Vector Addition
When a ship sails east and then north-east, the combined effect is a single resultant journey. That combined displacement is found using vector addition. The triangle law, the parallelogram law, and Chasles’ relation $\vec{AB} + \vec{BC} = \vec{AC}$ are three faces of the same geometric truth — master them here.
You walk 3 km east then 4 km north. Without using a formula — what single vector would describe your total journey? How far are you from your starting point, and in what direction?
When two or more vectors are applied in sequence, the resultant is the single vector that produces the same combined effect. Two geometric rules give the same result.
Triangle law: place the tail of $\vec{b}$ at the head of $\vec{a}$. The resultant goes from the tail of $\vec{a}$ to the head of $\vec{b}$.
Chasles' relation:
$\vec{AB} + \vec{BC} = \vec{AC}$
Key facts
- Triangle law: place tail of $\vec{b}$ at head of $\vec{a}$; resultant connects tail of $\vec{a}$ to head of $\vec{b}$
- Chasles' relation: $\vec{AB} + \vec{BC} = \vec{AC}$
- Vector addition is commutative and associative
Concepts
- Why the triangle law and parallelogram law give the same resultant
- How Chasles' relation follows directly from the triangle law
- The geometric meaning of the resultant as a combined displacement
Skills
- Add two or more vectors geometrically using the triangle or parallelogram law
- Simplify chains of vectors using Chasles' relation
- Add vectors in component form by adding corresponding components
To add two vectors $\vec{a}$ and $\vec{b}$ using the triangle law:
- Draw $\vec{a}$.
- Place the tail of $\vec{b}$ at the head of $\vec{a}$.
- The resultant $\vec{a} + \vec{b}$ goes from the tail of $\vec{a}$ to the head of $\vec{b}$.
In terms of points this is Chasles' relation:
The intermediate point $B$ cancels. The same idea extends to any chain: $\vec{AB} + \vec{BC} + \vec{CD} = \vec{AD}$.
Triangle law: place tail of $\vec{b}$ at head of $\vec{a}$; $\vec{a}+\vec{b}$ = closing vector. Chasles: $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$.
Pause — copy the triangle law and Chasles' relation: $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$ with a labelled diagram into your book.
Quick check: In triangle $XYZ$, which expression equals $\vec{XY} + \vec{YZ}$?
We just saw the triangle law: place $\vec{a}$ head-to-tail with $\vec{b}$; the sum $\vec{a}+\vec{b}$ is the vector closing the triangle. Chasles' relation states $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$ for any three points. That raises a question: the parallelogram law constructs the same sum differently — and also yields $\vec{a}-\vec{b}$ as the other diagonal — how does this relate to the triangle law? This card answers it → both diagonals of the parallelogram are $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$.
Alternatively, draw both vectors from the same starting point and complete the parallelogram. The diagonal from the shared point is $\vec{a} + \vec{b}$. This gives exactly the same result as the triangle law.
In component form, vector addition is simply component-wise:
The commutative property $\vec{a} + \vec{b} = \vec{b} + \vec{a}$ is immediately visible in the parallelogram: both diagonals point to the same resultant regardless of which side you label as $\vec{a}$.
Alternatively, draw both vectors from the same starting point and complete the parallelogram. The diagonal from the shared point is $\vec{a} + \vec{b}$. This gives exactly the same result as the triangle law.
Pause — copy the parallelogram law: draw $\vec{a}$ and $\vec{b}$ from the same point; diagonal = $\vec{a}+\vec{b}$; other diagonal = $\vec{a}-\vec{b}$ into your book.
Did you get this? True or false: $\begin{pmatrix} 3 \\ -1 \end{pmatrix} + \begin{pmatrix} -5 \\ 4 \end{pmatrix} = \begin{pmatrix} -2 \\ 3 \end{pmatrix}$.
Worked examples · 3 in a row, reveal as you go
In triangle $ABC$, $D$ is the midpoint of $BC$. Simplify $\vec{AB} + \vec{BD}$.
Given $\vec{a} = \begin{pmatrix} 2 \\ -3 \end{pmatrix}$ and $\vec{b} = \begin{pmatrix} -5 \\ 1 \end{pmatrix}$, find $\vec{a} + \vec{b}$ and verify that $\vec{a}+\vec{b} = \vec{b}+\vec{a}$.
$ABCD$ is a quadrilateral. Simplify $\vec{AB} + \vec{BC} + \vec{CD}$.
Fill the gap: $\begin{pmatrix} -2 \\ 5 \end{pmatrix} + \begin{pmatrix} 7 \\ -3 \end{pmatrix} = \begin{pmatrix}$$\\$$\end{pmatrix}$.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: $\vec{AB} + \vec{CB}$ can be simplified using Chasles' relation directly without rewriting either vector.
Activities · practice with the ideas
Let $\vec{a} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}$ and $\vec{b} = \begin{pmatrix} -2 \\ 5 \end{pmatrix}$. Calculate $\vec{a}+\vec{b}$ and $\vec{b}+\vec{a}$ and verify they are equal.
In quadrilateral $PQRS$, simplify $\vec{PQ} + \vec{QR} + \vec{RS}$.
Show that $\vec{AB} + \vec{BC} + \vec{CA} = \mathbf{0}$ for any triangle $ABC$.
$\vec{p} = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$, $\vec{q} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$, $\vec{r} = \begin{pmatrix} -4 \\ 0 \end{pmatrix}$. Find $\vec{p}+\vec{q}+\vec{r}$.
Prove that $(\vec{a}+\vec{b})+\vec{c} = \vec{a}+(\vec{b}+\vec{c})$ using column vectors $\vec{a}=\begin{pmatrix}a_1\\a_2\end{pmatrix}$, $\vec{b}=\begin{pmatrix}b_1\\b_2\end{pmatrix}$, $\vec{c}=\begin{pmatrix}c_1\\c_2\end{pmatrix}$.
Odd one out: Three of these vector addition statements are correct. Which one is FALSE?
Earlier you estimated the resultant of walking 3 km east then 4 km north.
The answer: $\vec{a}+\vec{b} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$, with magnitude $\sqrt{9+16} = 5$ km — the famous 3-4-5 right triangle. Did you predict the 5 km distance? The key insight is that the resultant is found by adding the component vectors, not by adding their magnitudes (3 + 4 = 7 would be wrong). Geometry and algebra align perfectly when you add vectors component by component.
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. Let $\vec{u} = \begin{pmatrix} -1 \\ 3 \end{pmatrix}$ and $\vec{v} = \begin{pmatrix} 4 \\ -2 \end{pmatrix}$. Find $\vec{u} + \vec{v}$. (1 mark)
Q2. In pentagon $ABCDE$, simplify $\vec{AB} + \vec{BC} + \vec{CD} + \vec{DE}$. State which law you are using. (2 marks)
Q3. $OABC$ is a parallelogram with $\vec{OA} = \vec{a}$ and $\vec{OC} = \vec{c}$. Express $\vec{OB}$ in terms of $\vec{a}$ and $\vec{c}$, and explain why $\vec{a}+\vec{c} = \vec{c}+\vec{a}$ geometrically. (3 marks)
Comprehensive answers (click to reveal)
Activity answers: 1. $\begin{pmatrix} 2 \\ 6 \end{pmatrix}$ (both ways) · 2. $\vec{PS}$ · 3. $\vec{AB}+\vec{BC}+\vec{CA} = \vec{AC}+\vec{CA} = \vec{AA} = \mathbf{0}$ · 4. $\begin{pmatrix} 0 \\ 2 \end{pmatrix}$ · 5. Both sides $= \begin{pmatrix} a_1+b_1+c_1 \\ a_2+b_2+c_2 \end{pmatrix}$ ✓
Q1 (1 mark): $\vec{u}+\vec{v} = \begin{pmatrix} -1+4 \\ 3+(-2) \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}$ [1].
Q2 (2 marks): Using Chasles' relation (triangle law) [1]: $\vec{AB}+\vec{BC}+\vec{CD}+\vec{DE} = \vec{AE}$ [1]. Each intermediate point cancels.
Q3 (3 marks): In parallelogram $OABC$, $\vec{AB} = \vec{OC} = \vec{c}$ (opposite sides) [1]. So $\vec{OB} = \vec{OA}+\vec{AB} = \vec{a}+\vec{c}$ [1]. Alternatively, via $C$: $\vec{OB} = \vec{OC}+\vec{CB} = \vec{c}+\vec{a}$. Both paths reach the same diagonal $B$, demonstrating geometrically that $\vec{a}+\vec{c} = \vec{c}+\vec{a}$ [1].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering vector addition questions. Lighter alternative to the boss.
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