Equation of a Line in Space

Line Through a Point, Parallel to a Vector

A line in space is pinned down by (i) a point on it plus a direction, or (ii) two points on it.

Let $A$ (position vector $\vec{a}$) lie on line $l$, and let $l$ be parallel to $\vec{b}$. For any point $P$ (position vector $\vec{r}$) on the line, $\overrightarrow{AP}$ is parallel to $\vec{b}$: $\vec{r} - \vec{a} = \lambda\vec{b}$. Hence the vector equation:

$$\vec{r} = \vec{a} + \lambda\vec{b}, \qquad \lambda \in \mathbb{R}$$

Each value of the parameter $\lambda$ produces exactly one point of the line — $\lambda = 0$ gives $A$ itself; sweeping $\lambda$ sweeps the whole line.

Key Point: if $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$, then $a, b, c$ are direction ratios of the line — and conversely, direction ratios of a line hand you a parallel vector for free. The letter $b$ here names a component, not the magnitude $|\vec{b}|$.

Cartesian (symmetric) form

Writing $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$, $A = (x_1, y_1, z_1)$ and equating components gives the parametric equations $x = x_1 + \lambda a$, $y = y_1 + \lambda b$, $z = z_1 + \lambda c$; eliminating $\lambda$:

$$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$$

If direction cosines $l, m, n$ are used as denominators, the common value of the three fractions is the signed distance along the line from $(x_1, y_1, z_1)$.

Line Through Two Points, and Form-Keeping Rules

Two-point form

The line through points $A(\vec{a})$ and $B(\vec{b})$ has direction $\vec{b} - \vec{a}$, so:

$$\vec{r} = \vec{a} + \lambda\left(\vec{b} - \vec{a}\right)$$

and in Cartesian form (direction ratios = coordinate differences):

$$\frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1}$$

$\lambda = 0$ lands on $A$, $\lambda = 1$ on $B$, and $0 < \lambda < 1$ traces the segment between them.

Reading equations correctly — the three traps

1. Numerators must be $x - x_1$, never $x_1 - x$ or $3 - x$. An equation like $\dfrac{3 - x}{5} = \dfrac{y + 4}{7} = \dfrac{2z - 6}{4}$ must be standardised before reading anything off: $\dfrac{x - 3}{-5} = \dfrac{y - (-4)}{7} = \dfrac{z - 3}{2}$ — direction ratios $(-5, 7, 2)$ and point $(3, -4, 3)$, not $(5, 7, 4)$ and $(3, -4, 6)$.
2. Coefficients of $x, y, z$ must be $1$: divide out any $2z$-style factor (and its denominator) first.
3. Testing whether a point lies on a line: substitute into all three fractions — the point is on the line only if all three give the same value of $\lambda$. Two matching and one differing means NOT on the line.

Key Point (conversions both ways): vector → Cartesian: read the point from $\vec{a}$ and denominators from $\vec{b}$. Cartesian → vector: the point comes from the numerators, the parallel vector from the denominators. Every exam conversion is these two sentences.

Solved Examples

Example 1: Both forms from a point and a vector

Find the vector and Cartesian equations of the line through $(5, 2, -4)$ parallel to $3\hat{i} + 2\hat{j} - 8\hat{k}$.

Solution:

1. Vector form: $\vec{r} = \left(5\hat{i} + 2\hat{j} - 4\hat{k}\right) + \lambda\left(3\hat{i} + 2\hat{j} - 8\hat{k}\right)$.
2. Cartesian form (point over ratios): $\dfrac{x - 5}{3} = \dfrac{y - 2}{2} = \dfrac{z + 4}{-8}$.

Answer: as above — note $z + 4$ is $z - (-4)$, and the $-8$ stays in the denominator with its sign.

Example 2: Cartesian to vector

Write the vector equation of the line $\dfrac{x + 3}{2} = \dfrac{y - 5}{4} = \dfrac{z - 6}{2}$.

Solution:

1. Point from numerators: $(-3, 5, 6)$.
2. Direction from denominators: $2\hat{i} + 4\hat{j} + 2\hat{k}$.

Answer: $\vec{r} = \left(-3\hat{i} + 5\hat{j} + 6\hat{k}\right) + \lambda\left(2\hat{i} + 4\hat{j} + 2\hat{k}\right)$ — equivalently with the simplified direction $\hat{i} + 2\hat{j} + \hat{k}$, since direction ratios scale freely.

Example 3: Through two points

Find the vector and Cartesian equations of the line through $A(-1, 0, 2)$ and $B(3, 4, 6)$.

Solution:

1. Direction: $\vec{b} - \vec{a} = 4\hat{i} + 4\hat{j} + 4\hat{k}$.
2. Vector form: $\vec{r} = \left(-\hat{i} + 2\hat{k}\right) + \lambda\left(4\hat{i} + 4\hat{j} + 4\hat{k}\right)$.
3. Cartesian form: $\dfrac{x + 1}{4} = \dfrac{y}{4} = \dfrac{z - 2}{4}$ (or with the reduced ratios $1, 1, 1$).

Answer: as above; the reduced direction $(1, 1, 1)$ shows the line is equally inclined to all three axes.

Example 4: Standardising a disguised equation

Find the direction ratios and a point of the line $\dfrac{3 - x}{5} = \dfrac{y + 4}{7} = \dfrac{2z - 6}{4}$.

Solution:

1. Fix the $x$-numerator: $\dfrac{3 - x}{5} = \dfrac{x - 3}{-5}$.
2. Fix the $z$-term: $\dfrac{2z - 6}{4} = \dfrac{2(z - 3)}{4} = \dfrac{z - 3}{2}$.
3. Read off the standard form $\dfrac{x - 3}{-5} = \dfrac{y + 4}{7} = \dfrac{z - 3}{2}$.

Answer: direction ratios $(-5, 7, 2)$; point $(3, -4, 3)$. Reading the raw equation without standardising gives wrong ratios and a wrong point — this single trap fills board papers every year.

Example 5: Is the point on the line?

Does the point $(5, 8, 11)$ lie on the line $\vec{r} = \left(\hat{i} + 2\hat{j} + 3\hat{k}\right) + \lambda\left(2\hat{i} + 3\hat{j} + 4\hat{k}\right)$?

Solution:

1. Equate components: $5 = 1 + 2\lambda$, $8 = 2 + 3\lambda$, $11 = 3 + 4\lambda$.
2. Solve each: $\lambda = 2$, $\lambda = 2$, $\lambda = 2$ — all three agree.

Answer: yes, it is the point with $\lambda = 2$. Had even one equation disagreed, the answer would be no — all three must produce the same $\lambda$.

Example 6: A line through the origin

Write both forms of the line through the origin with direction ratios $2, -1, 3$.

Solution:

1. Vector form ($\vec{a} = \vec{0}$): $\vec{r} = \lambda\left(2\hat{i} - \hat{j} + 3\hat{k}\right)$.
2. Cartesian form: $\dfrac{x}{2} = \dfrac{y}{-1} = \dfrac{z}{3}$.

Answer: as above — through the origin, the numerators are simply $x, y, z$.