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Module 6 · L5 of 20 ~35 min ⚡ +95 XP available

Vector Subtraction

A boat crosses a river with current — what is its velocity relative to the water? That is a subtraction problem. $\vec{a} - \vec{b}$ is simply $\vec{a} + (-\vec{b})$, where $-\vec{b}$ reverses direction. Master this operation and you unlock displacement vectors, relative velocity, and the geometry of parallelograms.

Today's hook — A boat travels with velocity $\vec{v}_b = \begin{pmatrix} 4 \\ 3 \end{pmatrix}$ m/s. A river current flows at $\vec{v}_c = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ m/s. Before you learn the method, guess: what is the boat's velocity relative to the water? Jot your instinct, then check it in card 05.

0/5QUESTS

Recall — your gut answer first

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A boat has velocity $\vec{v}_b = \begin{pmatrix} 4 \\ 3 \end{pmatrix}$ m/s and the river current has velocity $\vec{v}_c = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ m/s. Without using a method yet — what do you think the boat's velocity relative to the water would be? Write your reasoning below.

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The core idea

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Vector subtraction is defined as adding the negative vector:

The negative of $\vec{b}$, written $-\vec{b}$, has the same magnitude as $\vec{b}$ but points in the opposite direction. So:

$\vec{a} - \vec{b} = \vec{a} + (-\vec{b})$

This single rewrite lets you treat all vector subtraction problems as addition problems — a useful trick.

$\vec{a} - \vec{b} = \vec{a} + (-\vec{b})$

Negative vector

$-\vec{b}$ reverses the arrow. The magnitude $|-\vec{b}| = |\vec{b}|$ is unchanged — only direction flips.

Parallelogram rule

Drawing $\vec{a}$ and $\vec{b}$ from the same point, $\vec{a} - \vec{b}$ is the diagonal that does NOT represent the sum.

Head-to-head rule

$\vec{AB} - \vec{AC} = \vec{CB}$. When two vectors share a tail at A, the difference goes from head of $\vec{b}$ to head of $\vec{a}$.

What you'll master

Know

Key facts

$\vec{a} - \vec{b} = \vec{a} + (-\vec{b})$
$-\vec{b}$ has the same magnitude as $\vec{b}$ but opposite direction
$\vec{AB} - \vec{AC} = \vec{CB}$ (head-to-head gives the third side)

Understand

Concepts

Why subtraction is geometrically the "other diagonal" of a parallelogram
How displacement and relative velocity use vector subtraction
The difference between $\vec{a} - \vec{b}$ and $\vec{b} - \vec{a}$

Can do

Skills

Subtract vectors both geometrically and by components
Find relative velocity and displacement using subtraction
Interpret $\vec{AB} = \vec{OB} - \vec{OA}$ as position vector difference

Key terms

Negative vector $(-\vec{b})$A vector with the same magnitude as $\vec{b}$ but pointing in the opposite direction. Geometrically, it reverses the arrow.

Vector subtraction$\vec{a} - \vec{b} = \vec{a} + (-\vec{b})$. Subtract by adding the negative of the second vector.

Displacement vector$\vec{AB} = \vec{OB} - \vec{OA}$. The vector from point A to point B, found by subtracting position vectors.

Relative velocityThe velocity of object X relative to object Y is $\vec{v}_X - \vec{v}_Y$. Tells you how X appears to move from Y's perspective.

Parallelogram of vectorsWhen $\vec{a}$ and $\vec{b}$ share a tail, the sum $\vec{a}+\vec{b}$ is one diagonal and the difference $\vec{a}-\vec{b}$ is the other.

$\vec{AB} - \vec{AC} = \vec{CB}$When two vectors share the same starting point A, the difference goes from the head of the subtracted vector to the head of the first.

Defining vector subtraction

core concept

Vector subtraction is not a separate operation — it is addition of the negative:

$$\vec{a} - \vec{b} = \vec{a} + (-\vec{b})$$

The negative vector $-\vec{b}$ is obtained by reversing the direction of $\vec{b}$. Its magnitude is unchanged: $|-\vec{b}| = |\vec{b}|$.

Component form: If $\vec{a} = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix}$ and $\vec{b} = \begin{pmatrix} b_1 \\ b_2 \end{pmatrix}$, then:

$$\vec{a} - \vec{b} = \begin{pmatrix} a_1 - b_1 \\ a_2 - b_2 \end{pmatrix}$$

Subtract corresponding components. This extends naturally to 3D.

Position vector formula. If $A$ and $B$ have position vectors $\vec{OA}$ and $\vec{OB}$ respectively, then the displacement from $A$ to $B$ is $\vec{AB} = \vec{OB} - \vec{OA}$. This is the most common use of subtraction in coordinate geometry.

$\vec{a}-\vec{b}=\vec{a}+(-\vec{b})$. Geometry: with both from same point, $\vec{a}-\vec{b}$ goes from tip of $\vec{b}$ to tip of $\vec{a}$.

Pause — copy the definition: $\vec{a}-\vec{b}=\vec{a}+(-\vec{b})$ where $-\vec{b}$ has the same length as $\vec{b}$ but opposite direction, with a worked component example into your book.

Quick check: If $\vec{a} = \begin{pmatrix} 5 \\ 2 \end{pmatrix}$ and $\vec{b} = \begin{pmatrix} 3 \\ -1 \end{pmatrix}$, what is $\vec{a} - \vec{b}$?

Geometric interpretation

core concept

We just saw that $\vec{a}-\vec{b}=\vec{a}+(-\vec{b})$, so subtraction is just adding the negative (reversed) vector. That raises a question: geometrically, when $\vec{a}$ and $\vec{b}$ are placed from the same point, where does $\vec{a}-\vec{b}$ appear in the diagram, and how does this differ from the sum? This card answers it → $\vec{a}+\vec{b}$ is one diagonal of the parallelogram; $\vec{a}-\vec{b}$ points from the tip of $\vec{b}$ to the tip of $\vec{a}$ (the other diagonal).

To find $\vec{a} - \vec{b}$ geometrically, place $\vec{a}$ and $\vec{b}$ with the same starting point. The result $\vec{a} - \vec{b}$ is the vector that goes from the head of $\vec{b}$ to the head of $\vec{a}$.

This is one diagonal of the parallelogram formed by $\vec{a}$ and $\vec{b}$.
The other diagonal is the sum $\vec{a} + \vec{b}$.
The two diagonals are $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$.

Useful identity: $\vec{AB} - \vec{AC} = \vec{CB}$. When two vectors share tail $A$, the difference goes from the head of the second to the head of the first.

Note on direction. $\vec{a} - \vec{b}$ and $\vec{b} - \vec{a}$ are equal in magnitude but opposite in direction — they are negatives of each other. Be careful to subtract in the right order.

To find $\vec{a} - \vec{b}$ geometrically, place $\vec{a}$ and $\vec{b}$ with the same starting point . The result $\vec{a} - \vec{b}$ is the vector that goes from the head of $\vec{b}$ to the head of $\vec{a}$ .

Pause — copy the geometric interpretation: with $\vec{a}$ and $\vec{b}$ from the same point, $\vec{a}-\vec{b}$ is the vector from the tip of $\vec{b}$ to the tip of $\vec{a}$ into your book.

Did you get this? True or false: when $\vec{a}$ and $\vec{b}$ are drawn from the same point, the vector $\vec{a} - \vec{b}$ goes from the head of $\vec{a}$ to the head of $\vec{b}$.

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

PROBLEM 1 · COMPONENT SUBTRACTION

Given $\vec{a} = \begin{pmatrix} 7 \\ 4 \end{pmatrix}$ and $\vec{b} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$, find $\vec{a} - \vec{b}$ and $|\vec{a} - \vec{b}|$.

$\vec{a} - \vec{b} = \begin{pmatrix} 7 - 2 \\ 4 - (-1) \end{pmatrix} = \begin{pmatrix} 5 \\ 5 \end{pmatrix}$

Subtract corresponding components. Note $4 - (-1) = 5$, not 3 — watch the sign of the second component.

PROBLEM 2 · RELATIVE VELOCITY

A boat has velocity $\vec{v}_b = \begin{pmatrix} 4 \\ 3 \end{pmatrix}$ m/s and the river current has velocity $\vec{v}_c = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ m/s. Find the velocity of the boat relative to the water.

$\vec{v}_{\text{relative}} = \vec{v}_b - \vec{v}_c$

Relative velocity = velocity of object minus velocity of the reference (water). This is the hook problem from card 01!

PROBLEM 3 · DISPLACEMENT

Point $A$ has position vector $\vec{OA} = \begin{pmatrix} 2 \\ 5 \end{pmatrix}$ and point $B$ has position vector $\vec{OB} = \begin{pmatrix} 6 \\ 1 \end{pmatrix}$. Find the displacement vector $\vec{AB}$.

$\vec{AB} = \vec{OB} - \vec{OA}$

Displacement from A to B equals position of B minus position of A. Note the order: destination minus origin.

Fill the gap: If $\vec{OA} = \begin{pmatrix} 3 \\ 7 \end{pmatrix}$ and $\vec{OB} = \begin{pmatrix} 8 \\ 2 \end{pmatrix}$, then $\vec{AB} = \begin{pmatrix} 5 \\ \end{pmatrix}$ $\big)$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Reversing the subtraction order

$\vec{a} - \vec{b}$ and $\vec{b} - \vec{a}$ are NOT equal — they point in opposite directions. For displacement, $\vec{AB} = \vec{OB} - \vec{OA}$, not $\vec{OA} - \vec{OB}$. Always write "destination minus origin."

Trap 02

Sign errors with negative components

When subtracting a vector with a negative component, e.g. $\begin{pmatrix} 4 \\ 3 \end{pmatrix} - \begin{pmatrix} 1 \\ -2 \end{pmatrix}$, the second component becomes $3 - (-2) = 5$, not 1. Expand the subtraction as addition of the negative first.

Trap 03

Confusing $\vec{a}-\vec{b}$ direction geometrically

$\vec{a} - \vec{b}$ goes from the head of $\vec{b}$ to the head of $\vec{a}$ (NOT from head of $\vec{a}$ to head of $\vec{b}$). Drawing the parallelogram and checking both diagonals is the safest verification.

Did you get this? True or false: $\vec{a} - \vec{b}$ and $\vec{b} - \vec{a}$ always have the same magnitude but point in opposite directions.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Given $\vec{p} = \begin{pmatrix} 6 \\ -2 \end{pmatrix}$ and $\vec{q} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$, find $\vec{p} - \vec{q}$ and $\vec{q} - \vec{p}$.

Points $P$ and $Q$ have position vectors $\vec{OP} = \begin{pmatrix} -1 \\ 4 \end{pmatrix}$ and $\vec{OQ} = \begin{pmatrix} 5 \\ 0 \end{pmatrix}$. Find $\vec{PQ}$ and its magnitude.

Car A has velocity $\begin{pmatrix} 60 \\ 0 \end{pmatrix}$ km/h and Car B has velocity $\begin{pmatrix} 40 \\ 30 \end{pmatrix}$ km/h. Find the velocity of Car A relative to Car B.

If $\vec{a} - \vec{b} = \begin{pmatrix} 3 \\ -5 \end{pmatrix}$ and $\vec{a} = \begin{pmatrix} 7 \\ 2 \end{pmatrix}$, find $\vec{b}$.

In triangle OAB, $\vec{OA} = \vec{a}$ and $\vec{OB} = \vec{b}$. Express $\vec{AB}$ and $\vec{BA}$ in terms of $\vec{a}$ and $\vec{b}$.

Odd one out: Three of these statements about vector subtraction are correct. Which one is NOT?

Revisit your thinking

Earlier you estimated the boat's velocity relative to the water given $\vec{v}_b = \begin{pmatrix} 4 \\ 3 \end{pmatrix}$ m/s and $\vec{v}_c = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ m/s.

The answer is $\vec{v}_b - \vec{v}_c = \begin{pmatrix} 3 \\ 3 \end{pmatrix}$ m/s. The current's eastward component of 1 m/s is "removed" from the boat's velocity to reveal how fast the boat would appear to move to someone drifting with the water. Did your intuition match this? The key insight is that subtraction removes the reference motion.

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Multiple choice

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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 31 mark

Q1. Given $\vec{a} = \begin{pmatrix} 9 \\ 4 \end{pmatrix}$ and $\vec{b} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}$, find $\vec{a} - \vec{b}$. (1 mark)

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ApplyBand 42 marks

Q2. Points $M$ and $N$ have position vectors $\vec{OM} = \begin{pmatrix} 1 \\ -3 \end{pmatrix}$ and $\vec{ON} = \begin{pmatrix} 7 \\ 5 \end{pmatrix}$. Find $\vec{MN}$ and the distance $MN$. (2 marks)

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AnalyseBand 53 marks

Q3. In triangle $OAB$, $M$ is the midpoint of $AB$. Given $\vec{OA} = \vec{a}$ and $\vec{OB} = \vec{b}$, express $\vec{OM}$ in terms of $\vec{a}$ and $\vec{b}$. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $\vec{p}-\vec{q} = \begin{pmatrix} 3 \\ -6 \end{pmatrix}$; $\vec{q}-\vec{p} = \begin{pmatrix} -3 \\ 6 \end{pmatrix}$

2. $\vec{PQ} = \begin{pmatrix} 6 \\ -4 \end{pmatrix}$; $|\vec{PQ}| = \sqrt{36+16} = \sqrt{52} = 2\sqrt{13}$

3. $\vec{v}_A - \vec{v}_B = \begin{pmatrix} 20 \\ -30 \end{pmatrix}$ km/h

4. $\vec{b} = \vec{a} - (\vec{a}-\vec{b}) = \begin{pmatrix} 7 \\ 2 \end{pmatrix} - \begin{pmatrix} 3 \\ -5 \end{pmatrix} = \begin{pmatrix} 4 \\ 7 \end{pmatrix}$

5. $\vec{AB} = \vec{b} - \vec{a}$; $\vec{BA} = \vec{a} - \vec{b}$

Q1 (1 mark): $\vec{a} - \vec{b} = \begin{pmatrix} 9-3 \\ 4-(-2) \end{pmatrix} = \begin{pmatrix} 6 \\ 6 \end{pmatrix}$ [1].

Q2 (2 marks): $\vec{MN} = \vec{ON} - \vec{OM} = \begin{pmatrix} 7-1 \\ 5-(-3) \end{pmatrix} = \begin{pmatrix} 6 \\ 8 \end{pmatrix}$ [1]. $MN = \sqrt{36+64} = \sqrt{100} = 10$ [1].

Q3 (3 marks): $\vec{AB} = \vec{OB} - \vec{OA} = \vec{b} - \vec{a}$ [1]. $\vec{AM} = \frac{1}{2}\vec{AB} = \frac{1}{2}(\vec{b}-\vec{a})$ [1]. $\vec{OM} = \vec{OA} + \vec{AM} = \vec{a} + \frac{1}{2}(\vec{b}-\vec{a}) = \frac{1}{2}(\vec{a}+\vec{b})$ [1].

Boss battle · The Vector Subtractor

earn bronze · silver · gold

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

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Science Jump · platform challenge

Climb platforms by answering vector subtraction questions. Lighter alternative to the boss.

Mark lesson as complete

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← Lesson 4 · Vector Addition Lesson 6 · Scalar Multiplication →

Module overview · Extension 1 · Checkpoint 1