Topic 5 / 17intermediate

Lines & Planes

Write parametric and symmetric equations for lines, equations for planes via point-and-normal, find intersections, distances, and classify pairs as parallel, orthogonal, or skew.

A line in $R^{3}$ is determined by a point on it and a direction vector. A plane is determined by a point and a normal vector. Almost every line/plane question reduces to finding the right point and the right direction or normal — then plugging into a standard form.

Lines: parametric, vector, and symmetric forms

Given a point

P_{0} (x_{0}, y_{0}, z_{0})

on the line and a direction vector

v = ⟨ a, b, c ⟩

: - **Vector**:

r (t) = r_{0} + t v

. - **Parametric**:

x = x_{0} + a t, y = y_{0} + b t, z = z_{0} + c t

. - **Symmetric**:

\frac{x - x _{0}}{a} = \frac{y - y _{0}}{b} = \frac{z - z _{0}}{c}

(provided

a, b, c \neq = 0

Key takeaways

Point + direction is enough to write any line.
Symmetric form fails when a direction component is zero — split into pieces in that case.

Planes: point + normal vector

A plane through

P_{0} (x_{0}, y_{0}, z_{0})

with normal

n = ⟨ a, b, c ⟩

satisfies

a (x - x_{0}) + b (y - y_{0}) + c (z - z_{0}) = 0

, equivalently

a x + b y + cz = d

where

d = a x_{0} + b y_{0} + c z_{0}

. The coefficients are the components of

n

Strategy tips

If you have two direction vectors **in the plane**, take their cross product to get $n$ .
Three points $P, Q, R$ in a plane: form $P Q, P R$ and cross.

Determining lines and planes from data

Given two points: direction

= P_{2} - P_{1}

. Through three points

P, Q, R

: form two edges from

P

and cross. Parallel to a given plane: same normal

n

. Orthogonal to a given plane: direction

= n

of plane. Containing two intersecting lines: take cross product of their direction vectors for the normal.

Parallel, orthogonal, and skew tests

Lines are **parallel** iff their direction vectors are scalar multiples; **orthogonal** iff direction vectors have dot product 0. Two lines that are not parallel and don't intersect are **skew**. Planes are parallel iff normals are parallel; orthogonal iff normals are orthogonal.

Strategy tips

To check if two lines intersect, set their parametric forms equal and solve. If no consistent $(t, s)$ exists and they aren't parallel, they're skew.

Distances

Point to plane

a x + b y + cz = d

D = \frac{∣ a x _{0} + b y _{0} + c z _{0} - d ∣}{a ^{2} + b ^{2} + c ^{2}} .

Point to line through

P_{0}

with direction

v

D = \frac{∣ ( P - P _{0} ) \times v ∣}{∣ v ∣} .

Intersections

Line meets plane: substitute

x (t), y (t), z (t)

into the plane equation, solve for

t

, plug back in. Two planes intersect in a line; find the line by solving the two equations simultaneously and parameterizing — the line direction is

n_{1} \times n_{2}

Worked examples

Example 1

Find parametric equations of the line through

P (1, 2, - 1)

parallel to

⟨ 3, - 1, 4 ⟩

1
Take the point and direction directly.
$x = 1 + 3 t, y = 2 - t, z = - 1 + 4 t$

Answer

x = 1 + 3 t, y = 2 - t, z = - 1 + 4 t

Example 2

Find an equation of the plane through

P (2, 1, 0)

parallel to the plane

x + 4 y - 3 z = 1

1
Parallel planes share a normal: $n = ⟨ 1, 4, - 3 ⟩$ .
2
Plug into point-normal form.
$(x - 2) + 4 (y - 1) - 3 (z - 0) = 0$
3
Simplify.
$x + 4 y - 3 z = 6$

Answer

x + 4 y - 3 z = 6

Example 3

Find the intersection of the line

x = 2 - t, y = 1 + 3 t, z = 4 t

with the plane

2 x - y + z = 2

1
Substitute the parametric pieces into the plane equation.
$2 (2 - t) - (1 + 3 t) + (4 t) = 2$
2
Simplify.
$4 - 2 t - 1 - 3 t + 4 t = 2 \Rightarrow 3 - t = 2 \Rightarrow t = 1$
3
Plug $t = 1$ back into the line.
$(2 - 1, 1 + 3, 4) = (1, 4, 4)$

Answer

Intersection point:

(1, 4, 4)

Interactive visualizations

Loading visualization…

a1.00

b2.00

c2.00

d4.00

1.0 x + 2.0 y + 2.0 z = 4.0, n = ⟨ 1.0, 2.0, 2.0 ⟩

Slide

a, b, c, d

to change the plane

a x + b y + cz = d

. The red arrow is the normal

⟨ a, b, c ⟩

Formulas in this topic

Line: vector form

r (t) = r_{0} + t v

Parametric line through r0 with direction v.

Line: symmetric form

\frac{x - x _{0}}{a} = \frac{y - y _{0}}{b} = \frac{z - z _{0}}{c}

Equate ratios after eliminating t.

Plane: point-normal form

a (x - x_{0}) + b (y - y_{0}) + c (z - z_{0}) = 0

(a,b,c) is the normal vector.

Plane: standard form

a x + b y + cz = d

d = a x0 + b y0 + c z0.

Distance: point to plane

D = \frac{∣ a x _{0} + b y _{0} + c z _{0} - d ∣}{a ^{2} + b ^{2} + c ^{2}}

Standard formula.

Distance: point to line

D = \frac{∣ ( P - P _{0} ) \times v ∣}{∣ v ∣}

Uses cross product with direction.

Lines: parametric, vector, and symmetric forms

Planes: point + normal vector

Determining lines and planes from data

Parallel, orthogonal, and skew tests

Distances

Intersections

Worked examples

Interactive visualizations

Formulas in this topic

Related topics