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Module 6 · L16 of 20 ~40 min ⚡ +100 XP available

Vector Equations of Lines in 3D

The flight path of an aircraft is not confined to a flat map — it moves in three dimensions. The vector equation $\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}$ extends perfectly to 3D: just add a $z$-component. You already know the 2D form from Lesson 15. In this lesson you'll parameterise lines in 3D space, write symmetric (Cartesian) equations with three fractions, and determine whether points lie on 3D lines.

Today's hook — An aircraft is at position $(1, 2, 3)$ km and flying in the direction $\begin{pmatrix}4\\-1\\2\end{pmatrix}$ km per minute. Before reading further, where will it be after 3 minutes? Write your answer in coordinates now.

0/5QUESTS

Recall — your gut answer first

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An aircraft is at position $(1, 2, 3)$ km and flying in direction $\begin{pmatrix}4\\-1\\2\end{pmatrix}$ km per minute. Before using a formula — where will it be after 3 minutes? After $t$ minutes?

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Extending to 3D — one extra component

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The 2D vector equation $\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}$ extends to 3D by simply adding a $z$-component to every vector. The structure is identical — only the dimension changes.

In 3D, position vectors have three components. If the line passes through point $\mathbf{a} = \begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}$ with direction $\mathbf{d} = \begin{pmatrix}d_1\\d_2\\d_3\end{pmatrix}$, then:

$\mathbf{r} = \begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix} + \lambda\begin{pmatrix}d_1\\d_2\\d_3\end{pmatrix}, \quad \lambda \in \mathbb{R}$

$\dfrac{x-a_1}{d_1} = \dfrac{y-a_2}{d_2} = \dfrac{z-a_3}{d_3}$

Three parametric equations

In 3D you get $x = a_1+\lambda d_1$, $y = a_2+\lambda d_2$, $z = a_3+\lambda d_3$ — one per coordinate. All share the same $\lambda$.

Symmetric form needs $d_i \neq 0$

If any $d_i = 0$, that component is constant (e.g. $z = a_3$). Write the constant equation separately and use symmetric form only for the remaining two fractions.

No gradient in 3D

In 3D there is no single "slope" — the direction vector $\mathbf{d}$ encodes all directional information. This is why vector form is so powerful in higher dimensions.

What you'll master

Know

Key facts

3D vector equation: $\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}$, with $\mathbf{a}, \mathbf{d} \in \mathbb{R}^3$
Parametric form: $x = a_1+\lambda d_1$, $y = a_2+\lambda d_2$, $z = a_3+\lambda d_3$
Symmetric form: $\dfrac{x-a_1}{d_1} = \dfrac{y-a_2}{d_2} = \dfrac{z-a_3}{d_3}$

Understand

Concepts

Why the 2D structure extends naturally to 3D without new ideas
The geometric meaning of each component of $\mathbf{d}$ in 3D space
How to handle the case when one direction component is zero

Can do

Skills

Write the vector and parametric equations of a 3D line
Convert to symmetric (Cartesian) form
Determine whether a point lies on a 3D line

Key terms

3D vector equation$\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}$ where all vectors have three components. Same structure as 2D, one dimension added.

Direction vector $\mathbf{d}$ in 3D$\mathbf{d} = \begin{pmatrix}d_1\\d_2\\d_3\end{pmatrix}$ — a non-zero vector parallel to the line. Often found as the difference of two known points on the line.

Parametric equations (3D)Three equations: $x = a_1+\lambda d_1$, $y = a_2+\lambda d_2$, $z = a_3+\lambda d_3$, all sharing the same parameter $\lambda$.

Symmetric (Cartesian) form$\dfrac{x-a_1}{d_1} = \dfrac{y-a_2}{d_2} = \dfrac{z-a_3}{d_3}$ — obtained by solving each parametric equation for $\lambda$ and equating.

$\mathbb{R}^3$Three-dimensional real space. Points are ordered triples $(x, y, z)$; vectors have three components.

Point test (3D)To test $(p,q,r)$: find $\lambda$ from the $x$-equation, then verify the same $\lambda$ satisfies both $y$ and $z$ equations.

The vector equation in 3D

core concept

The derivation is identical to the 2D case. For a line through $\mathbf{a}$ with direction $\mathbf{d}$ in $\mathbb{R}^3$:

$$\mathbf{r} = \mathbf{a} + \lambda\mathbf{d} = \begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix} + \lambda\begin{pmatrix}d_1\\d_2\\d_3\end{pmatrix}, \quad \lambda \in \mathbb{R}$$

This gives three parametric equations:

$$x = a_1 + \lambda d_1, \quad y = a_2 + \lambda d_2, \quad z = a_3 + \lambda d_3$$

Example — the aircraft problem from the hook: Position $\mathbf{a} = \begin{pmatrix}1\\2\\3\end{pmatrix}$, direction $\mathbf{d} = \begin{pmatrix}4\\-1\\2\end{pmatrix}$:

$\mathbf{r} = \begin{pmatrix}1\\2\\3\end{pmatrix} + \lambda\begin{pmatrix}4\\-1\\2\end{pmatrix}$

After 3 minutes ($\lambda = 3$): position $= \begin{pmatrix}1+12\\2-3\\3+6\end{pmatrix} = \begin{pmatrix}13\\-1\\9\end{pmatrix}$ km. How did your estimate compare?

Two-point form in 3D. If the line passes through $A$ and $B$, the direction is $\mathbf{d} = \mathbf{b} - \mathbf{a}$. The equation is $\mathbf{r} = \mathbf{a} + \lambda(\mathbf{b}-\mathbf{a})$. Setting $\lambda = 0$ gives $A$; $\lambda = 1$ gives $B$.

The derivation is identical to the 2D case. For a line through $\mathbf{a}$ with direction $\mathbf{d}$ in $\mathbb{R}^3$:

Pause — copy the 3D vector equation and its three parametric equations: $x=a_1+\lambda d_1$, $y=a_2+\lambda d_2$, $z=a_3+\lambda d_3$ into your book.

Quick check: A line in 3D passes through $(0, 1, -2)$ with direction $\begin{pmatrix}3\\0\\-1\end{pmatrix}$. What is the $z$-coordinate of the point at $\lambda = 4$?

Symmetric (Cartesian) form in 3D

core concept

We just saw the 3D vector equation $\mathbf{r}=\mathbf{a}+\lambda\mathbf{d}$ with three parametric equations $x=a_1+\lambda d_1$, $y=a_2+\lambda d_2$, $z=a_3+\lambda d_3$. That raises a question: the symmetric form eliminates $\lambda$ by solving each equation for $\lambda$ and setting them equal — what is the resulting three-way equation, and when can't you write it? This card answers it → $\frac{x-a_1}{d_1}=\frac{y-a_2}{d_2}=\frac{z-a_3}{d_3}$; not writable when any $d_i=0$ (write that component's equation separately).

Solve each parametric equation for $\lambda$ and set all three equal:

$$\frac{x-a_1}{d_1} = \frac{y-a_2}{d_2} = \frac{z-a_3}{d_3} = \lambda$$

This is the symmetric form. It encodes the same line but uses only $x, y, z$ — the parameter $\lambda$ has been eliminated.

Special case — zero direction component: If $d_3 = 0$, the $z$-component gives $z = a_3 + 0 = a_3$ (constant). Write $z = a_3$ separately and use only the first two fractions: $\dfrac{x-a_1}{d_1} = \dfrac{y-a_2}{d_2}$ plus $z = a_3$.

Example: For $\mathbf{r} = \begin{pmatrix}2\\-1\\4\end{pmatrix} + \lambda\begin{pmatrix}3\\5\\-2\end{pmatrix}$:

Symmetric form: $\dfrac{x-2}{3} = \dfrac{y+1}{5} = \dfrac{z-4}{-2}$

Checking your symmetric form. Substitute the anchor point $\mathbf{a}$ back in — all three fractions should equal 0. Substitute a second known point — all three fractions should give the same non-zero $\lambda$.

Solve each parametric equation for $\lambda$ and set all three equal:

Pause — copy the symmetric form: $\frac{x-a_1}{d_1}=\frac{y-a_2}{d_2}=\frac{z-a_3}{d_3}$ and the exception when $d_i=0$ into your book.

Did you get this? True or false: The symmetric form of $\mathbf{r} = \begin{pmatrix}1\\0\\3\end{pmatrix} + \lambda\begin{pmatrix}2\\4\\-1\end{pmatrix}$ is $\dfrac{x-1}{2} = \dfrac{y}{4} = \dfrac{z-3}{-1}$.

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

PROBLEM 1 · WRITE THE 3D VECTOR EQUATION

Find the vector equation of the line through $A(1, -2, 4)$ and $B(3, 0, 1)$. Write in both vector and parametric form.

$\mathbf{a} = \begin{pmatrix}1\\-2\\4\end{pmatrix}$; direction $\mathbf{d} = \mathbf{b} - \mathbf{a} = \begin{pmatrix}3-1\\0-(-2)\\1-4\end{pmatrix} = \begin{pmatrix}2\\2\\-3\end{pmatrix}$

Subtract position vectors to find the direction. Either point can be the anchor.

PROBLEM 2 · SYMMETRIC FORM

Write the symmetric form of the line $\mathbf{r} = \begin{pmatrix}-3\\1\\0\end{pmatrix} + \lambda\begin{pmatrix}4\\-2\\5\end{pmatrix}$.

Parametric: $x = -3+4\lambda$, $y = 1-2\lambda$, $z = 0+5\lambda$

Write out all three component equations.

PROBLEM 3 · POINT TEST IN 3D

Does the point $P(5, -4, 9)$ lie on the line $\mathbf{r} = \begin{pmatrix}1\\0\\3\end{pmatrix} + \lambda\begin{pmatrix}2\\-2\\3\end{pmatrix}$?

$x$-equation: $5 = 1+2\lambda \Rightarrow \lambda = 2$

Set $x=5$ and solve for $\lambda$.

Fill the gap: To test whether a point lies on a 3D line, you find $\lambda$ from the $x$-equation, then verify that the same $\lambda$ satisfies both the $y$ and equations.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Only checking two of three equations

In 3D, the point test requires ALL THREE parametric equations to give the same $\lambda$. Checking only $x$ and $y$ is not sufficient — a point could satisfy two equations without lying on the line. Always verify all three.

Trap 02

Zero direction component in symmetric form

If $d_2 = 0$, for example, you cannot write $\dfrac{y-a_2}{0}$ (undefined). Instead, write $y = a_2$ as a separate equation and only use fractions for the non-zero direction components. This is a common error in HSC responses.

Trap 03

Confusing 2D and 3D gradient

In 2D the Cartesian form $y = mx+c$ uses a scalar gradient. In 3D there is no such thing as a single scalar gradient — you must use the full direction vector $\mathbf{d}$ to describe orientation. Never try to write a 3D line as $z = mx + ny + c$ as this describes a plane, not a line.

Did you get this? True or false: A 3D line with direction $\begin{pmatrix}2\\0\\-3\end{pmatrix}$ through $(1, 4, 5)$ has symmetric form $\dfrac{x-1}{2} = \dfrac{y-4}{0} = \dfrac{z-5}{-3}$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Write the vector equation of the line through $A(0, 3, -1)$ and $B(4, 1, 5)$.

Convert $\mathbf{r} = \begin{pmatrix}2\\-1\\4\end{pmatrix} + \lambda\begin{pmatrix}3\\2\\-1\end{pmatrix}$ to symmetric (Cartesian) form.

Does the point $(8, -5, 2)$ lie on the line $\mathbf{r} = \begin{pmatrix}2\\1\\-4\end{pmatrix} + \lambda\begin{pmatrix}2\\-2\\2\end{pmatrix}$?

A line has symmetric equation $\dfrac{x-1}{2} = \dfrac{y+3}{-1} = \dfrac{z}{4}$. Write it as a vector equation and state one other point on the line.

Write the vector equation of the line through $(3, 1, -2)$ with direction $\begin{pmatrix}1\\0\\5\end{pmatrix}$. Write the symmetric form, handling the zero direction component correctly.

Odd one out: A line passes through $(2, 0, -1)$ with direction $\begin{pmatrix}1\\3\\2\end{pmatrix}$. Three of the following statements are true. Which one is FALSE?

Revisit your thinking

Earlier you were asked where an aircraft at $(1,2,3)$ flying in direction $\begin{pmatrix}4\\-1\\2\end{pmatrix}$ would be after 3 minutes.

The answer: $\mathbf{r}(3) = \begin{pmatrix}1+12\\2-3\\3+6\end{pmatrix} = (13, -1, 9)$. The vector equation is $\mathbf{r} = \begin{pmatrix}1\\2\\3\end{pmatrix} + \lambda\begin{pmatrix}4\\-1\\2\end{pmatrix}$ — exactly the same structure as in 2D, just with three components. That's the beauty of the vector approach.

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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.

Short answer

ApplyBand 32 marks

Q1. Find the vector equation of the line passing through $A(0, 1, -3)$ and $B(2, 5, 1)$. Write in parametric form. (2 marks)

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ApplyBand 43 marks

Q2. Write the symmetric form of the line $\mathbf{r} = \begin{pmatrix}4\\-2\\1\end{pmatrix} + \lambda\begin{pmatrix}-1\\3\\2\end{pmatrix}$. Hence find the value of $t$ if the point $(t, 4, 5)$ lies on the line. (3 marks)

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AnalyseBand 53 marks

Q3. The line $\ell$ passes through $P(2, -1, 3)$ and has direction vector $\mathbf{d} = \begin{pmatrix}1\\2\\-1\end{pmatrix}$. Show that the point $Q(5, 5, 0)$ lies on $\ell$, and find the value of the parameter at $Q$. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $\mathbf{d} = \begin{pmatrix}4\\-2\\6\end{pmatrix}$; $\mathbf{r} = \begin{pmatrix}0\\3\\-1\end{pmatrix} + \lambda\begin{pmatrix}4\\-2\\6\end{pmatrix}$

2. Symmetric: $\dfrac{x-2}{3} = \dfrac{y+1}{2} = \dfrac{z-4}{-1}$

3. $x$-equation: $8 = 2+2\lambda \Rightarrow \lambda=3$. $y$: $1-2(3) = -5$ ✓. $z$: $-4+2(3)=2$ ✓. Yes, lies on the line.

4. $\mathbf{a} = \begin{pmatrix}1\\-3\\0\end{pmatrix}$, $\mathbf{d} = \begin{pmatrix}2\\-1\\4\end{pmatrix}$; $\mathbf{r} = \begin{pmatrix}1\\-3\\0\end{pmatrix} + \lambda\begin{pmatrix}2\\-1\\4\end{pmatrix}$. At $\lambda=1$: $(3,-4,4)$.

5. $\mathbf{r} = \begin{pmatrix}3\\1\\-2\end{pmatrix} + \lambda\begin{pmatrix}1\\0\\5\end{pmatrix}$. Symmetric: $\dfrac{x-3}{1} = \dfrac{z+2}{5}$ and $y = 1$.

Q1 (2 marks): $\mathbf{d} = \begin{pmatrix}2\\4\\4\end{pmatrix}$ [1]; $x=2\lambda$, $y=1+4\lambda$, $z=-3+4\lambda$ [1].

Q2 (3 marks): Symmetric: $\dfrac{x-4}{-1} = \dfrac{y+2}{3} = \dfrac{z-1}{2}$ [1]. From $y$: $\dfrac{4+2}{3} = 2$, so $\lambda=2$ [1]. From $x$: $t = 4+(-1)(2) = 2$ [1].

Q3 (3 marks): From $x$: $5 = 2+\lambda \Rightarrow \lambda=3$ [1]. $y$: $-1+2(3) = 5$ ✓ [1]. $z$: $3-3 = 0$ ✓. $\therefore Q$ lies on $\ell$ at $\lambda=3$ [1].

Boss battle · The 3D Navigator

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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 3D lines questions. Lighter alternative to the boss.

Mark lesson as complete

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← Lesson 15 · Vector Equations of Lines in 2D Lesson 17 · Intersection and Parallel Lines →

Module overview · Extension 1 · Checkpoint 4