Direction and Bearing
A ship sails north-east — but precisely how far north and how far east? A vector's direction can be described as an angle measured from the positive $x$-axis, or as a true bearing measured clockwise from north. In this lesson you'll find direction angles using $\theta = \arctan\!\left(\tfrac{a_2}{a_1}\right)$, convert between component form and polar form, and solve bearing problems in context.
A ship sails along the vector $\mathbf{d} = \mathbf{i} + \mathbf{j}$ (one unit east and one unit north). Without using a formula — estimate its true bearing. Is it closer to north-east, due east, or somewhere else? What angle does this vector make with the positive $x$-axis?
Every direction problem comes down to two decisions: choose the reference direction (positive $x$-axis or north), then apply the correct inverse tangent and adjust for quadrant.
For $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j}$, the direction angle $\theta$ from the positive $x$-axis satisfies:
$\tan\theta = \dfrac{a_2}{a_1}$
A true bearing is measured clockwise from north (positive $y$-axis). For a vector in the first quadrant ($a_1 > 0$, $a_2 > 0$):
bearing $= 90° - \theta$
Key facts
- Direction angle: $\tan\theta = \dfrac{a_2}{a_1}$ measured from positive $x$-axis
- True bearing measured clockwise from north, three-figure notation
- Polar–component conversion: $a_1 = r\cos\theta$, $a_2 = r\sin\theta$
Concepts
- Why $\arctan$ needs quadrant adjustment for vectors not in Q1
- The relationship between bearing and standard angle: bearing $= 90° - \theta$ (Q1)
- How magnitude and direction fully specify a vector (polar form)
Skills
- Find the direction angle of any 2D vector
- Convert between component form and polar form
- Find the true bearing of a displacement vector
For a 2D vector $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j}$, the direction angle $\theta$ (measured anticlockwise from the positive $x$-axis) satisfies:
Because $\arctan$ only returns values in $(-90°, 90°)$, you must adjust for the quadrant:
- Q1 ($a_1 > 0$, $a_2 > 0$): $\theta = \arctan\!\left(\tfrac{a_2}{a_1}\right)$
- Q2 ($a_1 < 0$, $a_2 > 0$): $\theta = 180° + \arctan\!\left(\tfrac{a_2}{a_1}\right)$
- Q3 ($a_1 < 0$, $a_2 < 0$): $\theta = 180° + \arctan\!\left(\tfrac{a_2}{a_1}\right)$
- Q4 ($a_1 > 0$, $a_2 < 0$): $\theta = 360° + \arctan\!\left(\tfrac{a_2}{a_1}\right)$
For the true bearing $\beta$ (clockwise from north), use the connection between bearing and direction angle. In Q1:
Hook example: $\mathbf{d} = \mathbf{i} + \mathbf{j}$. Then $\tan\theta = \tfrac{1}{1} = 1$, so $\theta = 45°$. Bearing $= 90° - 45° = \mathbf{045°T}$. Check your estimate from card 01!
For a 2D vector $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j}$, the direction angle $\theta$ (measured anticlockwise from the positive $x$-axis) satisfies:
Pause — copy the direction-angle method: compute $\alpha=\tan^{-1}|a_2/a_1|$, then adjust using the quadrant of $(a_1,a_2)$ to get the standard angle $\theta\in[0^\circ,360^\circ)$ into your book.
Quick check: What is the direction angle of $\mathbf{v} = \sqrt{3}\,\mathbf{i} + 1\,\mathbf{j}$?
We just saw that for $\vec{a}=a_1\mathbf{i}+a_2\mathbf{j}$, the direction angle $\theta$ satisfies $\tan\theta=a_2/a_1$; you compute $\alpha=\tan^{-1}|a_2/a_1|$ (acute angle), then adjust for quadrant. That raises a question: true bearings are measured clockwise from north, but direction angles are measured counterclockwise from east — what is the exact conversion formula? This card answers it → bearing $\beta=(90^\circ-\theta)\bmod360^\circ$ where $\theta$ is the standard direction angle.
A robust method for finding direction angle and bearing in any quadrant:
- Identify the quadrant from the signs of $a_1$ and $a_2$.
- Find the reference angle $\alpha = \arctan\!\left(\left|\tfrac{a_2}{a_1}\right|\right)$ — always positive, in $(0°, 90°)$.
- Adjust based on quadrant: Q1 → $\theta = \alpha$; Q2 → $\theta = 180° - \alpha$; Q3 → $\theta = 180° + \alpha$; Q4 → $\theta = 360° - \alpha$.
- Convert to bearing if required: bearing $= 90° - \theta$ (adjust to keep in $[0°, 360°)$).
For the reverse (bearing → components): if a vector has magnitude $r$ and true bearing $\beta$, then the direction angle is $\theta = 90° - \beta$ and:
$a_1 = r\cos\theta = r\sin\beta$, $a_2 = r\sin\theta = r\cos\beta$
A robust method for finding direction angle and bearing in any quadrant:
Pause — copy the bearing conversion: bearing $=90^\circ-\theta$ (adjust if negative by adding $360^\circ$); and the four quadrant adjustments into your book.
Did you get this? True or false: A vector in the second quadrant ($a_1 < 0$, $a_2 > 0$) has a direction angle between 90° and 180°.
Worked examples · 3 in a row, reveal as you go
Find the direction angle of $\mathbf{a} = 3\mathbf{i} + 3\mathbf{j}$, and state its true bearing.
Find the direction angle of $\mathbf{b} = -1\mathbf{i} + \sqrt{3}\mathbf{j}$.
A vector has magnitude 8 and true bearing 150°T. Express it in component form.
Fill the gap: A vector at direction angle 60° in Q1 has true bearing °T (three figures).
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: the direction angle of $-\mathbf{i} - \mathbf{j}$ is $225°$ (not $45°$).
Activities · practice with the ideas
Find the direction angle of $\mathbf{v} = 1\mathbf{i} + \sqrt{3}\mathbf{j}$ and its true bearing.
Find the direction angle of $\mathbf{w} = -4\mathbf{i} - 4\mathbf{j}$. Is this vector pointing south-west?
A vector has magnitude 10 and bearing 060°T. Write it in component form ($\mathbf{i}$ and $\mathbf{j}$ components, exact values).
A ship sails on bearing 120°T for a distance of 6 km. How far east and how far south has it travelled?
Convert $\mathbf{u} = 5\mathbf{i} - 5\mathbf{j}$ to polar form $(r, \theta)$ and state its true bearing.
Odd one out: Three of these statements are correct. Which one is NOT?
Earlier you estimated the true bearing of $\mathbf{d} = \mathbf{i} + \mathbf{j}$.
The exact answer is $\tan\theta = \tfrac{1}{1} = 1$, so $\theta = 45°$, giving bearing $= 90° - 45° = \mathbf{045°T}$. The equal east and north components place the vector exactly on the north-east diagonal — a useful reference point to memorise.
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. Find the direction angle of $\mathbf{p} = 1\mathbf{i} + \sqrt{3}\mathbf{j}$. (1 mark)
Q2. A velocity vector has magnitude 20 and bearing 330°T. Write it in component form using exact values. (2 marks)
Q3. Two vectors $\mathbf{u} = 3\mathbf{i} + 4\mathbf{j}$ and $\mathbf{v} = 4\mathbf{i} + 3\mathbf{j}$ have the same magnitude. Do they have the same direction? Justify your answer. (2 marks)
Comprehensive answers (click to reveal)
Activity answers: 1. $\tan\theta = \sqrt{3}$, $\theta = 60°$, bearing = 030°T · 2. Q3, $\alpha = 45°$, $\theta = 225°$ (pointing SW) · 3. $a_1 = 10\sin60° = 5\sqrt{3}\,\mathbf{i}$, $a_2 = 10\cos60° = 5\,\mathbf{j}$ · 4. East $= 6\sin120° = 3\sqrt{3}$ km; South $= -6\cos120° = 3$ km south · 5. $r = 5\sqrt{2}$, $\theta = 315°$, bearing = 135°T
Q1 (1 mark): $\tan\theta = \dfrac{\sqrt{3}}{1} = \sqrt{3}$, so $\theta = \arctan(\sqrt{3}) = \mathbf{60°}$ [1]. (Q1 adjustment not needed since $a_1 > 0$, $a_2 > 0$.)
Q2 (2 marks): $a_1 = 20\sin330° = 20\times(-\tfrac{1}{2}) = -10$ [1]; $a_2 = 20\cos330° = 20\times\tfrac{\sqrt{3}}{2} = 10\sqrt{3}$ [1]. So $\mathbf{v} = -10\mathbf{i} + 10\sqrt{3}\,\mathbf{j}$. (The bearing is in the north-west quadrant, so $a_1 < 0$, $a_2 > 0$ — Q2, consistent.)
Q3 (2 marks): $|\mathbf{u}| = |\mathbf{v}| = \sqrt{9+16} = 5$ — same magnitude [1]. Direction of $\mathbf{u}$: $\theta_u = \arctan(\tfrac{4}{3}) \approx 53.1°$; direction of $\mathbf{v}$: $\theta_v = \arctan(\tfrac{3}{4}) \approx 36.9°$. Since $\theta_u \neq \theta_v$, the vectors have different directions [1]. (They are not equal vectors.)
Five timed questions on direction angles and bearings. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering direction and bearing questions. A lighter alternative to the boss.
Mark lesson as complete
Tick when you've finished the practice and review.