The Central Idea
A geometric object can often be described by a starting position together with allowed directions of motion.
For a line:
\[ \boxed{ \mathbf{r} = \mathbf{r}_0+t\mathbf{v} } \]where:
- \(\mathbf{r}_0\) is the position vector of a known point;
- \(\mathbf{v}\) is a nonzero direction vector;
- \(t\) is a scalar parameter.
Why One Direction Generates a Line
Start from a fixed point with position vector
\[ \mathbf{r}_0. \]Multiplying a direction vector \(\mathbf{v}\) by every real value of \(t\) produces all scalar multiples of \(\mathbf{v}\).
Adding those displacements to \(\mathbf{r}_0\) moves through every point on the line.
The Parameter \(t\)
The parameter
\[ t \]controls how far and in which direction we move from the known point.
- \(t=0\) gives the starting point.
- \(t>0\) moves in the direction of \(\mathbf{v}\).
- \(t<0\) moves in the opposite direction.
Example: A Line in the Plane
Suppose a line passes through
\[ P=(2,1) \]and has direction vector
\[ \mathbf{v}=\langle 3,4\rangle. \]The position vector of \(P\) is
\[ \mathbf{r}_0=\langle 2,1\rangle. \]Therefore,
\[ \boxed{ \mathbf{r} = \langle 2,1\rangle + t\langle 3,4\rangle }. \]Expanding the Vector Equation
From
\[ \mathbf{r} = \langle 2,1\rangle + t\langle 3,4\rangle, \]scalar multiplication gives
\[ t\langle 3,4\rangle = \langle 3t,4t\rangle. \]Therefore,
\[ \mathbf{r} = \langle 2+3t,\;1+4t\rangle. \]Parametric Equations of a Line
If
\[ \mathbf{r} = \langle x,y\rangle, \]then
\[ \langle x,y\rangle = \langle 2+3t,\;1+4t\rangle. \]Matching components gives
\[ \boxed{ x=2+3t } \]and
\[ \boxed{ y=1+4t }. \]These are the parametric equations of the line.
Vector Form and Parametric Form Say the Same Thing
The vector equation
\[ \mathbf{r} = \mathbf{r}_0+t\mathbf{v} \]packages all coordinate equations into one statement.
The parametric form separates that statement into one scalar equation for each coordinate.
General Line Form in Two Dimensions
If a line passes through
\[ P=(x_0,y_0) \]with direction vector
\[ \mathbf{v}=\langle a,b\rangle, \]then
\[ \boxed{ \mathbf{r} = \langle x_0,y_0\rangle + t\langle a,b\rangle }. \]Its parametric equations are
\[ \boxed{ x=x_0+at,\qquad y=y_0+bt }. \]General Line Form in Three Dimensions
If a line passes through
\[ P=(x_0,y_0,z_0) \]with nonzero direction vector
\[ \mathbf{v}=\langle a,b,c\rangle, \]then
\[ \boxed{ \mathbf{r} = \langle x_0,y_0,z_0\rangle + t\langle a,b,c\rangle }. \]Three-Dimensional Parametric Equations
The vector equation
\[ \mathbf{r} = \langle x_0,y_0,z_0\rangle + t\langle a,b,c\rangle \]is equivalent to
\[ \boxed{ x=x_0+at } \] \[ \boxed{ y=y_0+bt } \] \[ \boxed{ z=z_0+ct }. \]Example: A Line in Three Dimensions
A line passes through
\[ P=(1,-2,4) \]with direction vector
\[ \mathbf{v}=\langle 2,3,-1\rangle. \]Its vector equation is
\[ \boxed{ \mathbf{r} = \langle 1,-2,4\rangle + t\langle 2,3,-1\rangle }. \]Parametric Form of the Example
Expanding component-wise gives
\[ \boxed{ x=1+2t } \] \[ \boxed{ y=-2+3t } \] \[ \boxed{ z=4-t }. \]Finding a Direction Vector from Two Points
If a line passes through
\[ A=(x_1,y_1,z_1) \]and
\[ B=(x_2,y_2,z_2), \]then a direction vector is
\[ \boxed{ \overrightarrow{AB} = \langle x_2-x_1,\; y_2-y_1,\; z_2-z_1 \rangle }. \]Example: Line through Two Points
Let
\[ A=(1,2,3) \]and
\[ B=(5,0,4). \]Then
\[ \overrightarrow{AB} = \langle 4,-2,1\rangle. \]A vector equation is
\[ \boxed{ \mathbf{r} = \langle 1,2,3\rangle + t\langle 4,-2,1\rangle }. \]A Direction Vector Is Not Unique
If
\[ \mathbf{v} \]is a direction vector for a line, then any nonzero scalar multiple
\[ k\mathbf{v} \]is also a direction vector for the same line.
For example,
\[ \langle 4,-2,1\rangle \]and
\[ \langle 8,-4,2\rangle \]determine the same line direction.
A Point on the Line Is Not Unique Either
Any point already lying on the line may be used as
\[ \mathbf{r}_0. \]Different point-direction pairs can therefore produce different-looking vector equations for the same geometric line.
Checking Whether a Point Lies on a Line
Suppose
\[ \mathbf{r} = \langle 1,2,3\rangle + t\langle 4,-2,1\rangle. \]To test a point, substitute its coordinates into the parametric equations and determine whether one common value of \(t\) satisfies all coordinates.
Example: Check a Point on a Line
Test
\[ Q=(9,-2,5) \]on
\[ x=1+4t,\qquad y=2-2t,\qquad z=3+t. \]From the first coordinate,
\[ 9=1+4t \]gives
\[ t=2. \]Complete the Point Check
Using \(t=2\):
\[ y=2-2(2)=-2 \]and
\[ z=3+2=5. \]Both coordinates match.
Therefore,
\[ \boxed{ Q=(9,-2,5) \text{ lies on the line} }. \]Symmetric Form of a Line
If
\[ x=x_0+at,\qquad y=y_0+bt,\qquad z=z_0+ct \]and \(a\), \(b\), and \(c\) are all nonzero, solve each equation for \(t\):
\[ \frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}. \]Thus,
\[ \boxed{ \frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c} }. \]Example: Convert to Symmetric Form
Suppose
\[ x=1+2t, \qquad y=-2+3t, \qquad z=4-t. \]Then
\[ t=\frac{x-1}{2}, \] \[ t=\frac{y+2}{3}, \] \[ t=4-z. \]Therefore,
\[ \boxed{ \frac{x-1}{2} = \frac{y+2}{3} = 4-z }. \]Zero Components in a Direction Vector
Symmetric form must be handled carefully when a direction component is zero.
If
\[ \mathbf{v}=\langle 2,0,-3\rangle, \]then the line has
\[ y=y_0 \]as a constant coordinate.
We do not divide by the zero direction component.
Parallel Lines
Two lines are parallel when their direction vectors are scalar multiples.
If
\[ \mathbf{v}_2=k\mathbf{v}_1 \]for some nonzero scalar \(k\), then the line directions are parallel.
The lines may be distinct or may coincide.
Coincident Lines
Parallel direction vectors are not enough to prove that two lines are the same line.
To be coincident, one known point from one line must also lie on the other line.
Perpendicular Line Directions
Direction vectors
\[ \mathbf{u} \]and
\[ \mathbf{v} \]are perpendicular when
\[ \boxed{ \mathbf{u}\cdot\mathbf{v}=0 }. \]The dot product therefore provides a direct test for perpendicular line directions.
From Lines to Planes
One direction is enough to generate a line.
A plane requires two independent directions.
Starting from a fixed point, we may move by any scalar multiple of one direction and any scalar multiple of another.
Vector Equation of a Plane
Let
\[ \mathbf{r}_0 \]be the position vector of a point on a plane.
Let
\[ \mathbf{u} \]and
\[ \mathbf{v} \]be two nonparallel direction vectors lying in the plane.
Then
\[ \boxed{ \mathbf{r} = \mathbf{r}_0 + s\mathbf{u} + t\mathbf{v} }. \]Why a Plane Needs Two Parameters
The parameter \(s\) controls motion in the direction of \(\mathbf{u}\).
The parameter \(t\) controls motion in the direction of \(\mathbf{v}\).
Because the two directions are not parallel, varying both parameters fills a two-dimensional plane.
Example: Vector Equation of a Plane
Suppose a plane passes through
\[ P=(1,2,3) \]and contains the directions
\[ \mathbf{u}=\langle 1,0,2\rangle \]and
\[ \mathbf{v}=\langle 0,3,-1\rangle. \]Then
\[ \boxed{ \mathbf{r} = \langle 1,2,3\rangle + s\langle 1,0,2\rangle + t\langle 0,3,-1\rangle }. \]Parametric Equations of the Plane
Expand:
\[ \mathbf{r} = \langle 1+s,\; 2+3t,\; 3+2s-t \rangle. \]Therefore,
\[ \boxed{x=1+s} \] \[ \boxed{y=2+3t} \] \[ \boxed{z=3+2s-t}. \]The Two Plane Directions Must Be Nonparallel
If
\[ \mathbf{v}=k\mathbf{u}, \]then
\[ \mathbf{r} = \mathbf{r}_0+s\mathbf{u}+t\mathbf{v} \]reduces to motion along only one direction.
The result is a line, not a plane.
A Plane through Three Points
Suppose a plane passes through three noncollinear points
\[ A,\qquad B,\qquad C. \]Two direction vectors in the plane are
\[ \overrightarrow{AB} \]and
\[ \overrightarrow{AC}. \]Therefore,
\[ \boxed{ \mathbf{r} = \mathbf{a} + s\overrightarrow{AB} + t\overrightarrow{AC} }. \]Example: Plane through Three Points
Let
\[ A=(1,0,2), \qquad B=(3,1,2), \qquad C=(1,2,5). \]Then
\[ \overrightarrow{AB} = \langle 2,1,0\rangle \]and
\[ \overrightarrow{AC} = \langle 0,2,3\rangle. \]Complete the Three-Point Plane Equation
Using \(A\) as the base point,
\[ \boxed{ \mathbf{r} = \langle 1,0,2\rangle + s\langle 2,1,0\rangle + t\langle 0,2,3\rangle }. \]Because the two direction vectors are not scalar multiples, they span a plane.
Normal Vectors
A normal vector to a plane is a nonzero vector perpendicular to every direction lying in the plane.
We often write the normal vector as
\[ \mathbf{n}. \]If
\[ \mathbf{n}=\langle A,B,C\rangle, \]then \(\mathbf{n}\) determines the plane's orientation.
Point-Normal Form of a Plane
Suppose a plane passes through a point with position vector
\[ \mathbf{r}_0 \]and has normal vector
\[ \mathbf{n}. \]For any point \(\mathbf{r}\) on the plane, the displacement
\[ \mathbf{r}-\mathbf{r}_0 \]lies in the plane and is therefore perpendicular to \(\mathbf{n}\).
Hence,
\[ \boxed{ \mathbf{n}\cdot(\mathbf{r}-\mathbf{r}_0)=0 }. \]Coordinate Form of the Point-Normal Equation
Let
\[ \mathbf{n}=\langle A,B,C\rangle, \] \[ \mathbf{r}=\langle x,y,z\rangle, \]and
\[ \mathbf{r}_0=\langle x_0,y_0,z_0\rangle. \]Then
\[ \boxed{ A(x-x_0) + B(y-y_0) + C(z-z_0) = 0 }. \]Standard Scalar Equation of a Plane
Expanding the point-normal form gives
\[ Ax+By+Cz=D \]for some constant \(D\).
Thus the standard scalar equation is
\[ \boxed{ Ax+By+Cz=D }. \]The vector
\[ \langle A,B,C\rangle \]is normal to the plane.
Example: Plane from a Point and Normal
Suppose a plane passes through
\[ P=(1,2,-1) \]with normal vector
\[ \mathbf{n}=\langle 2,-3,4\rangle. \]Point-normal form is
\[ 2(x-1)-3(y-2)+4(z+1)=0. \]Convert the Plane to Standard Form
Expand:
\[ 2x-2-3y+6+4z+4=0. \]Combine constants:
\[ 2x-3y+4z+8=0. \]Therefore,
\[ \boxed{ 2x-3y+4z=-8 }. \]Reading a Normal Vector from a Plane Equation
For
\[ 2x-3y+4z=-8, \]a normal vector is
\[ \boxed{ \mathbf{n} = \langle 2,-3,4\rangle }. \]Any nonzero scalar multiple of this normal is also a valid normal vector.
Finding a Normal from Two Plane Directions
If a plane is given by two nonparallel direction vectors
\[ \mathbf{u} \]and
\[ \mathbf{v}, \]then a vector perpendicular to both can serve as a normal vector.
In three dimensions, a cross product
\[ \mathbf{u}\times\mathbf{v} \]provides such a vector when the directions are not parallel.
Using a Known Normal without Computing a Cross Product
Sometimes a normal vector is already given or can be recognized directly.
If both plane directions satisfy
\[ \mathbf{n}\cdot\mathbf{u}=0 \]and
\[ \mathbf{n}\cdot\mathbf{v}=0, \]then \(\mathbf{n}\) is perpendicular to both directions and is therefore normal to the plane they span.
Checking Whether a Point Lies on a Plane
If a plane is written
\[ Ax+By+Cz=D, \]substitute the coordinates of the point.
If the equation is satisfied, the point lies on the plane.
Example: Check a Point on a Plane
Test
\[ Q=(2,0,-3) \]on
\[ 2x-3y+4z=-8. \]Substitute:
\[ 2(2)-3(0)+4(-3) = 4-12 = -8. \]Therefore, \(Q\) lies on the plane.
Parallel Planes
Two planes are parallel when their normal vectors are scalar multiples.
For example,
\[ 2x-y+3z=4 \]and
\[ 4x-2y+6z=7 \]have parallel normal vectors:
\[ \langle 2,-1,3\rangle \]and
\[ \langle 4,-2,6\rangle. \]Coincident Planes
Parallel normals do not automatically mean that the planes are distinct.
If one scalar equation is a nonzero scalar multiple of the other, then the equations describe the same plane.
Perpendicular Planes
Two planes are perpendicular when their normal vectors are perpendicular.
Thus, if
\[ \mathbf{n}_1\cdot\mathbf{n}_2=0, \]the planes meet at a right dihedral angle.
A Line Parallel to a Plane
Let a line have direction vector
\[ \mathbf{v} \]and let the plane have normal vector
\[ \mathbf{n}. \]If
\[ \boxed{ \mathbf{v}\cdot\mathbf{n}=0 }, \]then the line direction is parallel to the plane.
The line may lie in the plane or may be a distinct parallel line.
A Line Perpendicular to a Plane
A line is perpendicular to a plane when its direction vector is parallel to the plane's normal vector.
If
\[ \mathbf{v}=k\mathbf{n} \]for some nonzero scalar \(k\), the line meets the plane orthogonally.
Intersection of a Line and a Plane
A line and a plane may:
- intersect at one point;
- have no intersection because the line is parallel to the plane;
- have infinitely many common points because the line lies in the plane.
Substitution provides an algebraic way to determine which case occurs.
Example: Intersect a Line with a Plane
Consider the line
\[ \mathbf{r} = \langle 1,0,2\rangle + t\langle 1,2,-1\rangle \]and the plane
\[ x+y+z=6. \]The line has parametric equations
\[ x=1+t,\qquad y=2t,\qquad z=2-t. \]Complete the Line-Plane Intersection
Substitute into the plane:
\[ (1+t)+2t+(2-t)=6. \]Simplify:
\[ 3+2t=6. \]Therefore,
\[ t=\frac32. \]Substitute back:
\[ \boxed{ \left( \frac52,\; 3,\; \frac12 \right) }. \]Intersection of Two Planes
Two distinct nonparallel planes intersect in a line.
Their normal vectors determine the orientations of the planes.
A direction vector for their intersection line must be perpendicular to both normals.
In three dimensions, such a direction may be obtained from a cross product of the two normals.
Why Plane Equations Use Normals
A single direction vector cannot determine a plane because infinitely many planes contain that direction.
A normal vector behaves differently: once a point and a normal are fixed, the plane's orientation is determined.
Vector Form versus Normal Form
A plane can be described in two complementary ways.
| Form | Main Data | Geometric Idea |
|---|---|---|
| Vector form | Point + two in-plane directions | Generate the plane by moving within it |
| Point-normal form | Point + normal vector | Require every in-plane displacement to be perpendicular to the normal |
Common Error: Using a Point as a Direction Vector
A point identifies a location.
A direction vector identifies a directed change.
Although their coordinates may look similar, their mathematical roles are different.
Common Error: Using the Zero Vector as a Line Direction
A direction vector must be nonzero.
If
\[ \mathbf{v}=\mathbf{0}, \]then
\[ \mathbf{r} = \mathbf{r}_0+t\mathbf{0} = \mathbf{r}_0 \]describes only one point, not a line.
Common Error: Using Two Parallel Vectors to Define a Plane
A plane requires two independent in-plane directions.
If
\[ \mathbf{v}=k\mathbf{u}, \]then the two vectors provide only one direction.
Their linear combinations generate a line rather than a plane.
Common Error: Forgetting One Common Parameter for a Line
In
\[ x=x_0+at,\qquad y=y_0+bt,\qquad z=z_0+ct, \]the same value of \(t\) must satisfy all coordinate equations for a given point.
Using unrelated parameter values would describe a larger set than one line.
Common Error: Dividing by Zero in Symmetric Form
If a direction component is zero, do not write it in a denominator.
Instead, the corresponding coordinate remains constant.
For example, if
\[ y=y_0, \]retain that equation separately.
Common Error: Confusing Plane Directions with the Normal
Direction vectors in the plane lie along the plane.
The normal vector points perpendicular to the plane.
Therefore, a normal vector is not generally one of the plane's spanning directions.
Common Error: Reading the Wrong Normal from a Plane Equation
For
\[ Ax+By+Cz=D, \]the normal vector is
\[ \boxed{ \langle A,B,C\rangle }. \]The constant \(D\) is not a component of the normal vector.
Common Error: Assuming Parallel Normals Mean Distinct Planes
Parallel normals mean the planes are parallel in orientation.
The planes may still coincide.
Additional comparison of the equations or a point-membership test is needed to distinguish coincident from distinct parallel planes.
A Reliable Procedure for Writing a Line Equation
- Identify one point on the line.
- Find or identify a nonzero direction vector.
- Write the point as a position vector \(\mathbf{r}_0\).
- Use \(\mathbf{r}=\mathbf{r}_0+t\mathbf{v}\).
- Expand into parametric equations if needed.
- Verify that known points satisfy one common parameter value.
A Reliable Procedure for a Line through Two Points
- Choose one point as the base point.
- Subtract coordinates to form the direction vector between the points.
- Confirm that the direction vector is nonzero.
- Insert the point and direction into vector form.
- Check the second point using a parameter value.
A Reliable Procedure for Writing a Plane from a Point and Two Directions
- Identify one point on the plane.
- Identify two nonparallel direction vectors in the plane.
- Write the point as \(\mathbf{r}_0\).
- Use \(\mathbf{r}=\mathbf{r}_0+s\mathbf{u}+t\mathbf{v}\).
- Expand component-wise if parametric equations are required.
- Check that the two directions are not scalar multiples.
A Reliable Procedure for Writing a Plane from a Point and Normal
- Identify the point \((x_0,y_0,z_0)\).
- Identify the nonzero normal vector \(\langle A,B,C\rangle\).
- Write \(A(x-x_0)+B(y-y_0)+C(z-z_0)=0\).
- Expand if standard scalar form is requested.
- Substitute the known point to verify the equation.
Example: Full Line Workflow
Find a vector equation for the line through
\[ A=(-1,2,0) \]and
\[ B=(3,5,-2). \]First find the direction:
\[ \overrightarrow{AB} = \langle 3-(-1),\; 5-2,\; -2-0 \rangle. \]Therefore,
\[ \overrightarrow{AB} = \langle 4,3,-2\rangle. \]Complete the Full Line Workflow
Use \(A\) as the base point:
\[ \boxed{ \mathbf{r} = \langle -1,2,0\rangle + t\langle 4,3,-2\rangle }. \]The parametric equations are
\[ \boxed{ x=-1+4t,\qquad y=2+3t,\qquad z=-2t }. \]Example: Full Plane Workflow
Find a plane through
\[ P=(2,-1,3) \]with normal vector
\[ \mathbf{n}=\langle 1,2,-4\rangle. \]Point-normal form is
\[ (x-2)+2(y+1)-4(z-3)=0. \]Complete the Full Plane Workflow
Expand:
\[ x-2+2y+2-4z+12=0. \]Therefore,
\[ x+2y-4z+12=0. \]Equivalently,
\[ \boxed{ x+2y-4z=-12 }. \]Practice 1: Write a Line Equation
Write a vector equation for the line through
\[ P=(2,-1) \]with direction vector
\[ \mathbf{v}=\langle 4,3\rangle. \]Practice 2: Convert to Parametric Form
Convert
\[ \mathbf{r} = \langle 1,2,3\rangle + t\langle -2,4,5\rangle \]to parametric equations.
Practice 3: Line through Two Points
Write a vector equation for the line through
\[ A=(0,1,2) \]and
\[ B=(3,5,-1). \]Practice 4: Check a Point on a Line
Determine whether
\[ Q=(7,9) \]lies on
\[ x=1+2t,\qquad y=3+2t. \]Practice 5: Vector Equation of a Plane
Write a vector equation for the plane through
\[ P=(1,0,2) \]with direction vectors
\[ \mathbf{u}=\langle 1,1,0\rangle \]and
\[ \mathbf{v}=\langle 0,1,2\rangle. \]Practice 6: Plane from Point and Normal
Find the scalar equation of the plane through
\[ P=(2,1,-1) \]with normal vector
\[ \mathbf{n}=\langle 3,-2,1\rangle. \]Practice 7: Read a Normal Vector
Give a normal vector for the plane
\[ 5x+2y-7z=9. \]Practice 8: Check a Point on a Plane
Determine whether
\[ Q=(1,2,3) \]lies on
\[ x+2y-z=2. \]Practice 9: Parallel or Perpendicular Planes?
Compare the planes
\[ x+2y-z=4 \]and
\[ 2x+4y-2z=9. \]Determine their orientation relationship.
Practice 10: Identify the Error
A student writes that
\[ \mathbf{r} = \langle 1,2,3\rangle + s\langle 2,1,0\rangle + t\langle 4,2,0\rangle \]describes a plane.
Identify the problem.
Practice Answers
Practice 1
\[ \boxed{ \mathbf{r} = \langle 2,-1\rangle + t\langle 4,3\rangle }. \]Practice 2
\[ \boxed{ x=1-2t,\qquad y=2+4t,\qquad z=3+5t }. \]Practice 3
First find
\[ \overrightarrow{AB} = \langle 3,4,-3\rangle. \]Therefore,
\[ \boxed{ \mathbf{r} = \langle 0,1,2\rangle + t\langle 3,4,-3\rangle }. \]Practice 4
From
\[ 7=1+2t, \]we obtain
\[ t=3. \]Then
\[ y=3+2(3)=9. \]Therefore, \(Q=(7,9)\) lies on the line.
Practice 5
\[ \boxed{ \mathbf{r} = \langle 1,0,2\rangle + s\langle 1,1,0\rangle + t\langle 0,1,2\rangle }. \]Practice 6
Point-normal form is
\[ 3(x-2)-2(y-1)+(z+1)=0. \]Expanding gives
\[ 3x-2y+z-3=0. \]Therefore,
\[ \boxed{ 3x-2y+z=3 }. \]Practice 7
\[ \boxed{ \mathbf{n} = \langle 5,2,-7\rangle }. \]Practice 8
Substitute:
\[ 1+2(2)-3 = 2. \]Therefore, \(Q=(1,2,3)\) lies on the plane.
Practice 9
The normal vectors are
\[ \langle 1,2,-1\rangle \]and
\[ \langle 2,4,-2\rangle. \]The second is twice the first, so the planes are parallel in orientation.
They are distinct because multiplying the first equation by \(2\) would give right side \(8\), not \(9\).
Practice 10
The two direction vectors are parallel:
\[ \langle 4,2,0\rangle = 2\langle 2,1,0\rangle. \]Therefore, they provide only one independent direction.
The equation describes a line rather than a plane.
What to Remember
- A line can be written \(\mathbf{r}=\mathbf{r}_0+t\mathbf{v}\).
- A line needs one point and one nonzero direction vector.
- Parametric equations come from matching vector components.
- A direction vector can be found by subtracting coordinates of two points.
- A plane can be written \(\mathbf{r}=\mathbf{r}_0+s\mathbf{u}+t\mathbf{v}\).
- A plane requires two nonparallel in-plane directions.
- A normal vector is perpendicular to every direction in the plane.
- Point-normal form is \(\mathbf{n}\cdot(\mathbf{r}-\mathbf{r}_0)=0\).
- In \(Ax+By+Cz=D\), \(\langle A,B,C\rangle\) is a normal vector.
- Direction vectors test line relationships; normal vectors test plane relationships.
- Substitution can test membership and find line-plane intersections.
Where We Go Next
Vector equations turn lines and planes into algebraic families generated from points and directions. The next major step is to treat transformations themselves as algebraic operators.
The next lesson, Matrices as Transformation Operators, introduces matrices as structured rules that act on vectors to produce rotations, reflections, scalings, shears, and other spatial transformations.