01 · Length and unit vector
Find the magnitude and a same-direction unit vector for −5i + 12j.
Hint
Use 25 + 144 = 169.
Worked solution
|a| = 13
â = −(5/13)i + (12/13)j
Understand · explore · practise
Vector magnitude and direction
Find vector lengths, unit vectors and directions with the correct quadrant. Resolve components, find angles between vectors and calculate triangle or parallelogram areas.
Before you startPythagoras, trigonometric ratios, cosine rule and surds
01 / Magnitude
Use the two perpendicular components as triangle legs.
|a| = √(p² + q²), for a = pi + qj
For a = −8i + 6j, the magnitude is √(64 + 36) = 10. Squaring removes component signs, so changing direction may leave the length unchanged.
Calculate a combined vector before finding its magnitude. For a = 2i + j and b = −i + 3j, a + b = i + 4j and |a + b| = √17. In general |a + b| is not |a| + |b|: a bent route can be longer than its direct displacement.
Separate length from directionMove at your pace
a = (-4,3), magnitude ≈ 5. Anticlockwise direction from the positive x-axis ≈ 143.1301°. Same-direction unit vector ≈ (-0.8,0.6), shown in gold. Values are rounded to four decimal places.
Watch length and direction separate when a vector is rescaled
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 / Unit vectors
Divide a non-zero vector by its own length.
â = a / |a|, provided a ≠ 0
For a = −8i + 6j, its unit vector is −(4/5)i + (3/5)j. Its length is √(16/25 + 9/25) = 1 and its direction agrees with a.
A vector of length 7 in that direction is 7â = −(28/5)i + (21/5)j. For the opposite direction, use −7â. There is no unit vector “in the direction of” the zero vector because its direction is undefined and division by its magnitude would divide by zero.
03 / Direction angle
State where the angle starts and which way it turns.
Here a direction angle θ is measured anticlockwise from the positive x-axis, with 0° ≤ θ < 360°. Sketch the signs before applying inverse tangent.
For a = −8i + 6j, the vector lies in quadrant II. Its acute reference angle is α = tan⁻¹(6/8) ≈ 36.8699°, so θ = 180° − α ≈ 143.1°.
p > 0, q > 0: θ = α
p < 0, q > 0: θ = 180° − α
p < 0, q < 0: θ = 180° + α
p > 0, q < 0: θ = 360° − α
Here α = tan⁻¹(|q/p|) for non-zero p and q. Handle axis vectors directly: right 0°, up 90°, left 180°, down 270°. The zero vector has no θ.
04 / Angles with an axis
A smaller angle is different from a full-turn direction.
The smaller angle between a non-zero vector and i is between 0° and 180°. For a = −8i + 6j it is 143.1°. The smaller angle with j is 53.1°.
For b = 3i − 4j, the anticlockwise direction angle is 306.9°, but the smaller angle with i is 53.1°. The smaller angle with the undirected x-axis would lie between 0° and 90°; it would also treat leftward and rightward directions as the same line.
Use the convention requested by the question or indicated in its diagram. A bearing instead starts at north and turns clockwise. Do not transfer an inverse-tangent answer between these conventions without checking it.
05 / From length to components
Use the angle’s reference axis to choose sine and cosine.
If a vector has magnitude R and anticlockwise direction angle θ from the positive x-axis:
a = R cos θ i + R sin θ j
For R = 8 and θ = 150°, a = −4√3 i + 4j. The cosine is negative, matching the leftward component.
An angle measured from north
A vector of length 12 points 30° east of north. Its horizontal component is 12 sin 30° = 6 and its vertical component is 12 cos 30° = 6√3. If it instead points west of north, the horizontal component is −6. The phrase “30° with j” alone does not choose east or west.
When two directions fit
A vector of magnitude 13 has direction angle θ with sin θ = 5/13. Its vertical component is 5, while its horizontal component may be 12 or −12. Both (12,5) and (−12,5) fit until a quadrant or sign condition is supplied.
06 / Unknown components
A magnitude equation can have two roots.
If a = ki + 4j and |a| = 5, then k² + 16 = 25, so k = ±3. The two vectors have the same magnitude but different directions.
More generally, if |λa| = L for a non-zero vector a and L > 0, then |λ| = L/|a|. The magnitude alone gives λ = ±L/|a|. A same-direction condition keeps the positive sign; an opposite-direction condition keeps the negative one.
A constraint can be impossible
A vector ki + 4j cannot have magnitude 3 because k² + 16 ≥ 16. Its shortest possible length is 4, reached when k = 0.
07 / Angle between two vectors
Put the arrows at a common start, then use a triangle.
Let a = 4i + j and b = i + 3j start at O and end at A,B. The third side of triangle OAB is b − a = −3i + 2j. Its length is √13.
|a| = √17, |b| = √10
cos θ = (17 + 10 − 13)/(2√17√10)
= 7/√170
θ ≈ 57.5°
This gives the smaller angle, 0° ≤ θ ≤ 180°. It works even when the two arrows lie on different sides of an axis. Do not just subtract two acute reference angles without placing the directions correctly.
A useful component form
For a = (p,q) and b = (r,s), expanding |b − a|² in the cosine rule gives:
cos θ = (pr + qs) / [√(p² + q²)√(r² + s²)]
Both vectors must be non-zero. This follows from the cosine rule; no new geometric assumption is needed.
08 / Triangle and parallelogram areas
Area is non-negative, so keep the absolute value.
A parallelogram with adjacent vectors a = (p,q) and b = (r,s) has area |a||b| sin θ, where θ is their smaller angle. Combining that with the preceding cosine expression gives:
Area² = (p² + q²)(r² + s²) − (pr + qs)²
= (ps − qr)²
Parallelogram area = |ps − qr|
Triangle area = ½|ps − qr|
For a = 4i + j and b = i + 3j, the parallelogram area is |12 − 1| = 11 and the triangle area is 11/2. Swapping the two vectors changes the sign inside the absolute value but leaves the area unchanged.
If ps − qr = 0, the arrows are parallel or one is zero and the shape is degenerate, with area 0. Use squared units.
09 / Your turn
Separate magnitude, direction and the angle convention.
Give lengths exactly and non-exact angles to one decimal place.
02 · Add before measuring
a = 3i − j and b = −i + 2j. Compare |a + b| with |a| + |b|.
Hint
The resultant is 2i + j.
Worked solution
|a + b| = √5
|a| + |b| = √10 + √5
The magnitudes do not add in this case.
03 · A quadrant check
Find the anticlockwise direction angle of −i − √3 j.
Hint
It lies in quadrant III with reference angle 60°.
Worked solution
θ = 240°
04 · Axis conventions
For −j, give its anticlockwise direction angle, its smaller angle with i and its smaller angle with j.
Hint
The arrow points straight down.
Worked solution
Direction: 270°
Angle with i: 90°
Angle with j: 180°
05 · Resolve a vector
A vector has length 6 and anticlockwise direction angle 120°. Find its components.
Hint
Use cosine horizontally and sine vertically.
Worked solution
a = −3i + 3√3 j
06 · Two possible components
Find all k for which ki − 5j has magnitude 13.
Hint
k² + 25 = 169.
Worked solution
k = ±12
07 · The zero vector
Why does a/|a| fail for a = 0i + 0j?
Hint
Find its magnitude.
Worked solution
Its magnitude is zero, so the formula divides by zero. The zero vector has no direction to preserve.
08 · A right angle
Find the smaller angle between 3i + 2j and −2i + 3j.
Hint
Use the cosine-rule component numerator.
Worked solution
3(−2) + 2(3) = 0 ⇒ cos θ = 0
θ = 90°
09 · Area and orientation
A triangle has adjacent displacement vectors 3i − j and i + 4j. Find its area. What happens if you swap the vectors?
Hint
Use the absolute value of the determinant.
Worked solution
Area = ½|3 × 4 − (−1) × 1| = 13/2
Swapping gives −13 inside the absolute value; the area remains 13/2.
10 / Recap
Keep length and direction as separate questions.
- Magnitude is √(p² + q²).
- Normalise only a non-zero vector.
- A scalar’s sign controls direction; its absolute value controls length.
- State the angle convention and check the quadrant.
- Resolve components using the correct reference axis.
- Use the cosine rule for the smaller angle between two vectors.
- Area uses an absolute value and squared units.
Section 1 of 10 · Magnitude