Equations of Lines in 3-Space

In 2-space, a line could be described by a vector equation or by one scalar equation. In 3-space, the vector equation still works perfectly, but a single scalar equation no longer describes a line at all. This page shows how to write lines in 3-space, and how to describe one using two planes instead.

Key ideas

Why there’s no scalar equation of a line in 3-space

In 2-space, the scalar equation $Ax + By + C = 0$ works because there is only one direction perpendicular to a line (up to scalar multiples), so the normal $[A, B]$ pins down the line’s direction.

In 3-space, a whole flat fan of directions is perpendicular to any given vector. A single equation $Ax + By + Cz + D = 0$ has solutions forming a plane (see linear equations in 2-space and 3-space), not a line. So a line in 3-space needs either a vector (or parametric) equation, or two scalar equations, one for each of two planes that meet in the line.

Vector and parametric equations

Exactly as in 2-space, a line through $P_0(x_0, y_0, z_0)$ with direction vector $\vec{m} = [m_1, m_2, m_3]$ is

$$\vec{r} = \vec{r}_0 + t\vec{m}, \qquad t \in \mathbb{R}$$

where $\vec{r} = [x, y, z]$ and $\vec{r}_0 = [x_0, y_0, z_0]$. In parametric form:

$$x = x_0 + t m_1, \qquad y = y_0 + t m_2, \qquad z = z_0 + t m_3$$

Every point $P$ on the line is reached by going to $P_0$ (that’s $\vec{r}_0$), then $t$ copies of $\vec{m}$.

If you know two points $A$ and $B$ on the line, use $\vec{m} = \overrightarrow{AB}$ (or any non-zero multiple of it, to keep the numbers small) and either point for $\vec{r}_0$.

A zero component in $\vec{m}$ means that coordinate never changes. For example, the line $\vec{r} = [3, -1, 2] + t[2, 0, -5]$ has $y = -1$ for every point, so it lies in the plane $y = -1$. IB courses write the same line with column vectors, $\vec{r} = \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} + t\begin{pmatrix} 2 \\ 0 \\ -5 \end{pmatrix}$; it means exactly the same thing.

Symmetric equations (an aside)

If none of $m_1, m_2, m_3$ is zero, solve each parametric equation for $t$ and set the results equal:

$$\frac{x - x_0}{m_1} = \frac{y - y_0}{m_2} = \frac{z - z_0}{m_3}$$

These are the symmetric equations of the line. You’ll see them in some textbooks; they’re really two scalar equations in disguise (the first fraction equals the second, and the second equals the third).

Is a point on the line?

Substitute the point into the parametric equations and solve each one for $t$. The point is on the line only if all three give the same $t$.

A line as the intersection of two planes

To describe a line with scalar equations, find two planes that contain it:

• From parametric to planes: solve one parametric equation for $t$ and substitute into the other two. Each result is the equation of a plane containing the line.
• From planes to parametric: solve the two plane equations together, letting one variable be the parameter $t$ (as in linear equations in 2-space and 3-space).

A quick check for the second direction: the line lies in both planes, so its direction is perpendicular to both normals. The cross product $\vec{n}_1 \times \vec{n}_2$ is a direction vector for the line.

Worked examples

Example 1: From a point and a direction

Write vector and parametric equations of the line through $P(3, -1, 2)$ with direction vector $[2, 0, -5]$, and find the point where $t = 1$.

Solution.

$$\vec{r} = [3, -1, 2] + t[2, 0, -5], \quad t \in \mathbb{R}$$ $$x = 3 + 2t, \qquad y = -1, \qquad z = 2 - 5t$$

For $t = 1$: $(5, -1, -3)$. Notice $y$ is $-1$ for every point, because the direction vector has no $y$-component.

Example 2: Through two points, and testing points

Find a vector equation of the line through $A(1, 4, -2)$ and $B(3, 1, 2)$. Then decide whether $C(7, -5, 10)$ and $D(5, -2, 5)$ are on the line.

Solution. $\overrightarrow{AB} = [3 - 1,\ 1 - 4,\ 2 - (-2)] = [2, -3, 4]$, so

$$\vec{r} = [1, 4, -2] + t[2, -3, 4], \quad t \in \mathbb{R}$$

For $C(7, -5, 10)$:

$$1 + 2t = 7 \Rightarrow t = 3, \qquad 4 - 3t = -5 \Rightarrow t = 3, \qquad -2 + 4t = 10 \Rightarrow t = 3$$

All three agree, so $C$ is on the line.

For $D(5, -2, 5)$: $1 + 2t = 5$ gives $t = 2$, and $4 - 3(2) = -2$ ✓, but $-2 + 4(2) = 6 \ne 5$. $D$ is not on the line, even though two of the coordinates matched.

Example 3: From parametric equations to two planes

Write the line $x = 2 + t$, $y = -1 + 3t$, $z = 4 - 2t$ as the intersection of two planes.

Solution. From the first equation, $t = x - 2$. Substitute into the other two:

$$\begin{aligned} y &= -1 + 3(x - 2) = 3x - 7 && \Rightarrow\quad 3x - y - 7 = 0 \\ z &= 4 - 2(x - 2) = 8 - 2x && \Rightarrow\quad 2x + z - 8 = 0 \end{aligned}$$

The line is the intersection of the planes $3x - y - 7 = 0$ and $2x + z - 8 = 0$.

Check with $t = 1$, the point $(3, 2, 2)$: $9 - 2 - 7 = 0$ ✓ and $6 + 2 - 8 = 0$ ✓.

Example 4: From two planes to a vector equation

The line $L$ is the intersection of the planes $x - y + z = 1$ and $2x + y - z = 5$. Find a vector equation of $L$.

Solution. Add the equations to eliminate both $y$ and $z$:

$$3x = 6 \quad\Rightarrow\quad x = 2$$

Then the first equation gives $2 - y + z = 1$, so $z = y - 1$. Let $y = t$:

$$x = 2, \qquad y = t, \qquad z = -1 + t$$ $$\vec{r} = [2, 0, -1] + t[0, 1, 1], \quad t \in \mathbb{R}$$

Check with the cross product: $[1, -1, 1] \times [2, 1, -1] = [0, 3, 3] = 3[0, 1, 1]$ ✓. And the point $(2, 0, -1)$ satisfies both planes: $2 - 0 - 1 = 1$ ✓ and $4 + 0 + 1 = 5$ ✓.

Common mistakes

Writing a single scalar equation for a line in 3-space. An equation like $2x - y + 3z = 4$ is a plane. A line in 3-space needs a vector or parametric equation, or two scalar equations.

Testing only two coordinates. In Example 2, point $D$ matched the line in $x$ and $y$ but not in $z$. All three components must give the same $t$.

Using a position vector as the direction. The direction vector through $A$ and $B$ is $\overrightarrow{AB} = \vec{b} - \vec{a}$, not $\vec{a}$ or $\vec{b}$ themselves.

Dividing by zero in symmetric equations. If a direction component is $0$, that coordinate is constant. Write it separately (for example, $y = -1$) instead of putting a $0$ in a denominator.

Using the same parameter for two different lines. When you work with two lines at once, give them different parameters, like $t$ and $s$. Otherwise you’re forcing both lines to be at the “same time”, which is a different question.

Practice

1. (Warm-up) Write vector and parametric equations of the line through $(0, 5, -3)$ with direction vector $[1, -2, 4]$. Find the point where $t = -1$.

Solution

$$\vec{r} = [0, 5, -3] + t[1, -2, 4] \qquad x = t, \quad y = 5 - 2t, \quad z = -3 + 4t$$

For $t = -1$: $(-1,\ 5 + 2,\ -3 - 4) = (-1, 7, -7)$.

2. (Warm-up) State a point on the line and a direction vector: $x = 4 - t$, $y = 2t$, $z = -3 + 5t$.

Solution

At $t = 0$ the point is $(4, 0, -3)$. The coefficients of $t$ give the direction vector $[-1, 2, 5]$.

3. (Warm-up) Is the point $(6, -4, 7)$ on the line $\vec{r} = [2, 0, -1] + t[2, -2, 4]$?

Solution

$2 + 2t = 6$ gives $t = 2$. Then $y = 0 - 2(2) = -4$ ✓ and $z = -1 + 4(2) = 7$ ✓.

Yes, the point is on the line (at $t = 2$).

4. (Core) Find vector and symmetric equations of the line through $A(-2, 3, 1)$ and $B(4, 0, 7)$.

Solution

$\overrightarrow{AB} = [6, -3, 6] = 3[2, -1, 2]$, so use $\vec{m} = [2, -1, 2]$:

$$\vec{r} = [-2, 3, 1] + t[2, -1, 2]$$

Symmetric equations:

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

Check $B$: $t = 3$ gives $(-2 + 6,\ 3 - 3,\ 1 + 6) = (4, 0, 7)$ ✓.

5. (Core) Find a vector equation of the line through $(1, -1, 4)$ parallel to the line $\vec{r} = [0, 2, 5] + s[3, 1, -2]$. Are the two lines the same?

Solution

Parallel lines share a direction vector:

$$\vec{r} = [1, -1, 4] + t[3, 1, -2]$$

Is $(1, -1, 4)$ on the given line? $3s = 1$ gives $s = \dfrac{1}{3}$, but then $y = 2 + \dfrac{1}{3} \ne -1$. So the lines are parallel and distinct.

6. (Core)

• (a) Write a vector equation of the line through $(2, -3, 5)$ parallel to the $z$-axis.
• (b) Explain why the equation $2x - y = 7$ does not describe a single line in 3-space.

Solution

(a) The $z$-axis has direction $\vec{k} = [0, 0, 1]$:

$$\vec{r} = [2, -3, 5] + t[0, 0, 1]$$

(b) In 3-space, $z$ is free in $2x - y = 7$: every point $(x, y, z)$ with $2x - y = 7$ works, for any $z$. Those points form a plane (parallel to the $z$-axis), not a line.

7. (Core) Represent the line through $(3, 1, -2)$ and $(1, 1, 2)$ with a vector equation, with parametric equations, and as the intersection of two planes.

Solution

Direction: $[1 - 3,\ 1 - 1,\ 2 - (-2)] = [-2, 0, 4] = 2[-1, 0, 2]$.

$$\vec{r} = [3, 1, -2] + t[-1, 0, 2] \qquad x = 3 - t, \quad y = 1, \quad z = -2 + 2t$$

For the planes: $y = 1$ is already one. From $x = 3 - t$, $t = 3 - x$, so $z = -2 + 2(3 - x) = 4 - 2x$, which is $2x + z - 4 = 0$.

The line is the intersection of $y = 1$ and $2x + z - 4 = 0$.

Check $(1, 1, 2)$: $y = 1$ ✓ and $2 + 2 - 4 = 0$ ✓.

8. (Challenge) Where does the line $\vec{r} = [1, 2, -3] + t[2, -1, 3]$ cross the $xy$-plane? Where does it cross the $xz$-plane?

Solution

The $xy$-plane is $z = 0$: $-3 + 3t = 0$ gives $t = 1$, so the point is $(3, 1, 0)$.

The $xz$-plane is $y = 0$: $2 - t = 0$ gives $t = 2$, so the point is $(5, 0, 3)$.

9. (Challenge) Find a vector equation of the line of intersection of the planes $x + 2y - z = 4$ and $3x - y + z = 2$, using the cross product for the direction.

Solution

Direction: the line is perpendicular to both normals.

$$[1, 2, -1] \times [3, -1, 1] = [2(1) - (-1)(-1),\ (-1)(3) - 1(1),\ 1(-1) - 2(3)] = [1, -4, -7]$$

For a point, set $x = 0$: $2y - z = 4$ and $-y + z = 2$. Adding gives $y = 6$, then $z = 8$. The point is $(0, 6, 8)$.

$$\vec{r} = [0, 6, 8] + t[1, -4, -7]$$

Check with $t = 1$, the point $(1, 2, 1)$: $1 + 4 - 1 = 4$ ✓ and $3 - 2 + 1 = 2$ ✓.