Equation of a line in 3D

A line in three-dimensional space is determined by one point and one direction. Both data come together in the standard parametric / vector form.

Vector form

A line through the point with position vector $\vec a$ in the direction of the (non-zero) vector $\vec d$ consists of all points $\vec r = \vec a + t \vec d, \qquad t \in \mathbb R.$ The parameter $t$ traces the line: at $t = 0$ you're at $\vec a$; at $t = 1$ you're at $\vec a + \vec d$, etc.

The same line is described by many such equations. Replacing $\vec a$ by any other point $\vec a'$ on the line, or scaling $\vec d$ by any non-zero constant, gives an equally valid description.

Cartesian form

If $\vec a = (x_0, y_0, z_0)$ and $\vec d = (d_1, d_2, d_3)$, then writing $\vec r = (x, y, z)$: $x = x_0 + t d_1, \quad y = y_0 + t d_2, \quad z = z_0 + t d_3.$ Eliminate $t$: $\frac{x - x_0}{d_1} = \frac{y - y_0}{d_2} = \frac{z - z_0}{d_3}.$ This is the symmetric or Cartesian form. It is valid only when no $d_i = 0$; when some component of $\vec d$ vanishes, you write that coordinate as a constant equation. E.g., if $d_3 = 0$: $\dfrac{x - x_0}{d_1} = \dfrac{y - y_0}{d_2}, z = z_0$.

Line through two given points

A line through $A = (x_1, y_1, z_1)$ and $B = (x_2, y_2, z_2)$ has direction $\vec{AB} = (x_2 - x_1, y_2 - y_1, z_2 - z_1)$. So $\vec r = \vec a + t(\vec b - \vec a),$ or in Cartesian form $\frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1}.$ (With the same caveat about zero denominators.)

Converting between forms

Vector $\to$ Cartesian: read off the point $(x_0, y_0, z_0)$ and the direction $(d_1, d_2, d_3)$, then write the symmetric form.

Cartesian $\to$ Vector: from $\dfrac{x - x_0}{d_1} = \dfrac{y - y_0}{d_2} = \dfrac{z - z_0}{d_3}$, the line is $\vec r = (x_0, y_0, z_0) + t(d_1, d_2, d_3)$.

When is a given point on a line?

Substitute. In vector form: solve $\vec p = \vec a + t \vec d$ for $t$. If a consistent $t$ exists in all three coordinates, $\vec p$ is on the line. In Cartesian form: plug into the symmetric equation and check that all three ratios are equal.

Worked examples

Example 1. Find the vector and Cartesian equations of the line through $(2, -1, 3)$ in the direction $\hat i + 2\hat j - 2\hat k$.

Vector: $\vec r = (2, -1, 3) + t(1, 2, -2)$, i.e. $\vec r = (2 + t)\hat i + (-1 + 2t)\hat j + (3 - 2t)\hat k$.

Cartesian: $\dfrac{x - 2}{1} = \dfrac{y + 1}{2} = \dfrac{z - 3}{-2}$.

Example 2. Find the equation of the line through $(1, 0, 2)$ and $(4, 3, 8)$.

Direction: $(3, 3, 6)$, or simplified $(1, 1, 2)$. Vector form: $\vec r = (1, 0, 2) + t(1, 1, 2)$.

Cartesian: $\dfrac{x - 1}{1} = \dfrac{y}{1} = \dfrac{z - 2}{2}$.

Example 3. Find the equation of the $x$-axis.

Point $(0, 0, 0)$, direction $(1, 0, 0)$. Vector form: $\vec r = t\hat i$. Cartesian: $y = 0$, $z = 0$.

Example 4. Does the point $(3, 2, 1)$ lie on $\dfrac{x - 1}{1} = \dfrac{y + 1}{3} = \dfrac{z - 2}{-1}$?

Check: $\dfrac{3 - 1}{1} = 2$; $\dfrac{2 + 1}{3} = 1$; $\dfrac{1 - 2}{-1} = 1$. Not all equal , so no.

Example 5. Find $\lambda$ if the point $(\lambda, 1, 2)$ lies on the line $\dfrac{x - 1}{2} = \dfrac{y - 1}{0} = \dfrac{z - 2}{3}$.

The middle ratio $\dfrac{y - 1}{0}$ is interpreted as $y - 1 = 0$, i.e. $y = 1$ , which is satisfied by our point. So the other two ratios must agree: $\dfrac{\lambda - 1}{2} = \dfrac{2 - 2}{3} = 0$, so $\lambda = 1$.

Example 6. Find the foot of the perpendicular from $(1, 2, 3)$ to the line $\dfrac{x}{1} = \dfrac{y}{2} = \dfrac{z}{3}$.

Parametrise: $\vec r = t(1, 2, 3)$. Foot $F = (t, 2t, 3t)$ on the line is the one where $\vec{PF} \perp$ direction $(1, 2, 3)$. $\vec{PF} = (t - 1, 2t - 2, 3t - 3)$. Dot with $(1, 2, 3)$: $(t - 1) + (4t - 4) + (9t - 9) = 14t - 14 = 0$, so $t = 1$. Foot: $(1, 2, 3)$. (Interesting , the point itself is on the line.)

Try it yourself

1. Find the vector equation of the line through $(0, 1, 2)$ in direction $\hat i + \hat j + \hat k$.
2. Find the Cartesian equation of the line through $(1, 2, 3)$ and $(4, 5, 6)$.
3. Find direction ratios of the line $\dfrac{x - 1}{2} = \dfrac{y + 3}{-1} = \dfrac{z}{5}$.
4. Does the point $(2, -1, 5)$ lie on $\dfrac{x - 1}{1} = \dfrac{y + 2}{-1} = \dfrac{z - 4}{1}$?
5. Find the Cartesian equation of the line through $(1, 0, -1)$ parallel to the $y$-axis.
6. Find vector and Cartesian forms of the line through $(0, 0, 0)$ and $(2, 3, -1)$.
7. Find the equation of the line through $(2, 1, 4)$ perpendicular to the lines $\vec r = t(1, 1, 0)$ and $\vec r = s(0, 1, 1)$.
8. Find the foot of perpendicular from $(1, 2, 1)$ to the line $\dfrac{x}{1} = \dfrac{y - 1}{2} = \dfrac{z - 2}{3}$.
9. Find $\alpha, \beta$ if $(\alpha, \beta, 7)$ lies on $\dfrac{x}{1} = \dfrac{y}{2} = \dfrac{z}{7}$.
10. The line through $(1, -1, 3)$ with direction $(2, 1, 2)$ crosses the $xy$-plane at what point?
11. Find the image of $(1, 2, 3)$ in the line $\dfrac{x - 1}{1} = \dfrac{y - 1}{2} = \dfrac{z + 1}{1}$.
12. Find the distance between $(1, 0, 0)$ and the line $\dfrac{x}{1} = \dfrac{y}{2} = \dfrac{z}{3}$.
13. Convert vector form $\vec r = (1, 2, -1) + t(3, 1, 2)$ to Cartesian.
14. Convert Cartesian form $\dfrac{x + 1}{1} = \dfrac{y - 2}{-1} = \dfrac{z}{2}$ to vector.

Pitfalls and tricks

• Both forms describe the same line. Pick whichever is convenient for the problem.
• Zero denominators in Cartesian form are handled by writing the corresponding coordinate as a constant equation.
• A line in 3D needs two equations (two simultaneous constraints) in Cartesian form , symmetric form packages them as one chain.
• Parameter $t$ is arbitrary , it does not matter that it appears explicitly; eliminate it for Cartesian.
• To check whether a point is on a line, plug into the parametric form and solve for $t$.