Intersection of Lines in 3D
In two dimensions, two distinct lines either intersect at exactly one point or are parallel. In three dimensions, a third possibility opens up: the lines may be skew — non-parallel and yet never meeting. This lesson teaches you to test two lines given in parametric form, solve the simultaneous system in two parameters $\lambda$ and $\mu$, and classify the relationship with confidence.
Two lines in 2D either meet at a point or are parallel. List every possibility for two lines in 3D, and for each case describe how the parametric equations would behave when you try to solve them simultaneously.
Two habits make every intersection question routine: first compare the direction vectors — if one is a scalar multiple of the other, the lines are parallel (and you only need to test a single point to know whether they coincide). If they aren't parallel, set the position vectors equal, solve any two component equations for $\lambda$ and $\mu$, then test the third. Consistency means intersection; contradiction means skew.
The direction-first then consistency reading: (1) extract direction vectors $\mathbf{d}_1, \mathbf{d}_2$, (2) check $\mathbf{d}_1 \parallel \mathbf{d}_2$, (3) if not parallel, solve simultaneously and test the third component.
$\mathbf{r}_1 = \mathbf{a} + \lambda \mathbf{d}_1$ · $\mathbf{r}_2 = \mathbf{c} + \mu \mathbf{d}_2$ · meet $\Leftrightarrow \mathbf{a} + \lambda \mathbf{d}_1 = \mathbf{c} + \mu \mathbf{d}_2$
Key facts
- Three classifications: intersecting, parallel (including coincident), skew
- Two lines are parallel iff $\mathbf{d}_1 = k\mathbf{d}_2$ for some scalar $k$
- Skew lines exist only in $\mathbb{R}^3$ (or higher) — never in $\mathbb{R}^2$
- NESA MEX-V1: vectors in three dimensions and lines
Concepts
- Why two different parameters $\lambda, \mu$ are required
- Why the third component equation is the consistency test
- Why parallel and coincident must be distinguished by a point check
Skills
- Set up and solve the simultaneous parametric system
- Classify any pair of 3D lines as intersecting, parallel or skew
- Find the point of intersection when one exists
Let $\ell_1: \mathbf{r}_1 = \mathbf{a} + \lambda \mathbf{d}_1$ and $\ell_2: \mathbf{r}_2 = \mathbf{c} + \mu \mathbf{d}_2$ be two lines in $\mathbb{R}^3$. The classification proceeds in two steps.
- Compare directions. If $\mathbf{d}_1 = k\mathbf{d}_2$ for some scalar $k$, the lines are parallel. Substitute $\mathbf{a}$ into $\ell_2$: if it satisfies $\ell_2$, the lines are coincident; otherwise they are strictly parallel (no intersection).
- If not parallel, solve simultaneously. Set $\mathbf{a} + \lambda \mathbf{d}_1 = \mathbf{c} + \mu \mathbf{d}_2$. Component-wise this gives three equations in two unknowns. Solve any two for $\lambda$ and $\mu$, then substitute into the third. If it holds — the lines intersect, and the point is found by substituting $\lambda$ into $\ell_1$ (or $\mu$ into $\ell_2$). If the third equation fails — the lines are skew.
Three cases: intersect (unique $\lambda, \mu$), parallel ($\mathbf{d}_1 \parallel \mathbf{d}_2$), skew (not parallel, no solution) · Parallel + point on both line $\Rightarrow$ coincident · Use different parameters $\lambda, \mu$ for the two lines · Three equations, two unknowns: solve two, test the third
Pause — copy the three classification cases (intersect, parallel/coincident, skew), the rule to use different parameters $\lambda,\mu$ for the two lines, and the strategy of 3 equations in 2 unknowns into your book.
Quick check: Two lines in $\mathbb{R}^3$ have non-parallel direction vectors, and the simultaneous parametric system gives no solution. The lines are:
We just saw the three cases for two 3D lines: intersect (unique $\lambda,\mu$), parallel (directions proportional), skew (not parallel, no solution). That raises a question: how do we actually solve the simultaneous system to decide between intersection and skew? This card answers it → set position vectors equal for 3 equations in $\lambda,\mu$; solve any 2, substitute into the 3rd; if it holds → intersection, if not → skew.
Given $\mathbf{r}_1 = (a_1, a_2, a_3) + \lambda(p_1, p_2, p_3)$ and $\mathbf{r}_2 = (c_1, c_2, c_3) + \mu(q_1, q_2, q_3)$, equating components yields:
- $a_1 + \lambda p_1 = c_1 + \mu q_1$ (x-equation)
- $a_2 + \lambda p_2 = c_2 + \mu q_2$ (y-equation)
- $a_3 + \lambda p_3 = c_3 + \mu q_3$ (z-equation)
Strategy: pick the two equations with the simplest coefficients, rearrange into a linear system, and solve for $\lambda$ and $\mu$ (elimination or substitution). Then substitute both values into the third equation:
Set position vectors equal $\to$ 3 component equations · Solve any 2 for $\lambda$ and $\mu$ $\to$ candidate solution · Substitute into the 3rd: if it holds, intersection; if it fails, skew · Point of intersection: substitute $\lambda$ back into $\mathbf{r}_1$
Pause — copy the solution procedure (equate, solve 2 equations, test the 3rd), the intersection-point recovery ($\lambda^*$ back into $\mathbf{r}_1$), and the skew conclusion (3rd equation fails) into your book.
Did you get this? True or false: if two non-parallel lines in $\mathbb{R}^3$ are written as $\mathbf{r}_1 = \mathbf{a} + \lambda \mathbf{d}_1$ and $\mathbf{r}_2 = \mathbf{c} + \mu \mathbf{d}_2$, we may set $\lambda = \mu$ when testing for intersection.
Worked examples · 3 in a row, reveal as you go
Show that the lines $\mathbf{r}_1 = (1,2,3) + \lambda(2,-1,1)$ and $\mathbf{r}_2 = (4,0,5) + \mu(1,1,-1)$ intersect, and find the point of intersection.
x: $1 + 2\lambda = 4 + \mu$
y: $2 - \lambda = \mu$
z: $3 + \lambda = 5 - \mu$.
Determine the relationship between $\mathbf{r}_1 = (1,0,-1) + \lambda(2,4,-2)$ and $\mathbf{r}_2 = (3,5,1) + \mu(1,2,-1)$. Are they coincident or strictly parallel?
Show that $\mathbf{r}_1 = (1,0,2) + \lambda(2,1,-1)$ and $\mathbf{r}_2 = (0,1,3) + \mu(1,-1,2)$ are skew.
x: $1 + 2\lambda = \mu$
y: $\lambda = 1 - \mu$
z: $2 - \lambda = 3 + 2\mu$.
From y: $\lambda = 1 - \mu$. Substitute into x: $1 + 2(1 - \mu) = \mu \Rightarrow 3 - 2\mu = \mu \Rightarrow \mu = 1$, so $\lambda = 0$.
Fill the gap: Two lines in 3D are skew if their direction vectors are and the simultaneous parametric system has .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: two lines in $\mathbb{R}^3$ with direction vectors $(1, 2, 3)$ and $(2, 4, 6)$ are necessarily coincident.
Activities · practice with the ideas
Determine whether the lines $\mathbf{r}_1 = (2,1,0) + \lambda(1,1,1)$ and $\mathbf{r}_2 = (3,2,1) + \mu(2,-1,1)$ intersect. If so, give the point.
Show that $\mathbf{r}_1 = (0,1,2) + \lambda(3,-2,1)$ and $\mathbf{r}_2 = (1,0,3) + \mu(-6,4,-2)$ are parallel. Are they coincident?
Decide whether $\mathbf{r}_1 = (1,1,1) + \lambda(1,0,2)$ and $\mathbf{r}_2 = (0,2,0) + \mu(0,1,1)$ meet, are parallel or are skew.
Find the value of $k$ for which the lines $\mathbf{r}_1 = (1,2,k) + \lambda(1,1,1)$ and $\mathbf{r}_2 = (0,0,0) + \mu(2,1,3)$ intersect.
Two lines have the same direction vector but pass through different points. Explain why this is sufficient to know they cannot be skew.
Odd one out: Three of these describe situations where two lines in $\mathbb{R}^3$ have no point in common. Which one does NOT?
Earlier you analysed $\mathbf{r}_1 = (1,0,2) + \lambda(2,1,-1)$ and $\mathbf{r}_2 = (0,1,3) + \mu(1,-1,2)$ — do they meet?
From WE3: the direction vectors are not parallel, but solving the x and y component equations gives $\lambda = 0, \mu = 1$, and the z equation fails ($2 \neq 5$). So the lines are skew. The lesson lives or dies on remembering the third component test — without it, you might wrongly declare the lines intersecting on the strength of two equations alone.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Show that $\mathbf{r}_1 = (1,0,0) + \lambda(1,2,-1)$ and $\mathbf{r}_2 = (2,2,-1) + \mu(0,1,-1)$ intersect, and give the point. (2 marks)
Q2. Determine whether $\mathbf{r}_1 = (1,1,0) + \lambda(2,0,1)$ and $\mathbf{r}_2 = (0,2,1) + \mu(1,1,0)$ intersect, are parallel, or are skew. Justify your classification. (3 marks)
Q3. Find all values of the constant $a$ for which $\mathbf{r}_1 = (a,1,2) + \lambda(1,2,-1)$ and $\mathbf{r}_2 = (0,0,0) + \mu(2,1,1)$ intersect. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. Directions $(1,1,1), (2,-1,1)$ not parallel. x and z: $\lambda - 2\mu = 1$, $\lambda - \mu = 1 \Rightarrow \mu = 0, \lambda = 1$. y check: $1 + 1 = 2 - 0 = 2$ ✓. Point $(3, 2, 1)$.
2. $(-6, 4, -2) = -2(3, -2, 1)$, so parallel. Test $(0, 1, 2)$ on $\ell_2$: x gives $\mu = 1/6$, y gives $4/3 \neq 1$. Not coincident — strictly parallel.
3. Directions $(1, 0, 2), (0, 1, 1)$ not parallel. x: $\lambda = -1$; y: $\mu = -1$. z: $1 + 2(-1) = -1$ and $0 + (-1) = -1$ ✓. Intersect at $(0, 1, -1)$.
4. From x and y: $\mu = 1, \lambda = -1$. z: $k - 1 = 3 \Rightarrow k = 4$.
5. Skew requires non-parallel directions by definition. Equal direction vectors $\Rightarrow$ parallel; the lines are either coincident or strictly parallel — never skew.
Q1 (2 marks): x: $\lambda = 1$ [½]. y: $\mu = 0$ [½]. z check: $-1 = -1$ ✓; intersection at $(2, 2, -1)$ [1].
Q2 (3 marks): Directions not scalar multiples [1]. From y: $\mu = -1$; from x: $\lambda = -1$ [1]. z gives $-1 \neq 1$ — inconsistent; non-parallel + no solution $\Rightarrow$ skew [1].
Q3 (3 marks): y and z: $1 + 2\lambda = 2 - \lambda \Rightarrow \lambda = \tfrac{1}{3}, \mu = \tfrac{5}{3}$ [1]. x: $a + \tfrac{1}{3} = \tfrac{10}{3}$ [1]; $a = 3$ [1].
Five timed questions on classifying lines, solving the parametric system and identifying skew configurations. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by classifying line pairs quickly. Lighter alternative to the boss.
Mark lesson as complete
Tick when you've finished the practice and review.