The Main Idea 

In 2D, you describe lines using slope. In 3D, you use a direction vector parallel to the line instead.

Two nonzero vectors [latex]\mathbf{u}[/latex] and [latex]\mathbf{v}[/latex] are parallel if [latex]\mathbf{u} = k\mathbf{v}[/latex] for some scalar [latex]k[/latex]. When [latex]k > 0[/latex], they point the same way; when [latex]k < 0[/latex], they point opposite ways.

Three Forms of Line Equations

A line through point [latex]P(x_0, y_0, z_0)[/latex] with direction vector [latex]\mathbf{v} = \langle a, b, c \rangle[/latex]:

• Vector form: [latex]\mathbf{r} = \mathbf{r}_0 + t\mathbf{v}[/latex]
• Parametric form: [latex]x = x_0 + ta, \quad y = y_0 + tb, \quad z = z_0 + tc[/latex]
• Symmetric form: [latex]\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}[/latex] (when [latex]a, b, c \neq 0[/latex])

Finding the Direction Vector

Given two points on the line, form the direction vector by subtracting coordinates:

[latex]\mathbf{v} = \langle x_1 - x_0, y_1 - y_0, z_1 - z_0 \rangle[/latex].

Line Segments

To describe just the segment between points [latex]P[/latex] and [latex]Q[/latex], restrict the parameter: [latex]0 \leq t \leq 1[/latex].

• Vector form: [latex]\mathbf{r} = (1-t)\mathbf{p} + t\mathbf{q}, \quad 0 \leq t \leq 1[/latex]
• Parametric form: [latex]x = x_0 + t(x_1 - x_0)[/latex], [latex]y = y_0 + t(y_1 - y_0)[/latex], [latex]z = z_0 + t(z_1 - z_0)[/latex], [latex]0 \leq t \leq 1[/latex]

At [latex]t = 0[/latex], you’re at [latex]P[/latex]; at [latex]t = 1[/latex], you’re at [latex]Q[/latex].

Line equations aren’t unique—different direction vectors or reference points produce different-looking equations for the same line.

The Main Idea 

Finding the distance between a point and a line in 3D has practical applications—from construction setback requirements to airline flight path planning.

The distance from a point to a line is the length of the shortest segment connecting them. This shortest segment is always perpendicular to the line.

In 3D, there’s no obvious way to identify which point on the line creates the perpendicular segment. We solve this using the cross product.

The Formula

Let [latex]L[/latex] be a line through point [latex]P[/latex] with direction vector [latex]\mathbf{v}[/latex]. For any point [latex]M[/latex] not on the line:

[latex]d = \frac{||\overrightarrow{PM} \times \mathbf{v}||}{||\mathbf{v}||}[/latex]

Why This Works

Vectors [latex]\overrightarrow{PM}[/latex] and [latex]\mathbf{v}[/latex] form two sides of a parallelogram:

• Area of parallelogram: [latex]||\overrightarrow{PM} \times \mathbf{v}||[/latex]
• Also area: base × height = [latex]||\mathbf{v}|| \cdot d[/latex]

Setting these equal and solving for [latex]d[/latex] gives the formula. The cross product’s magnitude captures the parallelogram area, and dividing by the base [latex]||\mathbf{v}||[/latex] isolates the height—exactly the perpendicular distance we need.

Process:

1. Choose any point [latex]P[/latex] on the line
2. Form vector [latex]\overrightarrow{PM}[/latex]
3. Find the direction vector [latex]\mathbf{v}[/latex] for the line
4. Calculate [latex]\overrightarrow{PM} \times \mathbf{v}[/latex]
5. Apply the formula: [latex]d = \frac{||\overrightarrow{PM} \times \mathbf{v}||}{||\mathbf{v}||}[/latex]

The Main Idea 

In 2D, two lines are either identical, parallel, or intersecting. Three dimensions adds a fourth possibility that can’t exist in a plane.

Four Possible Relationships

1. Equal lines: Same direction, same points
2. Parallel lines: Same direction, never meet
3. Intersecting lines: Cross at exactly one point
4. Skew lines: Not parallel and don’t intersect (unique to 3D!)

Skew lines have different directions but never meet—like one line running east-west on a building’s ground floor while another runs north-south on the second floor. They “miss” each other in space. In 2D, non-parallel lines must intersect. In 3D, they can miss each other entirely—that’s what makes skew lines possible.

Classification Strategy

Answer two questions to determine the relationship:

1. Are direction vectors parallel? (Check if [latex]\mathbf{v}_1 = k\mathbf{v}_2[/latex] for some scalar [latex]k[/latex])
2. Do the lines share a point? (Set parametric equations equal and solve)

Decision Table:

• Parallel directions + shared point → Equal lines
• Parallel directions + no shared point → Parallel but distinct
• Non-parallel directions + shared point → Intersecting lines
• Non-parallel directions + no shared point → Skew lines

Testing for a Shared Point

Set the parametric equations equal and solve the system:

• If you find consistent values for both parameters, the lines intersect at that point
• If you get a contradiction, the lines are skew (or parallel if directions match)

The Main Idea 

A plane is uniquely determined by three non-collinear points, but the most powerful way to describe a plane uses a normal vector—a vector perpendicular to the plane.

A point [latex]Q[/latex] lies on a plane if and only if the vector [latex]\overrightarrow{PQ}[/latex] (from a fixed point [latex]P[/latex] on the plane to [latex]Q[/latex]) is perpendicular to the normal vector [latex]\mathbf{n}[/latex]. Since orthogonal vectors have dot product zero: [latex]\mathbf{n} \cdot \overrightarrow{PQ} = 0[/latex].

Three Forms of Plane Equations

Given point [latex]P = (x_0, y_0, z_0)[/latex] on the plane and normal vector [latex]\mathbf{n} = \langle a, b, c \rangle[/latex]:

• Vector equation: [latex]\mathbf{n} \cdot \overrightarrow{PQ} = 0[/latex]
• Scalar equation: [latex]a(x - x_0) + b(y - y_0) + c(z - z_0) = 0[/latex]
• General form: [latex]ax + by + cz + d = 0[/latex] where [latex]d = -ax_0 - by_0 - cz_0[/latex]

Finding a Plane Through Three Points

Use the cross product to find a normal vector:

1. Form two vectors from the three points (e.g., [latex]\overrightarrow{PQ}[/latex] and [latex]\overrightarrow{PR}[/latex])
2. Calculate [latex]\mathbf{n} = \overrightarrow{PQ} \times \overrightarrow{PR}[/latex]
3. Use [latex]\mathbf{n}[/latex] and any of the three points to write the plane equation

Distance from a Point to a Plane

The shortest distance from point [latex]P[/latex] to a plane is measured perpendicular to the plane. For a plane with normal vector [latex]\mathbf{n}[/latex] passing through point [latex]Q[/latex]:

[latex]d = \frac{|\overrightarrow{QP} \cdot \mathbf{n}|}{||\mathbf{n}||}[/latex]

This is the scalar projection of [latex]\overrightarrow{QP}[/latex] onto [latex]\mathbf{n}[/latex]—the component of [latex]\overrightarrow{QP}[/latex] perpendicular to the plane.

The Main Idea 

Unlike lines (which can be parallel, intersecting, or skew), planes have only two possible relationships: they’re either parallel or they intersect. There’s no such thing as skew planes.

Planes are parallel when their normal vectors are parallel: [latex]\mathbf{n}_1 = k\mathbf{n}_2[/latex] for some scalar [latex]k[/latex].

When planes aren’t parallel, they must intersect along a line. The intersection is always a straight line, never just a point or scattered points.

Finding the Line of Intersection

To find parametric equations for the intersection line:

1. Use both plane equations simultaneously
2. Eliminate one variable by adding/subtracting equations
3. Express two variables in terms of the third
4. Define the third variable as parameter [latex]t[/latex]

Angle Between Planes

The angle between two planes equals the angle between their normal vectors:

[latex]\cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{||\mathbf{n}_1||  ||\mathbf{n}_2||}[/latex]

We use absolute value to ensure [latex]0° \leq \theta \leq 90°[/latex].

Special Cases:

• Parallel: [latex]\mathbf{n}_1 \times \mathbf{n}_2 = \mathbf{0}[/latex] (cross product is zero)
• Perpendicular: [latex]\mathbf{n}_1 \cdot \mathbf{n}_2 = 0[/latex] (dot product is zero)

Distance Between Parallel Planes

For point [latex]P(x_0, y_0, z_0)[/latex] and plane [latex]ax + by + cz + k = 0[/latex]:

[latex]d = \frac{|ax_0 + by_0 + cz_0 + k|}{\sqrt{a^2 + b^2 + c^2}}[/latex]

If planes are [latex]ax + by + cz + k_1 = 0[/latex] and [latex]ax + by + cz + k_2 = 0[/latex], the distance is [latex]\frac{|k_1 - k_2|}{\sqrt{a^2 + b^2 + c^2}}[/latex].