Lines in 3D

Why a point and a direction determine a line

In 2D you have already seen this idea. The slope-intercept form

y = ax + b

encodes exactly two pieces of information:

The direction: the slope $a = \tan\theta$ , where $\theta$ is the angle the line makes with the $x$ -axis (see polar coordinates for the connection between slope and angle).
A point: the $y$ -intercept $(0, b)$ , which fixes where the line passes through the $y$ -axis.

Try the illustration below. The blue dot lives on the $y$ -axis at $(0, b)$ — drag it up or down to change the intercept while the slope stays fixed. The green dot sits on the line at $x = 3.5$ — drag it up or down to rotate the line while the intercept stays fixed.

y = ax + b — drag to change intercept (blue) or slope (green)

y = 1x + 0b = 0a = 1 (slope = tan θ)

—— current line (y = ax + b)- - - parallel family (same slope, different b)

💡 Drag the blue dot to slide the intercept · drag the green dot to rotate the line.

Notice what happens when you drag only the blue dot: the line translates vertically — and the four dashed lines (same slope, different intercepts) show the entire parallel family that shares the same direction. Every one of them has slope $a$ ; they are all valid if you only know the direction. Adding $b$ — fixing the $y$ -intercept — singles out exactly one line from this parallel family.

Conversely, if you drag only the green dot, the line rotates around the fixed intercept $(0, b)$ : all lines through a single point but with different slopes. The point alone is not enough; you need the direction too.

Change either ingredient and you get a different line. These two pieces of data — a point and a direction — are the minimal recipe for a line in any dimension.

In 3D, a slope no longer makes sense as a single number, but a direction vector $\vec{v}$ plays the same role. The $y$ -intercept generalises to an arbitrary known point $\vec{r}_0$ on the line.

Vector equation of a line

Let $\vec{r}_0$ be the position vector of a known point on the line, and let $\vec{v}$ be the direction vector of the line. Every point on the line can be reached by starting at $\vec{r}_0$ and walking some multiple of $\vec{v}$ :

\boxed{\vec{r}(t) = \vec{r}_0 + t\,\vec{v}, \quad t \in \mathbb{R}.}

$t = 0$ gives $\vec{r}_0$ itself.
As $t$ increases, we move in the direction $\vec{v}$ .
As $t$ decreases, we move in the opposite direction.

The graph below shows $\vec{r}_0$ (blue arrow from the origin) and $\vec{v}$ (green arrow, starting from the tip of $\vec{r}_0$ ). The grey dashed line is the full line generated by the equation. Drag either handle to explore.

r(t) = r₀ + tv — drag r₀ (blue) and v (green)

r₀ = (1, 2)v = (3, 1)

r(t) = (1, 2) + t·(3, 1)

💡 Drag the blue circle (tip of r₀) or green circle (tip of v) to adjust.

Parametric form: eliminating the parameter

The equation $\vec{r}(t) = \vec{r}_0 + t\,\vec{v}$ is itself a parametric equation — it uses a parameter $t$ to trace out a curve (see parametric curves for the general idea).

Writing $\vec{r}_0 = (x_0,y_0,z_0)$ and $\vec{v} = (a,b,c)$ , the vector equation expands component-wise to

\begin{cases} x = x_0 + at \\ y = y_0 + bt \\ z = z_0 + ct. \end{cases}

These are the parametric equations of the line in 3D.

Symmetric equations (no parameter)

If $a, b, c$ are all nonzero, we can solve each equation for $t$ and set them equal:

\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}.

These are the symmetric equations of the line. They eliminate the parameter $t$ entirely.

Find the parametric and symmetric equations of the line through $P_0 = (2, -1, 3)$ with direction $\vec{v} = (4, 1, -2)$ .

Parametric equations:

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

Symmetric equations:

\frac{x-2}{4} = \frac{y+1}{1} = \frac{z-3}{-2}.

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

Direction vector: $\vec{v} = B - A = (2,4,-3)$ .

Vector equation:

\vec{r}(t) = (1,0,2) + t(2,4,-3).

At $t=0$ we are at $A$ ; at $t=1$ we are at $B$ .

Line segments as linear combinations

Definition

The line $\vec{r}(t) = \vec{a} + t(\vec{b}-\vec{a})$ can be rewritten as

\vec{r}(t) = (1-t)\,\vec{a} + t\,\vec{b}.

This is a linear combination of $\vec{a}$ and $\vec{b}$ with weights $(1-t)$ and $t$ .

When $t$ is restricted to $[0,1]$ , the weights are non-negative and sum to 1. The result is exactly the line segment from $\vec{a}$ to $\vec{b}$ .

Range of $t$	Location
$t = 0$	at $\vec{a}$
$t = 1$	at $\vec{b}$
$0 < t < 1$	between $\vec{a}$ and $\vec{b}$ (the segment)
$t < 0$	beyond $\vec{a}$ , away from $\vec{b}$
$t > 1$	beyond $\vec{b}$ , away from $\vec{a}$

The interactive graph below shows both vectors $\vec{a}$ and $\vec{b}$ from the origin. The colored line illustrates the different regions. Drag either vector to explore.

r(t) = (1−t)a + tb — segment and extensions

a = (3, 1)b = (-1, 3)

r(t) = (1−t)·a + t·b

— t < 0— 0 ≤ t ≤ 1 (segment)— t > 1

💡 Drag the blue circle (a) or red circle (b) to adjust.

What value of $t$ gives the midpoint of the segment from $\vec{a}$ to $\vec{b}$ ?

Answer: At $t = \tfrac{1}{2}$ :

\vec{r}\!\left(\tfrac{1}{2}\right) = \tfrac{1}{2}\,\vec{a} + \tfrac{1}{2}\,\vec{b} = \frac{\vec{a}+\vec{b}}{2}.

Show that $Q = (2,1,0)$ lies on the segment from $A=(0,0,0)$ to $B=(4,2,0)$ , and find the value of $t$ .

Answer: $\vec{r}(t) = t(4,2,0) = (4t,2t,0)$ . Setting equal to $(2,1,0)$ gives $t = \tfrac{1}{2} \in [0,1]$ . ✓

Planes in 3D

A point and a normal direction

Just as a line is determined by a point and a direction it runs along, a plane is determined by a point and a direction it is perpendicular to — the normal direction $\vec{n}$ .

The interactive graph below shows a plane (purple surface) together with its normal vector (red arrow). Adjust the inputs to see how changing the normal or the base point rotates and translates the plane.

A plane is determined by a point r₀ and a normal n

n_x=n_y=n_z=

r₀_x=r₀_y=r₀_z=

Equation of a plane

Let $\vec{r}_0 = (x_0,y_0,z_0)$ be a known point on the plane and $\vec{n} = (a,b,c)$ be a normal vector. A point $\vec{r} = (x,y,z)$ lies on the plane if and only if the vector $\vec{r} - \vec{r}_0$ is perpendicular to $\vec{n}$ :

\boxed{\vec{n} \cdot (\vec{r} - \vec{r}_0) = 0.}

Expanding to component form

a(x-x_0) + b(y-y_0) + c(z-z_0) = 0.

Expanding and collecting the constant:

ax + by + cz + d = 0,

where $d = -(ax_0 + by_0 + cz_0)$ .

The coefficients $(a,b,c)$ of $x,y,z$ in this equation are exactly the normal vector of the plane.

Find the equation of the plane through $P_0 = (1,2,-1)$ with normal $\vec{n} = (3,-1,4)$ .

3(x-1) + (-1)(y-2) + 4(z+1) = 0 \implies 3x - y + 4z + 3 = 0.

What is the normal vector of the plane $2x - 5y + z = 7$ ?

Answer: $\vec{n} = (2,-5,1)$ .

Finding the normal via cross product

When a plane is described by two vectors lying in it (e.g., by three points), the normal can be found using the cross product (see cross product):

\vec{n} = \vec{u} \times \vec{v},

where $\vec{u}$ and $\vec{v}$ are two non-parallel vectors in the plane.

Procedure given three points $A$ , $B$ , $C$ :

Compute $\vec{u} = B - A$ and $\vec{v} = C - A$ .
Compute $\vec{n} = \vec{u} \times \vec{v}$ .
Use $A$ as the base point: $\vec{n} \cdot (\vec{r} - A) = 0$ .

Find the equation of the plane through $A=(1,0,0)$ , $B=(0,1,0)$ , $C=(0,0,1)$ .

$\vec{u} = B-A = (-1,1,0)$ , $\vec{v} = C-A = (-1,0,1)$ .

\vec{n} = \vec{u}\times\vec{v} = \begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\-1&1&0\\-1&0&1\end{vmatrix} = (1,1,1).

1(x-1)+1(y)+1(z)=0 \implies x+y+z=1.

Find the plane through $P=(2,0,-1)$ that contains both $\vec{u}=(1,2,0)$ and $\vec{v}=(0,1,3)$ .

\vec{n}=\vec{u}\times\vec{v}=(6,-3,1).

6(x-2)-3y+(z+1)=0 \implies 6x-3y+z-11=0.

Find the plane through $P=(1,1,1)$ perpendicular to the line $\vec{r}(t)=t(2,3,-1)$ .

The direction of the line is $\vec{v}=(2,3,-1)$ , which becomes the normal: $\vec{n}=(2,3,-1)$ .

2(x-1)+3(y-1)-(z-1)=0 \implies 2x+3y-z-4=0.

Find the plane through $P=(1,0,0)$ containing the line $x=t,\; y=t,\; z=2t$ .

Two vectors in the plane: direction of the line $\vec{u}=(1,1,2)$ and $\vec{v}=P-(0,0,0)=(1,0,0)$ .

\vec{n}=\vec{u}\times\vec{v}=(0,2,-1).

2y-z=0.

Find the plane through $A=(2,1,-1)$ , $B=(3,3,0)$ , $C=(0,2,1)$ .

$\vec{u}=(1,2,1)$ , $\vec{v}=(-2,1,2)$ .

\vec{n}=\vec{u}\times\vec{v}=(3,-4,5).

3(x-2)-4(y-1)+5(z+1)=0 \implies 3x-4y+5z+3=0.

Distance from a point to a plane

Geometric setup

Given a plane and a point $P$ not on it, the distance from $P$ to the plane is the length of the perpendicular segment from $P$ to the plane.

The graph below illustrates this: the plane is fixed and the point $P$ (blue) is draggable. The red dashed segment shows the perpendicular distance. Drag inside the graph to move $P$ in the $xy$ -direction; hold Shift to adjust the $z$ -coordinate.

Distance from point P to the plane

Plane: 0x + 1y + 2z = 0

P = (2, 3, 1)distance = 2.24

💡 Drag inside the 3D graph to move point P (xy-plane). Hold Shift to adjust z.

Distance formula via projection

We use scalar projection (see dot product for the projection formula).

Let $Q$ be any point on the plane and let $\vec{b} = P - Q$ . The distance from $P$ to the plane is the component of $\vec{b}$ along the unit normal $\hat{n} = \vec{n}/|\vec{n}|$ :

\text{distance} = |\,\vec{b} \cdot \hat{n}\,| = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{n}|}.

This equals $|\vec{b}|\cos\theta$ — the perpendicular drop from $P$ to the plane.

Formula for the plane $ax + by + cz + d = 0$

The normal is $\vec{n} = (a,b,c)$ . For any point $Q$ on the plane, $aQ_x + bQ_y + cQ_z + d = 0$ , so

\vec{b} \cdot \vec{n} = a(x_1 - Q_x) + b(y_1 - Q_y) + c(z_1 - Q_z) = ax_1 + by_1 + cz_1 + d.

Therefore:

\boxed{ \text{distance from } (x_1,y_1,z_1) \text{ to } ax+by+cz+d=0 \ =\ \frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}. }

Find the distance from $P=(1,2,3)$ to the plane $x - 2y + 2z - 4 = 0$ .

\text{distance} = \frac{|1 - 4 + 6 - 4|}{\sqrt{1+4+4}} = \frac{1}{3}.

Find the distance from the origin to the plane $3x + 4y - 12 = 0$ .

\text{distance} = \frac{|3(0)+4(0)-12|}{\sqrt{9+16}} = \frac{12}{5}.

Find the distance between $2x - y + 2z = 4$ and $2x - y + 2z = -2$ .

Pick any point on the first plane, say $(2,0,0)$ . Plug into the second plane's equation: $2(2)-0+2(0)+2=6$ .

\text{distance} = \frac{|6|}{\sqrt{4+1+4}} = \frac{6}{3} = 2.

The plane $x + 2y + 2z = k$ is at distance 3 from $(1,1,1)$ . Find $k$ .

\frac{|1+2+2-k|}{3} = 3 \implies |5-k| = 9 \implies k = -4 \text{ or } k = 14.

Lines and Planes