Your weak spots

Insights load after your first practice round.

Module 6 · L17 of 20 ~35 min ⚡ +95 XP available

Intersection and Parallel Lines

Two roads in a city either cross, run side by side, or — in 3D — pass over each other without ever meeting. Vector lines work the same way: they can be parallel, intersecting, or skew. In this lesson you'll classify any pair of lines and, when they do intersect, pinpoint the exact meeting point.

Today's hook — Line 1 passes through $(1,2,3)$ with direction $\mathbf{d}_1 = (2,1,-1)$. Line 2 passes through $(3,3,2)$ with direction $\mathbf{d}_2 = (4,2,-2)$. Before reading on — are these lines parallel, intersecting, or skew? Write your guess below.

0/5QUESTS

Recall — your gut answer first

+5 XP warm-up

Line 1 passes through $(1,2,3)$ with direction $\mathbf{d}_1 = (2,1,-1)$. Line 2 passes through $(3,3,2)$ with direction $\mathbf{d}_2 = (4,2,-2)$. Without any calculation — guess whether these lines are parallel, intersecting, or skew. Write your reasoning.

auto-saved

The classification strategy

+5 XP to read

Every pair of lines falls into exactly one of three categories. The decision tree is always the same:

Step 1 — Test for parallel: Are the direction vectors scalar multiples of each other? If $\mathbf{d}_2 = k\mathbf{d}_1$ for some scalar $k$, the lines are parallel (or coincident).

Step 2 — Test for intersection: Set the parametric equations equal and solve for the parameters $\lambda$ and $\mu$. If a consistent solution exists, the lines intersect.

Step 3 — Otherwise skew: If not parallel and the simultaneous equations are inconsistent, the lines are skew (3D only).

$\mathbf{r}_1 = \mathbf{a} + \lambda\mathbf{d}_1 \quad \mathbf{r}_2 = \mathbf{b} + \mu\mathbf{d}_2$

Parallel vs coincident

Parallel lines have proportional direction vectors but different position vectors. If the position vectors are also proportional (i.e. one point lies on both lines), the lines are the same — coincident.

Skew lines only in 3D

In 2D, two non-parallel lines always intersect. Skew lines exist only in 3D — they are not parallel and never meet because they lie in different planes.

Check the third equation

In 3D you get three equations from two unknowns ($\lambda$, $\mu$). Solve any two for $\lambda$ and $\mu$, then verify in the third. If it fails, the lines are skew.

What you'll master

Know

Key facts

Three possible relationships between lines: parallel, intersecting, skew
Skew lines exist only in three dimensions
Two direction vectors are parallel iff one is a scalar multiple of the other

Understand

Concepts

Why solving two of the three parametric equations and checking the third reveals skew
The difference between parallel and coincident lines
How the parameter values give the exact point of intersection

Can do

Skills

Classify a pair of lines as parallel, intersecting or skew
Find the coordinates of the point where two lines meet
Check that a claimed intersection is consistent in all three components

Key terms

Parallel linesLines with direction vectors that are scalar multiples: $\mathbf{d}_2 = k\mathbf{d}_1$. They never meet (unless coincident).

Coincident linesLines that are identical — every point on one also lies on the other. Direction vectors are parallel and position vectors satisfy the line equation.

Intersecting linesLines that cross at exactly one point. The parametric equations have a unique consistent solution $(\lambda_0, \mu_0)$.

Skew linesLines in 3D that are not parallel and do not intersect. The system of parametric equations is inconsistent.

ParameterA scalar ($\lambda$ or $\mu$) that moves a point along a line. Setting $\lambda = \lambda_0$ gives the specific point on that line.

Scalar multipleVector $\mathbf{u}$ is a scalar multiple of $\mathbf{v}$ if $\mathbf{u} = k\mathbf{v}$ for some $k \in \mathbb{R}$. Equivalently, $\dfrac{u_1}{v_1} = \dfrac{u_2}{v_2} = \dfrac{u_3}{v_3}$.

Testing for parallel lines

core concept

Two lines $\ell_1 : \mathbf{r} = \mathbf{a} + \lambda\mathbf{d}_1$ and $\ell_2 : \mathbf{r} = \mathbf{b} + \mu\mathbf{d}_2$ are parallel if and only if their direction vectors are scalar multiples:

$$\mathbf{d}_2 = k\,\mathbf{d}_1 \quad \text{for some } k \in \mathbb{R}$$

To check this, form the ratios of corresponding components. If all three ratios are equal, the vectors are parallel:

$$\frac{d_{2x}}{d_{1x}} = \frac{d_{2y}}{d_{1y}} = \frac{d_{2z}}{d_{1z}} = k$$

Example (from card 01): $\mathbf{d}_1 = (2,1,-1)$ and $\mathbf{d}_2 = (4,2,-2)$. Check: $4/2 = 2$, $2/1 = 2$, $-2/(-1) = 2$. All equal — so the lines are parallel.

Now check coincident: does $(3,3,2)$ satisfy $\mathbf{r} = (1,2,3) + \lambda(2,1,-1)$? Solve: $1+2\lambda=3 \Rightarrow \lambda=1$; check $y$: $2+1=3$ ✓; check $z$: $3-1=2$ ✓. The point lies on $\ell_1$, so the lines are coincident.

Answer to the hook. The lines from card 01 are actually the same line (coincident). The hook was designed to remind you that "parallel direction vectors" means you must always run the extra coincident check.

Two lines $\ell_1 : \mathbf{r} = \mathbf{a} + \lambda\mathbf{d}_1$ and $\ell_2 : \mathbf{r} = \mathbf{b} + \mu\mathbf{d}_2$ are parallel if and only if their direction vectors are scalar multiples:

Pause — copy the parallel-lines test: $\mathbf{d}_1\parallel\mathbf{d}_2\iff\mathbf{d}_2=k\mathbf{d}_1$; and note that parallel ≠ identical (check if a point on one line lies on the other) into your book.

Quick check: Lines $\ell_1$ and $\ell_2$ have direction vectors $\mathbf{d}_1 = (3,-6,9)$ and $\mathbf{d}_2 = (1,-2,3)$. What is the relationship between the direction vectors?

Finding the point of intersection

core concept

We just saw that two lines $\mathbf{r}=\mathbf{a}+\lambda\mathbf{d}_1$ and $\mathbf{r}=\mathbf{b}+\mu\mathbf{d}_2$ are parallel iff $\mathbf{d}_1=k\mathbf{d}_2$ for some scalar $k$ (i.e.\ direction vectors are scalar multiples). That raises a question: when direction vectors are NOT parallel, the lines might intersect or might be skew — how do you find the intersection point, and how do you confirm the lines actually meet? This card answers it → equate components to get a simultaneous system in $\lambda$ and $\mu$; solve; substitute both values back to verify they give the same point.

If the direction vectors are not parallel, set the two vector equations equal and equate components to get a system of equations in $\lambda$ and $\mu$:

$$\mathbf{a} + \lambda\mathbf{d}_1 = \mathbf{b} + \mu\mathbf{d}_2$$

This gives three scalar equations (in 3D) with two unknowns. The strategy:

Choose any two of the three equations and solve for $\lambda$ and $\mu$.
Substitute those values into the third equation to check consistency.
If consistent, compute the intersection point using either parametric equation.
If inconsistent, the lines are skew.

In 2D you only have two equations and two unknowns — a unique solution always exists (if not parallel).

Why check the third equation? In 3D, two planes always intersect in a line. The third equation is the additional constraint that forces both lines to share a point. Without it, you could solve two equations and find "intersection" values that are geometrically inconsistent.

If the direction vectors are not parallel, set the two vector equations equal and equate components to get a system of equations in $\lambda$ and $\mu$:

Pause — copy the intersection method: equate $\mathbf{a}+\lambda\mathbf{d}_1=\mathbf{b}+\mu\mathbf{d}_2$, solve the 2×2 system, substitute back to verify into your book.

Did you get this? True or false: In 3D, if two non-parallel lines give a consistent solution when you solve any two of the three parametric equations, you still need to check the third equation before concluding the lines intersect.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · PARALLEL TEST IN 3D

Determine whether $\ell_1: \mathbf{r} = (1,0,-2)+\lambda(3,1,2)$ and $\ell_2: \mathbf{r} = (4,1,0)+\mu(6,2,4)$ are parallel, intersecting or skew.

Check if $\mathbf{d}_2 = k\mathbf{d}_1$: $\dfrac{6}{3}=2$, $\dfrac{2}{1}=2$, $\dfrac{4}{2}=2$. All equal $k=2$.

Form component ratios. All equal, so direction vectors are parallel.

PROBLEM 2 · INTERSECTION IN 2D

Find the point of intersection of $\ell_1: \mathbf{r}=(1,2)+\lambda(1,3)$ and $\ell_2: \mathbf{r}=(3,0)+\mu(2,1)$.

Direction vectors $(1,3)$ and $(2,1)$: $1/2 \neq 3/1$, so not parallel. Set equations equal: $1+\lambda = 3+2\mu$ and $2+3\lambda = 0+\mu$.

Since not parallel, proceed to solve the system.

PROBLEM 3 · SKEW LINES IN 3D

Determine whether $\ell_1: \mathbf{r}=(0,1,0)+\lambda(1,1,0)$ and $\ell_2: \mathbf{r}=(1,0,1)+\mu(0,1,1)$ are parallel, intersecting or skew.

Check parallel: $(1,1,0)$ vs $(0,1,1)$. No scalar $k$ satisfies all three ratios simultaneously (e.g. $1/0$ is undefined). Not parallel.

One ratio is undefined ($k$ would need to be $\infty$), so direction vectors are not parallel.

Fill the gap: Two lines in 3D are classified as skew if they are not and their parametric equations are .

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Forgetting the coincident check

If direction vectors are parallel, you must still check whether the lines are identical or just parallel. Substitute one position point into the other line's equation. If it works, the lines are coincident — not merely parallel.

Trap 02

Skipping the third equation

Solving only two of the three parametric equations and declaring "they intersect" is a common error. The third equation is the consistency check. If you skip it, you risk labelling skew lines as intersecting.

Trap 03

Different parameters = different lines

The two lines use different parameters ($\lambda$ and $\mu$). They do not have to equal each other at intersection — only the position vectors need to match. Using the same parameter letter for both lines is a recipe for errors.

Did you get this? True or false: Two distinct non-parallel lines in 2D can be skew.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Line $\ell_1$ has direction $(2,-4,6)$ and line $\ell_2$ has direction $(-1,2,-3)$. Are these lines necessarily parallel? Explain.

Find the point of intersection of $\ell_1: \mathbf{r}=(2,1)+\lambda(1,2)$ and $\ell_2: \mathbf{r}=(4,3)+\mu(-1,1)$.

Classify the lines $\ell_1: \mathbf{r}=(1,0,1)+\lambda(2,1,0)$ and $\ell_2: \mathbf{r}=(0,1,2)+\mu(1,0,2)$. If they intersect, find the point.

Lines $\ell_1: \mathbf{r}=(0,0,0)+\lambda(1,2,3)$ and $\ell_2: \mathbf{r}=(1,1,0)+\mu(1,2,3)$. Classify and justify.

Show that $\ell_1: \mathbf{r}=(0,0,0)+\lambda(1,0,1)$ and $\ell_2: \mathbf{r}=(1,1,0)+\mu(0,1,1)$ are skew.

Odd one out: Three of these statements about lines in 3D are true. Which one is FALSE?

Revisit your thinking

Earlier you guessed whether the lines in card 01 were parallel, intersecting or skew.

The answer was coincident — the same line. The direction vector $(4,2,-2)$ is exactly twice $(2,1,-1)$, and the point $(3,3,2)$ satisfies the first line's equation at $\lambda = 1$. The key insight is that "parallel direction vectors" does not automatically mean the lines are distinct — always run the coincident check.

auto-saved

Multiple choice

+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 32 marks

Q1. Determine whether the lines $\ell_1: \mathbf{r}=(1,2,-1)+\lambda(2,-1,3)$ and $\ell_2: \mathbf{r}=(3,1,2)+\mu(4,-2,6)$ are parallel, intersecting or skew. (2 marks)

auto-saved

ApplyBand 43 marks

Q2. Find the coordinates of the point where $\ell_1: \mathbf{r}=(0,3)+\lambda(2,1)$ and $\ell_2: \mathbf{r}=(4,1)+\mu(-1,2)$ intersect. (3 marks)

auto-saved

AnalyseBand 53 marks

Q3. Prove that $\ell_1: \mathbf{r}=(2,1,3)+\lambda(1,1,2)$ and $\ell_2: \mathbf{r}=(4,0,1)+\mu(1,-1,0)$ are skew. (3 marks)

auto-saved

Comprehensive answers (click to reveal)

Activity answers:
1. $(-1,2,-3) = -\tfrac{1}{2}(2,-4,6)$ — direction vectors are parallel (scalar multiple $k=-\tfrac{1}{2}$), so the lines are at least parallel.
2. $1+\lambda=4-\mu$ and $1+2\lambda=3+\mu$. From eq 2: $\mu=2\lambda-2$. Into eq 1: $1+\lambda=4-(2\lambda-2)=6-2\lambda \Rightarrow 3\lambda=5 \Rightarrow \lambda=\tfrac{5}{3}$, $\mu=\tfrac{4}{3}$. Point: $(2,1)+\tfrac{5}{3}(1,2)=(\tfrac{11}{3},\tfrac{13}{3})$.
3. Not parallel ($\tfrac{2}{1}\neq\tfrac{1}{0}$). Eq $x$: $1+2\lambda=\mu$; eq $z$: $1=2+2\mu\Rightarrow\mu=-\tfrac{1}{2}$, $\lambda=-\tfrac{3}{4}$. Check $y$: $0+\lambda=0-\tfrac{3}{4}$ vs $1+0=-\tfrac{1}{2}$: inconsistent — skew.
4. Direction vectors identical $(1,2,3)$. Check if $(1,1,0)$ on $\ell_1$: $\lambda=1$ from $x$; $y:2\neq 1$ — inconsistent, so truly parallel (not coincident).
5. Not parallel. $x$: $\lambda=1$; $z$: $\mu=1$. Check $y$: $0\neq 1+1=2$ — inconsistent, skew.

Q1 (2 marks): $\mathbf{d}_2=(4,-2,6)=2(2,-1,3)=2\mathbf{d}_1$ [1]. Check if $(3,1,2)$ lies on $\ell_1$: $\lambda=1$ from $x$; $y:2-1=1$ ✓; $z:-1+3=2$ ✓. Lines are coincident [1].

Q2 (3 marks): $x$: $2\lambda+\mu=4$ [1]; $y$: $\lambda-2\mu=-2$. From $y$: $\lambda=2\mu-2$; into $x$: $4\mu-4+\mu=4\Rightarrow\mu=\tfrac{8}{5}$, $\lambda=\tfrac{6}{5}$ [1]. Point: $(0,3)+\tfrac{6}{5}(2,1)=\left(\tfrac{12}{5},\tfrac{21}{5}\right)$ [1].

Q3 (3 marks): $\tfrac{1}{1}\neq\tfrac{1}{1}=\tfrac{2}{0}$ (undefined) — not parallel [1]. $x$: $2+\lambda=4+\mu\Rightarrow\lambda-\mu=2$; $z$: $3+2\lambda=1\Rightarrow\lambda=-1$, so $\mu=-3$ [1]. Check $y$: $1+(-1)=0$ but $0-(-3)=3$; $0\neq3$ — inconsistent, lines are skew [1].

Boss battle · The Line Classifier

earn bronze · silver · gold

Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering questions about parallel and intersecting lines. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

← Lesson 16 · Vector Equations of Lines in 3D Lesson 18 · Applications of Vectors →

Module overview · Extension 1 · Checkpoint 4