Equation of a line in 3D — vector, symmetric, parametric forms: Worked examples

Before anything, one reminder built from zero. A direction ratio (DR) is just one of the three numbers $(a, b, c)$ that tells you "for every step of the parameter, move $a$ in $x$ , $b$ in $y$ , $c$ in $z$ ." The parameter is the ==scalar $λ$ ==: think of it as how many steps you take (negative = walk backwards). See Direction ratios and direction cosines for why any scalar multiple of $(a, b, c)$ names the same direction.

The scenario matrix

Every problem this topic can throw at you falls into one of these cells. The worked examples below are tagged with the cell they cover.

Cell	What makes it special	Example
A. Clean	all three DRs nonzero, build all forms	Ex 1
B. One zero DR	line frozen in one coordinate (lies in a plane)	Ex 2
C. Two zero DRs	line parallel to a coordinate axis	Ex 3
D. Negative DRs / sign traps	reading a point back from symmetric form	Ex 4
E. Point-on-line test	same $λ$ must fit all three	Ex 5
F. Intersection with a plane	limiting/pierce point — find the one $λ$	Ex 6
G. Two lines, same or different?	parallel vs identical vs distinct	Ex 7
H. Real-world word problem	drone/ray path, physical anchor + heading	Ex 8

The degenerate/limiting behaviours (zero DRs, axis-parallel lines, a point sitting exactly at the anchor $λ = 0$ ) are covered inside cells B, C, and E — nothing is left unshown.

Cell A — the clean build

Worked example Example 1 — all three forms, every DR nonzero

Find the vector, parametric and symmetric forms of the line through $A (1, - 2, 4)$ parallel to $b = 3 \hat{i} - \hat{j} + 2 \hat{k}$ .

Forecast: Before reading on, guess what number will sit under the $y$ -fraction in the symmetric form. (It is the $y$ -component of $b$ — is it $+ 1$ or $- 1$ ?)

Step 1 — write the vector form. Why this step? The master equation is anchor $+ λ \cdot$ direction; nothing else defines a line. $r = (1 \hat{i} - 2 \hat{j} + 4 \hat{k}) + λ (3 \hat{i} - \hat{j} + 2 \hat{k}) .$

Step 2 — split into coordinates. Why this step? Each coordinate of $r$ moves independently but is driven by the same $λ$ . $x = 1 + 3 λ, y = - 2 - λ, z = 4 + 2 λ .$

Step 3 — eliminate $λ$ . Why this step? $λ$ is scaffolding; solving each equation for it and equating removes it. $λ = \frac{x - 1}{3} = \frac{y + 2}{- 1} = \frac{z - 4}{2} .$

Verify: Put $λ = 2$ . Parametric gives $(1 + 6, - 2 - 2, 4 + 4) = (7, - 4, 8)$ . Check in symmetric: $\frac{7 - 1}{3} = 2$ , $\frac{- 4 + 2}{- 1} = 2$ , $\frac{8 - 4}{2} = 2$ . All equal $2$ ✓. (Forecast answer: the $y$ -denominator is $- 1$ .)

Notice how the anchor lives in the numerators and the direction in the denominators — see the figure: the arrow is $b$ , the fixed dot is $A$ .

Figure — Equation of a line in 3D — vector, symmetric, parametric forms

Cell B — one zero DR (line trapped in a plane)

Worked example Example 2 — the

$z$ -coordinate is frozen Find the symmetric form of the line through $P (2, 1, 5)$ and $Q (6, 4, 5)$ .

Forecast: Both points have $z = 5$ . As you walk along the line, does $z$ ever change? Guess before computing.

Step 1 — get the direction. Why this step? The direction of a two-point line is the displacement $P Q = q - p$ (see Vectors — scalar multiplication & parallel vectors). $P Q = (6 - 2, 4 - 1, 5 - 5) = (4, 3, 0) .$

Step 2 — spot the zero DR. Why this step? A denominator of $0$ is meaningless. A zero DR means that coordinate never changes — it is frozen at its anchor value. Here $c = 0$ , so $z$ stays at $5$ forever.

Step 3 — write the correct symmetric form. Why this step? Drop the impossible fraction and state the frozen coordinate as a separate equation. $\frac{x - 2}{4} = \frac{y - 1}{3}, z = 5.$

Verify: $λ = 1$ should land on $Q$ . Parametric: $x = 2 + 4 = 6$ , $y = 1 + 3 = 4$ , $z = 5$ → $(6, 4, 5) = Q$ ✓. (Forecast: $z$ never changes — the line lives entirely in the flat plane $z = 5$ .)

Cell C — two zero DRs (parallel to an axis)

Worked example Example 3 — a line parallel to the

$x$ -axis Write the equation of the line through $A (3, - 2, 7)$ with direction $b = (5, 0, 0)$ .

Forecast: Two of the DRs are zero. Which coordinates are frozen and which is free? Guess.

Step 1 — parametric form. Why this step? Read the components straight off the master equation. $x = 3 + 5 λ, y = - 2, z = 7.$

Step 2 — interpret. Why this step? Two zero DRs freeze two coordinates; only $x$ moves. The line is a horizontal skewer parallel to the $x$ -axis.

Step 3 — symmetric form. Why this step? Only one coordinate has a nonzero denominator, so only one fraction survives; the other two become equalities. $y = - 2, z = 7 (x free) .$

Verify: any $λ$ keeps $y = - 2, z = 7$ ; the point always has the form $(*, - 2, 7)$ . At $λ = 0$ we sit on the anchor $(3, - 2, 7)$ ✓. (Forecast: $y$ and $z$ frozen, $x$ free.)

Cell D — negative DRs & sign traps

Worked example Example 4 — read the point back correctly

Given the line $\frac{x + 3}{2} = \frac{y - 1}{- 4} = \frac{z + 5}{- 1}$ , state its anchor point and direction ratios.

Forecast: Is the $z$ -coordinate of the anchor $+ 5$ or $- 5$ ? Guess before reading.

Step 1 — force each numerator into the shape $x - x_{1}$ . Why this step? The point is only readable when the numerator literally looks like " $x$ minus something." $x + 3 = x - (- 3), y - 1 = y - (1), z + 5 = z - (- 5) .$

Step 2 — read the anchor. Why this step? Now the subtracted numbers are exactly the coordinates. $(x_{1}, y_{1}, z_{1}) = (- 3, 1, - 5) .$

Step 3 — read the direction. Why this step? Denominators are the DRs, signs and all. Negative just means "walk the other way" along that axis. $(a, b, c) = (2, - 4, - 1) .$

Verify: plug the anchor: $\frac{- 3 + 3}{2} = 0$ , $\frac{1 - 1}{- 4} = 0$ , $\frac{- 5 + 5}{- 1} = 0$ — all zero, so the anchor satisfies the line (this is the $λ = 0$ point) ✓. (Forecast: the anchor's $z$ is $- 5$ , not $+ 5$ .)

Cell E — does a point lie on the line?

Worked example Example 5 — the one-

$λ$ -fits-all test Does $R (8, - 3, 1)$ lie on $r = (2, 1, 4) + λ (3, - 2, - 1)$ ?

Forecast: Guess yes or no before computing — then see if your instinct survives.

Step 1 — solve for $λ$ from the $x$ -equation. Why this step? One coordinate fixes the candidate parameter; the same $λ$ must then work everywhere. $8 = 2 + 3 λ \Rightarrow λ = 2.$

Step 2 — test that $λ$ in $y$ . Why this step? All three coordinates share one $λ$ ; a mismatch anywhere kills the membership. $y = 1 + (- 2) (2) = 1 - 4 = - 3. Matches R_{y} = - 3 ✓$

Step 3 — test $λ$ in $z$ . Why this step? We must check the last coordinate too — passing two is not enough. $z = 4 + (- 1) (2) = 2. But R_{z} = 1. \times$

Verify: since $z$ fails, $R$ is not on the line, even though $x$ and $y$ agreed. This is exactly the classic trap — always check all three. (Contrast: the point $(8, - 3, 2)$ would lie on it, at $λ = 2$ .)

Cell F — where a line pierces a plane (limiting/pierce point)

Worked example Example 6 — intersection of a line and a plane

Where does the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 2}{- 1}$ meet the plane $x + y + z = 6$ ?

Forecast: Will there be one pierce point, none, or a whole line of them? Guess.

Step 1 — go parametric. Why this step? A single parameter lets us feed the whole line into the plane at once. $x = 1 + 2 λ, y = - 1 + 3 λ, z = 2 - λ .$

Step 2 — substitute into the plane. Why this step? The pierce point is the one $λ$ where the moving point satisfies the plane's equation. (See Equation of a plane in 3D.) $(1 + 2 λ) + (- 1 + 3 λ) + (2 - λ) = 6.$

Step 3 — solve the single linear equation. Why this step? Collect terms; a unique $λ$ means one clean crossing. $2 + 4 λ = 6 \Rightarrow λ = 1.$

Step 4 — put $λ = 1$ back. Why this step? The parameter labels the point; feed it back to get coordinates. $(x, y, z) = (1 + 2, - 1 + 3, 2 - 1) = (3, 2, 1) .$

Verify: does $(3, 2, 1)$ satisfy the plane? $3 + 2 + 1 = 6$ ✓. And it satisfies the line at $λ = 1$ by construction ✓. (Forecast: exactly one pierce point, because the direction is not parallel to the plane.)

Cell G — two lines: same, parallel, or distinct?

Worked example Example 7 — identical, parallel, or crossing?

Line $L_{1} : r = (1, 0, 2) + λ (2, 4, - 2)$ and line $L_{2} : r = (2, 2, 1) + μ (1, 2, - 1)$ . Are they the same line, parallel-but-distinct, or neither?

Forecast: Their directions look suspiciously proportional. Guess the verdict.

Step 1 — compare directions. Why this step? Two lines are parallel exactly when their directions are scalar multiples (Vectors — scalar multiplication & parallel vectors). $(2, 4, - 2) = 2 \times (1, 2, - 1) . Parallel ✓$

Step 2 — test if $L_{1}$ 's anchor lies on $L_{2}$ . Why this step? Parallel lines are the same line only if they share a point; otherwise they are distinct. Solve $(1, 0, 2) = (2, 2, 1) + μ (1, 2, - 1)$ . From $x$ : $1 = 2 + μ \Rightarrow μ = - 1$ . Check $y$ : $2 + 2 (- 1) = 0$ ✓. Check $z$ : $1 - (- 1) = 2$ ✓.

Step 3 — conclude. Why this step? All three coordinates agreed for one $μ$ , so the anchor of $L_{1}$ sits on $L_{2}$ . The lines are identical.

Verify: $L_{1}$ 's anchor $(1, 0, 2)$ reproduced on $L_{2}$ at $μ = - 1$ : $(2 - 1, 2 - 2, 1 + 1) = (1, 0, 2)$ ✓. Same direction + shared point ⇒ one and the same line.

Cell H — real-world word problem

Worked example Example 8 — a drone's straight flight

A drone launches from $(0, 0, 10)$ metres and flies in a straight line with constant velocity $v = (4, 3, - 2)$ m/s. (a) Where is it after $t = 3$ s? (b) At what time does it reach ground level $z = 0$ ?

Forecast: Guess whether the drone actually reaches the ground (does $z$ decrease?) before computing.

Step 1 — model the path as a line. Why this step? Constant velocity means position is anchor $+ t \cdot$ velocity — the parameter $t$ is time here, with real units (seconds). $r (t) = (0, 0, 10) + t (4, 3, - 2) .$

Step 2 — (a) evaluate at $t = 3$ . Why this step? Plug the value of the parameter to get the point. $r (3) = (12, 9, 10 - 6) = (12, 9, 4) m .$

Step 3 — (b) set $z = 0$ and solve. Why this step? Ground level is the plane $z = 0$ ; find the one $t$ where the $z$ -coordinate hits it. $10 - 2 t = 0 \Rightarrow t = 5 s .$

Verify: at $t = 5$ , $z = 10 - 2 (5) = 0$ ✓ — units: $[m] - [m/s] [s] = [m]$ , consistent. The horizontal position then is $(20, 15, 0)$ . (Forecast: yes, it lands, because the $z$ -velocity $- 2$ is negative so $z$ steadily drops.)

Recall Which cell? (click to reveal)

A line through two points both having $y = 4$ ::: Cell B — one zero DR, frozen $y$ , lives in plane $y = 4$ . Direction $(0, 0, 7)$ ::: Cell C — two zero DRs, parallel to the $z$ -axis. Symmetric form with a negative denominator ::: Cell D — sign trap, direction just points the other way. "Does this point lie on the line?" ::: Cell E — one $λ$ must satisfy all three equations. "Where does the line hit the plane?" ::: Cell F — substitute parametric into the plane, solve one $λ$ .

Connections

Equation of a line in 3D — vector, symmetric, parametric forms (parent)
Direction ratios and direction cosines
Vectors — scalar multiplication & parallel vectors
Angle between two lines in 3D
Equation of a plane in 3D
Coplanarity of two lines
Distance of a point from a line in 3D
Shortest distance between two skew lines