A swimmer points straight across a river, yet ends up far downstream. A plane aims north but drifts east. In both cases two vectors combine into one resultant, and the only reliable way to find it is to draw the arrows correctly. This lesson turns vectors from symbols into pictures you can add and subtract by hand.
You walk $3\ \text{m}$ east, then $4\ \text{m}$ north.
Before reading: how far are you from your starting point in a straight line, and is that more or less than the $7\ \text{m}$ you actually walked? Write your guess and one sentence on your reasoning.
A vector becomes a picture: length for magnitude, direction for the arrow
To represent any vector you draw an arrow whose length is set by a scale and whose direction matches the real-world direction, the same recipe works for displacement, velocity and acceleration.
Choose a scale (say $1\ \text{cm} = 1\ \text{m s}^{-1}$), draw the arrow that long, and point it the right way. The tail is the start, the head is the arrow tip. A velocity of $4\ \text{m s}^{-1}$ east and an acceleration of $4\ \text{m s}^{-2}$ east are drawn the same way, only the labelled quantity differs.
Represent any vector as an arrow: choose a scale, make the length match the magnitude, and point it in the real direction. The tail is the start and the head (arrowhead) shows direction. Displacement, velocity and acceleration are all drawn this way.
Pause, copy the highlighted statement into your book before moving on.
Did you get this? True or false: on a scaled vector diagram, a longer arrow always represents a larger magnitude.
Combining two vectors into a single resultant
We just saw how to draw one vector as an arrow. That raises a question: how do two arrows combine into one? This card answers it → place them head-to-tail and the resultant runs from the first tail to the last head.
Slide the second arrow so its tail sits on the first arrow's head; the resultant is the single arrow from the very first tail to the very last head.
For two perpendicular vectors, the resultant's magnitude comes from Pythagoras and its direction from trigonometry:
$$R = \sqrt{A^2 + B^2} \qquad\qquad \theta = \tan^{-1}\!\left(\frac{B}{A}\right)$$
A boat heads straight across a river at $3.0\ \text{m s}^{-1}$. The current flows downstream at $4.0\ \text{m s}^{-1}$, perpendicular to the boat's heading. Find the boat's resultant velocity (magnitude and direction relative to its heading).
Add vectors head-to-tail: the resultant runs from the first tail to the last head. For two perpendicular vectors, $R = \sqrt{A^2 + B^2}$ and the direction is $\theta = \tan^{-1}(B/A)$.
Add the highlighted method to your notes before the check below.
Quick check: A person walks $6\ \text{m}$ east then $8\ \text{m}$ north. What is the magnitude of their resultant displacement?
Pattern A, Construct then calculate
A hiker walks $5\ \text{km}$ north, then $12\ \text{km}$ east. In your book: (1) choose a scale and draw both legs head-to-tail; (2) draw the resultant displacement; (3) calculate its magnitude with Pythagoras and its bearing with $\tan^{-1}$.
Two truths, one lie: which statement about adding vectors in 2D is the lie?
Dot point: Add and subtract vectors in 2 dimensions
Why subtraction is just adding the reverse arrow
We just learned to add vectors head-to-tail. That raises a question: how do we handle a change in a vector, like a change in velocity? This card answers it → subtracting a vector means adding its reverse.
To subtract $\vec{B}$, flip it around to get $-\vec{B}$ (same length, opposite direction) and then add it head-to-tail just like before.
So $\vec{A} - \vec{B} = \vec{A} + (-\vec{B})$. This matters because a change in velocity is $\Delta\vec{v} = \vec{v}_{\text{final}} - \vec{v}_{\text{initial}}$, and acceleration points in the direction of $\Delta\vec{v}$. For a car that turns a corner without changing speed, $\vec{v}_{\text{final}}$ and $\vec{v}_{\text{initial}}$ have the same length but different directions, so $\Delta\vec{v}$ is non-zero, that is the acceleration.
To subtract, reverse the second vector and add: $\vec{A} - \vec{B} = \vec{A} + (-\vec{B})$, where $-\vec{B}$ has the same magnitude as $\vec{B}$ but the opposite direction. A change in velocity $\Delta\vec{v} = \vec{v}_f - \vec{v}_i$ is found this way, and acceleration points along $\Delta\vec{v}$.
Pause, copy the highlighted rule into your book before the check below.
Last check: Decide if the statement is true or false.
To work out $\vec{A} - \vec{B}$ on a diagram, you reverse $\vec{B}$ and add it head-to-tail to $\vec{A}$.
Odd one out: three of these belong on a correctly drawn scale vector diagram. Which one does not?
Dot point: Use vector diagrams to represent the displacement, velocity and acceleration of an object
A fresh set drawn from this lesson's question bank, feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence, that tells the system what to drill next.
ApplyBand 5(4 marks) 1. A plane flies $90\ \text{km}$ north, then $120\ \text{km}$ east. Using a head-to-tail vector diagram, determine the magnitude and direction of the plane's resultant displacement. Show your working.
1 mark: head-to-tail diagram with resultant drawn · 1 mark: $R = \sqrt{90^2 + 120^2}$ · 1 mark: $R = 150\ \text{km}$ · 1 mark: direction $= \tan^{-1}(120/90) \approx 53°$ east of north
ApplyBand 5(4 marks) 2. A hiker walks $6.0\ \text{km}$ north, then $8.0\ \text{km}$ east. Using a head-to-tail vector diagram, determine the magnitude and direction of the resultant displacement.
1 mark: head-to-tail diagram with both legs and the resultant · 1 mark: magnitude by Pythagoras · 1 mark: angle by trigonometry · 1 mark: direction stated correctly as a bearing or as "east of north"
Dot point: Add and subtract vectors in 2 dimensions
UnderstandBand 4(3 marks) 3. Explain how subtracting a vector $\vec{B}$ from a vector $\vec{A}$ is carried out on a diagram, and state why $\vec{A} - \vec{B}$ is generally not the same as $\vec{B} - \vec{A}$.
1 mark: reverse $\vec{B}$ to get $-\vec{B}$ · 1 mark: add $\vec{A}$ and $-\vec{B}$ head-to-tail · 1 mark: the two subtractions have equal magnitude but opposite direction
Dot point: Add and subtract vectors in 2 dimensions
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set from the lesson bank.
Q1 (4 marks): Drawing the $90\ \text{km}$ north leg, then placing the tail of the $120\ \text{km}$ east leg at its head, the resultant runs from the start to the final position. The two legs are perpendicular, so by Pythagoras $R = \sqrt{90^2 + 120^2} = \sqrt{8100 + 14400} = \sqrt{22500} = 150\ \text{km}$. The direction is measured from north toward east: $\theta = \tan^{-1}\!\left(\frac{120}{90}\right) \approx 53°$. The resultant displacement is therefore $150\ \text{km}$ at a bearing of about $053°$ (i.e. $53°$ east of north).
Q2 (4 marks): Drawing the $6.0\ \text{km}$ north leg first and the $8.0\ \text{km}$ east leg from its head, the resultant runs from the start to the final head. The legs are perpendicular, so $$s = \sqrt{6.0^2 + 8.0^2} = \sqrt{100} = 10.0\ \text{km}$$ The angle east of north is $$\theta = \tan^{-1}\!\left(\frac{8.0}{6.0}\right) = 53^\circ$$ so the resultant displacement is $10.0\ \text{km}$ at $053^\circ$, that is $53^\circ$ east of north.
Q3 (3 marks): To find $\vec{A} - \vec{B}$, first reverse $\vec{B}$ to obtain $-\vec{B}$, which has the same length and points the opposite way. Then add $\vec{A}$ and $-\vec{B}$ head-to-tail in the usual way; the resultant drawn from the first tail to the last head is $\vec{A} - \vec{B}$. Reversing the order gives $\vec{B} - \vec{A}$, which is the reverse of the first result: the same magnitude but the opposite direction. Because direction is part of a vector, the two are not the same vector, so vector subtraction is not commutative even though vector addition is.
You can now draw vectors to scale, add them head-to-tail to find a resultant, and subtract them by reversing. Next lesson does the reverse operation, breaking a single vector into perpendicular components.
↩ Back to focus-area hubAt the start you walked $3\ \text{m}$ east then $4\ \text{m}$ north. The straight-line distance is $5\ \text{m}$, found from $\sqrt{3^2 + 4^2}$, less than the $7\ \text{m}$ you walked, because the path was not straight. If you guessed $7\ \text{m}$, revisit Card 2 and the head-to-tail diagram.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
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