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Parametric Forms of Lines and Planes

Lines and planes can be described using parametric equations. This representation is especially powerful and concise in higher dimensions. We will also use parametric form to describe the solution set of a system of linear equations.

Parametric Form of a Line

Definition (Parametric Form of a Line):
A line through point p⃗\vec{p} in the direction of vector v⃗\vec{v} is given by:

x⃗(t)=p⃗+tv⃗,t∈R\vec{x}(t) = \vec{p} + t\vec{v}, \quad t \in \mathbb{R}

where tt is called the parameter.

Why Does This Represent a Line?

Think of p⃗\vec{p} as a starting point and v⃗\vec{v} as the direction to walk. Then, x(t)x(t) is our position at time tt:

As tt varies over all real numbers, x⃗(t)\vec{x}(t) traces out all points on the line.

Parametric Line: x⃗(t) = p⃗ + tv⃗

p⃗ =(,)(point on line)
v⃗ =(,)(direction vector)
Line: x⃗(t) = p⃗ + tv⃗
💡 Tip: Drag the arrow heads to change p⃗ (point on line) and v⃗ (direction). The line extends as x⃗(t) = p⃗ + tv⃗.
t=-1t=0t=1t=2p⃗(-1.0, 1.0)v⃗(3.0, 1.0)
  • Blue vector p⃗: A point on the line (starts at origin)
  • Green vector v⃗: The direction vector (starts at the end of p⃗)
  • Dashed line: The infinite line extending through these vectors
  • Labeled points: Sample positions for t = -1, 0, 1, 2

Find the parametric equation of the line through [23]\begin{bmatrix} 2 \\ 3 \end{bmatrix} and [57]\begin{bmatrix} 5 \\ 7 \end{bmatrix}.

Solution:

  1. Choose p⃗=[23]\vec{p} = \begin{bmatrix} 2 \\ 3 \end{bmatrix} as the point on the line
  2. Find direction vector: v⃗=[57]−[23]=[34]\vec{v} = \begin{bmatrix} 5 \\ 7 \end{bmatrix} - \begin{bmatrix} 2 \\ 3 \end{bmatrix} = \begin{bmatrix} 3 \\ 4 \end{bmatrix}
  3. Parametric form: x⃗(t)=[23]+t[34]\vec{x}(t) = \begin{bmatrix} 2 \\ 3 \end{bmatrix} + t\begin{bmatrix} 3 \\ 4 \end{bmatrix}

Alternatively, we could also use

  1. p⃗=[57]\vec{p} = \begin{bmatrix} 5 \\ 7 \end{bmatrix}
  2. v⃗=[−3−4]\vec{v} = \begin{bmatrix} -3 \\ -4 \end{bmatrix} (pointing the opposite direction).
  3. Parametric form: x⃗(t)=[57]+t[−3−4]\vec{x}(t) = \begin{bmatrix} 5 \\ 7 \end{bmatrix} + t\begin{bmatrix} -3 \\ -4 \end{bmatrix}

This represents the same line but parametrized differently. The key point: parametric form is not unique, but all valid forms describe the same geometric object.

Why Parametric Form in Higher Dimensions?

In R2\mathbb{R}^2, we can describe a line with a single equation like y=2x+1y = 2x + 1. However, in R3\mathbb{R}^3 and higher, representing a line will require more equations.

In R3\mathbb{R}^3:

  • A single equation like x+y+2z=6x + y + 2z = 6 describes a plane, not a line
  • A line in R3\mathbb{R}^3 is the intersection of two planes, requiring two equations.

We consider the following system of linear equations representing the intersection of two planes:

{x+y+2z=6x−y=0\begin{cases} x + y + 2z = 6 \\ x - y = 0 \end{cases}

Row Reduction Calculator

Step 0 of 3

Based on the row-reduced echelon form, the solution set is given by

{x=3−zy=3−zz is free\begin{cases} x = 3 - z \\ y = 3 - z \\ z \text{ is free} \\ \end{cases}

We can express this solution set in parametric vector form. Since zz is free, let z=tz = t. Then

x⃗=[xyz]=[3−t3−tt]=[330]+[−1−11]t\vec{x} = \begin{bmatrix}x \\ y \\ z\end{bmatrix} = \begin{bmatrix}3-t \\ 3-t \\ t\end{bmatrix} = \begin{bmatrix}3 \\ 3 \\ 0\end{bmatrix} + \begin{bmatrix}-1 \\ -1 \\ 1\end{bmatrix}t

In this form, it is clear that the solution set (the intersection of two planes) is a line passing through the point (3,3,0)(3, 3, 0) with direction vector (−1,−1,1)(-1,-1,1).

More generally, in Rn\mathbb{R}^n, a line requires (n−1)(n-1) equations to describe it implicitly, but only one parameter in parametric form.For this reason, the parametric form is often simpler and more intuitive.

Parametric Form of a Plane

Definition (Parametric Form of a Plane):
A plane through point p⃗\vec{p} spanned by two not parallel vectors u⃗\vec{u} and v⃗\vec{v} can be described as:

p⃗+su⃗+tv⃗,s,t∈R\vec{p} + s\vec{u} + t\vec{v}, \quad s, t \in \mathbb{R}

where ss and tt are parameters.

Geometric Interpretation

Think of building the plane by:

  1. Start at point p⃗\vec{p}
  2. Walk ss steps in the u⃗\vec{u} direction
  3. Walk tt steps in the v⃗\vec{v} direction
  4. You reach point x⃗\vec{x}

By varying ss and tt independently, we sweep out the entire plane.

  • Line: 1 parameter → 1-dimensional object
  • Plane: 2 parameters → 2-dimensional object
  • Space: 3 parameters → 3-dimensional object

The number of parameters equals the dimension of the object. A plane is 2-dimensional, so we need 2 independent directions (u⃗\vec{u} and v⃗\vec{v}) and 2 parameters (ss and tt).

Example 3: Plane in R3\mathbb{R}^3

Find the parametric equation of the plane through points: P=[100],Q=[010],R=[001]P = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \quad Q = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \quad R = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}

Solution:

  1. Choose p⃗=[100]\vec{p} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}

  2. Find two direction vectors: u⃗=Q−P=[−110],v⃗=R−P=[−101]\vec{u} = Q - P = \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix}, \quad \vec{v} = R - P = \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}

  3. Parametric form: x⃗=[100]+s[−110]+t[−101]\vec{x} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + s\begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix} + t\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}

Component form: {x=1−s−ty=sz=t\begin{cases} x = 1 - s - t \\ y = s \\ z = t \end{cases}

Let's check that PP, QQ, and RR are on this plane:

  • Point PP: Set s=0,t=0s = 0, t = 0: x⃗=[100]✓\vec{x} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \checkmark

  • Point QQ: Set s=1,t=0s = 1, t = 0: x⃗=[100]+[−110]=[010]✓\vec{x} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \checkmark

  • Point RR: Set s=0,t=1s = 0, t = 1: x⃗=[100]+[−101]=[001]✓\vec{x} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \checkmark

Example 4: Another Plane

The plane through origin with direction vectors u⃗=[101]\vec{u} = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} and v⃗=[011]\vec{v} = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} is:

x⃗=s[101]+t[011]=[sts+t]\vec{x} = s\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} + t\begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} s \\ t \\ s + t \end{bmatrix}

Connection to Linear Combinations

Notice the pattern:

When p⃗=0⃗\vec{p} = \vec{0}:

This connects to our earlier study of linear combinations!

Practice Problems

Find the parametric equation of the line through [123]\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} and [415]\begin{bmatrix} 4 \\ 1 \\ 5 \end{bmatrix} in R3\mathbb{R}^3.

Use one point as p⃗\vec{p} and the difference as the direction vector v⃗\vec{v}.

Step 1: Choose p⃗=[123]\vec{p} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}

Step 2: Direction vector: v⃗=[415]−[123]=[3−12]\vec{v} = \begin{bmatrix} 4 \\ 1 \\ 5 \end{bmatrix} - \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} = \begin{bmatrix} 3 \\ -1 \\ 2 \end{bmatrix}

Step 3: Parametric form: x⃗=[123]+t[3−12]\vec{x} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} + t\begin{bmatrix} 3 \\ -1 \\ 2 \end{bmatrix}

Component form: {x=1+3ty=2−tz=3+2t\begin{cases} x = 1 + 3t \\ y = 2 - t \\ z = 3 + 2t \end{cases}

Consider the line x⃗=[21]+t[34]\vec{x} = \begin{bmatrix} 2 \\ 1 \end{bmatrix} + t\begin{bmatrix} 3 \\ 4 \end{bmatrix}.

Is the point [1113]\begin{bmatrix} 11 \\ 13 \end{bmatrix} on this line? If so, what is the value of tt?

Set up the equation and solve for tt. Check that both components give the same tt value.

We need: [21]+t[34]=[1113]\begin{bmatrix} 2 \\ 1 \end{bmatrix} + t\begin{bmatrix} 3 \\ 4 \end{bmatrix} = \begin{bmatrix} 11 \\ 13 \end{bmatrix}

This gives: {2+3t=111+4t=13\begin{cases} 2 + 3t = 11 \\ 1 + 4t = 13 \end{cases}

From the first equation: 3t=93t = 9, so t=3t = 3

Check with second equation: 1+4(3)=1+12=131 + 4(3) = 1 + 12 = 13 ✓

Answer: Yes, the point is on the line with t=3t = 3.

Find the parametric equation of the plane through: A=[111],B=[201],C=[102]A = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}, \quad B = \begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix}, \quad C = \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix}

Use AA as your base point p⃗\vec{p}. Find two direction vectors using B−AB - A and C−AC - A.

Step 1: Choose p⃗=[111]\vec{p} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}

Step 2: Direction vectors: u⃗=B−A=[1−10],v⃗=C−A=[0−11]\vec{u} = B - A = \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}, \quad \vec{v} = C - A = \begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix}

Step 3: Parametric form: x⃗=[111]+s[1−10]+t[0−11]\vec{x} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} + s\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} + t\begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix}

Component form: {x=1+sy=1−s−tz=1+t\begin{cases} x = 1 + s \\ y = 1 - s - t \\ z = 1 + t \end{cases}

The line x⃗=[12]+t[34]\vec{x} = \begin{bmatrix} 1 \\ 2 \end{bmatrix} + t\begin{bmatrix} 3 \\ 4 \end{bmatrix} can be written as a single equation in R2\mathbb{R}^2. Find this equation.

Write out the component equations, solve one for tt, and substitute into the other.

Component form: {x=1+3ty=2+4t\begin{cases} x = 1 + 3t \\ y = 2 + 4t \end{cases}

From the first equation: t=x−13t = \frac{x - 1}{3}

Substitute into the second: y=2+4⋅x−13=2+4(x−1)3=2+4x−43y = 2 + 4 \cdot \frac{x - 1}{3} = 2 + \frac{4(x-1)}{3} = 2 + \frac{4x - 4}{3}

3y=6+4x−4=4x+23y = 6 + 4x - 4 = 4x + 2

4x−3y+2=04x - 3y + 2 = 0

Or in slope-intercept form: y=43x+23y = \frac{4}{3}x + \frac{2}{3}