result satisfies the original equations or geometric conditions.

Key Takeaways

• A line in three-dimensional space can be represented using a known point and a nonzero direction vector.

• The vector equation of a line is r = a + td.

• The parameter t determines the position of a point on the line.

• A line through two distinct points uses their coordinate difference as a direction vector.

• Parametric equations describe each coordinate separately.

• Symmetric equations can be used when the relevant direction components are nonzero.

• A plane is a two-dimensional surface in three-dimensional space.

• A plane can be described using a point and a normal vector.

• The Cartesian equation of a plane is ax + by + cz = d.

• The vector (a, b, c) is normal to the plane.

• A plane can also be described using a point and two nonparallel direction vectors.

• Three non-collinear points determine a unique plane.

• The cross product can be used to find a normal vector to a plane.

• Two lines have parallel directions when their direction vectors are scalar multiples.

• Two intersecting lines are perpendicular when their direction vectors have a zero dot product.

• Skew lines are nonparallel lines that do not intersect.

• Two planes have parallel orientations when their normal vectors are parallel.

• Two planes are perpendicular when their normal vectors have a zero dot product.

• A line is perpendicular to a plane when its direction vector is parallel to the plane's normal vector.

• A line has direction parallel to a plane when its direction vector is perpendicular to the plane's normal vector.

• Line-plane intersections are found by substituting parametric equations into the plane equation.

• Two distinct nonparallel planes intersect along a line.

• The cross product of two plane normals gives a direction vector for their intersection line.

• Distance formulas allow us to calculate perpendicular distances from points to planes and between parallel planes.

• Dot products can be used to calculate angles between lines and planes.

• Lines and planes are important mathematical models in engineering, architecture, navigation, robotics, and computer graphics.

• The central skill is using vector equations, dot products, cross products, and algebraic methods to describe and analyse geometric relationships in three dimensions.

Check Your Understanding

Recall and Understanding

1. Explain what information is needed to write the vector equation of a line.

2. Describe the difference between a direction vector and a normal vector.

3. State the vector equation of a line and explain the meaning of each symbol.

4. Explain the difference between parallel, perpendicular, and skew lines.

5. Describe how the cross product can be used to determine the equation of a plane.

Application

6. Write the vector equation of a line passing through P = (2, 1, 3) with direction vector d = (4, −1, 2).

7. Find the vector equation of the line passing through A = (1, 2, 0) and B = (4, 6, 3).

8. Determine whether the point P = (7, 5, 9) lies on the line r = (1, 2, 3) + t(2, 1, 2).

9. Find the equation of a plane passing through P = (1, 2, 3) with normal vector n = (2, −1, 4).

10. Determine whether the planes 2x + 4y − 2z = 6 and x + 2y − z = 5 are parallel, perpendicular, or coincident.

Analysis and Challenge

11. Find the intersection of the line r = (1, 0, 2) + t(2, 1, −1) with the plane x + y + z = 8.

12. Determine whether the lines r = (0, 0, 0) + t(1, 2, 3) and r = (1, 1, 1) + s(2, 4, 6) are parallel, coincident, or intersecting.

13. Find the equation of the plane through P = (1, 0, 0), Q = (0, 2, 0), and R = (0, 0, 4).

14. Calculate the perpendicular distance from P = (2, 3, 1) to the plane 2x − y + 2z − 7 = 0.

15. Find the line of intersection of the planes x + y + z = 6 and x − y + z = 2. Express your answer as a vector equation.

Final Summary

Lines and planes form the foundation of three-dimensional coordinate geometry.

Vectors provide a systematic method for describing their positions, directions, and orientations.

A line is defined using a known point and a direction vector, while a plane can be defined using a point and a normal vector or two nonparallel direction vectors.

Dot products allow us to investigate perpendicular relationships and angles, while cross products allow us to calculate normal vectors and intersection directions.

Algebraic methods make it possible to find intersections between lines, between lines and planes, and between planes.

These techniques also allow us to calculate distances and solve practical geometric problems in three-dimensional space.

The central principle is that vector equations provide a powerful and consistent way to describe lines and planes, determine their relationships, and solve geometric problems involving position, direction, intersection, and distance in three dimensions.