Vectors and Applications

A vector records both a size and a direction. It is the natural language for a displacement, a velocity, a force, or any quantity for which “how much” alone is incomplete. This lesson assumes comfort with Pythagoras’ theorem, right-triangle trigonometry, and coordinate geometry; the related trigonometry and coordinate geometry lessons give useful preparation.

Scalars, Vectors, and Notation

A scalar has magnitude only: 12 kg, 20°C, and 5 s are scalars. A vector has magnitude and direction: 12 N east, or a displacement 5 m north. Vectors can be drawn as arrows; moving an arrow without turning or resizing it does not change the vector.

In two dimensions, write a vector with components as a = ⟨a_x, a_y⟩. The first component is horizontal (positive right/east) and the second is vertical (positive up/north). In typed work, a may also be written a or a⃗; state the convention once and keep it consistent.

Magnitude and Direction

The magnitude (or length) of a = ⟨x, y⟩ is written |a| and follows from Pythagoras:

|a| = √(x² + y²).

For a nonzero vector, its standard-position direction angle θ is measured counterclockwise from the positive x-axis. First find a reference angle using tan θ = y/x, then use the signs of both components to choose the correct quadrant. A calculator’s atan2(y, x), where available, does this quadrant check.

Worked example: find magnitude and direction

Let u = ⟨−6, 8⟩.

|u| = √[(−6)² + 8²]
    = √(36 + 64)
    = 10

Reference angle = tan⁻¹(|8/−6|) = tan⁻¹(4/3) ≈ 53.1°
The vector is in quadrant II, so
θ = 180° − 53.1° = 126.9°

Thus u has magnitude 10 and direction 126.9° from the positive x-axis. Do not use tan⁻¹(8/−6) = −53.1° without correcting its quadrant.

Components from a Magnitude and Direction

A vector of magnitude r and standard angle θ has components ⟨r cos θ, r sin θ⟩. The cosine supplies the horizontal component and the sine supplies the vertical component; their signs come from the quadrant.

Worked example: resolve a displacement

A hiker walks 18 km at 35° north of east. Taking east as positive x and north as positive y, find the displacement vector.

d = ⟨18 cos 35°, 18 sin 35°⟩
  ≈ ⟨14.7, 10.3⟩ km

The wording “north of east” starts at east and turns toward north,
so 35° is the standard direction angle.

Bearings use a different convention: they are measured clockwise from north. For a bearing B, components are ⟨r sin B, r cos B⟩ when east and north are positive.

Addition, Subtraction, and Scalar Multiplication

Add or subtract corresponding components:

⟨a, b⟩ + ⟨c, d⟩ = ⟨a + c, b + d⟩ and ⟨a, b⟩ − ⟨c, d⟩ = ⟨a − c, b − d⟩.

Multiplication by a scalar k gives k⟨a, b⟩ = ⟨ka, kb⟩. It scales the magnitude by |k|; a negative scalar reverses direction.

Worked example: combine two legs of a trip

First displacement:  p = ⟨3, −2⟩ km
Second displacement: q = ⟨−5, 7⟩ km

Resultant: p + q = ⟨3 + (−5), −2 + 7⟩
                 = ⟨−2, 5⟩ km
Distance from start = |p + q| = √[(−2)² + 5²] = √29 ≈ 5.39 km

The total distance travelled is not necessarily 5.39 km: it is |p| + |q|. A resultant displacement connects start to finish.

Dot Product and the Angle Between Vectors

The dot product of a = ⟨a_x, a_y⟩ and b = ⟨b_x, b_y⟩ is a scalar:

a · b = a_xb_x + a_yb_y = |a||b| cos θ,

where θ is the smaller angle, from 0° to 180°, between nonzero vectors. Therefore cos θ = (a · b)/(|a||b|). Perpendicular vectors have dot product zero. A positive dot product means an acute angle; a negative one means an obtuse angle.

Worked example: angle and perpendicularity

Find the angle between a = ⟨2, 1⟩ and b = ⟨1, −2⟩.

a · b = (2)(1) + (1)(−2) = 0

Because neither vector is zero and their dot product is zero,
cos θ = 0, so θ = 90°.

Projection: the Part in a Chosen Direction

The scalar projection of a onto a nonzero vector b is the signed length of a in b’s direction:

comp_b a = (a · b)/|b|.

The vector projection, the actual vector parallel to b, is

proj_b a = [(a · b)/(|b|²)]b.

Worked example: project onto a direction

a = ⟨4, 3⟩,    b = ⟨2, 1⟩
a · b = (4)(2) + (3)(1) = 11
|b|² = 2² + 1² = 5

proj_b a = (11/5)⟨2, 1⟩ = ⟨22/5, 11/5⟩

The leftover vector a − proj_b a is perpendicular to b. Projection is used to separate a force into a component along a ramp and a component perpendicular to it.

Forces, Relative Velocity, and Navigation

A force is a vector measured in newtons (N). For an object in equilibrium, the vector sum of all forces is ⟨0, 0⟩. Draw a free-body diagram, choose axes, resolve angled forces into components, and set the horizontal and vertical totals equal to zero.

Worked example: tension on a level pull

A rope pulls a crate with 50 N at 30° above horizontal. Find its horizontal and vertical components.

F = ⟨50 cos 30°, 50 sin 30°⟩
  = ⟨25√3, 25⟩ N
  ≈ ⟨43.3, 25.0⟩ N

If the crate moves level at constant velocity and this is the only horizontal pull, friction must be 43.3 N in the opposite direction. The vertical component changes the normal force; it is not automatically “cancelled” unless another force balances it.

For velocity, use a reference statement carefully. “Velocity of plane relative to ground” equals “velocity of plane relative to air” plus “velocity of air relative to ground”:

v_PG = v_PA + v_AG.

Worked example: wind and ground track

An aircraft’s airspeed is 200 km/h due east, and wind is 30 km/h due north. Find its ground velocity.

v_PA = ⟨200, 0⟩     v_AG = ⟨0, 30⟩
v_PG = ⟨200, 30⟩ km/h

Ground speed = √(200² + 30²) ≈ 202.2 km/h
Track angle north of east = tan⁻¹(30/200) ≈ 8.5°

For navigation, distinguish a vehicle’s heading through the air or water from its actual ground track. To maintain a desired track, choose a heading whose velocity counters the crosswind or current.

Common Mistakes

• Adding magnitudes instead of vectors: |a + b| is usually not |a| + |b|. Add components first.
• Mixing angle conventions: “40° north of west,” a standard angle, and a bearing are not interchangeable. Sketch axes and label the starting direction.
• Forgetting signs: west and south components are negative if east and north are positive.
• Using degrees in one line and radians in another: set the calculator mode to match the stated angle. In calculus, angles are commonly in radians.
• Calling a dot product a vector: a · b is one number; projection is a vector.
• Dividing by a zero vector: direction, angle, and projection onto 0 are undefined.

Practice Set

1. Classify each as scalar or vector: (a) 15 m/s southwest, (b) 15 m/s, (c) a force of 8 N upward.
2. For p = ⟨−9, 12⟩, find |p| and its standard direction angle to the nearest tenth of a degree.
3. Write the components of a 24 N force at 60° above the positive x-axis.
4. Let a = ⟨4, −3⟩ and b = ⟨−2, 5⟩. Find a + b, a − b, and −3a.
5. Find the dot product and the angle between u = ⟨3, 4⟩ and v = ⟨4, −3⟩.
6. Find proj_b a for a = ⟨5, 1⟩ and b = ⟨1, 2⟩.
7. A boat moves at ⟨12, 0⟩ m/s relative to the water while the current is ⟨0, −5⟩ m/s relative to the ground. Find the boat’s velocity relative to the ground, its speed, and its direction south of east.
8. Two forces on an object are ⟨18, −7⟩ N and ⟨−6, 15⟩ N. What equilibrant force makes the net force zero?

Answer Checks

1. (a) vector; (b) scalar, because no direction is given; (c) vector.
2. |p| = √(81 + 144) = 15. The reference angle is tan⁻¹(12/9) ≈ 53.1°; quadrant II gives 126.9°.
3. ⟨24 cos 60°, 24 sin 60°⟩ = ⟨12, 12√3⟩ N, approximately ⟨12, 20.8⟩ N.
4. a + b = ⟨2, 2⟩; a − b = ⟨6, −8⟩; −3a = ⟨−12, 9⟩.
5. u · v = 12 − 12 = 0, so the angle is 90°.
6. a · b = 7 and |b|² = 5. Thus proj_b a = (7/5)⟨1, 2⟩ = ⟨7/5, 14/5⟩.
7. v_BG = ⟨12, −5⟩ m/s. Its speed is √169 = 13 m/s, and its direction is tan⁻¹(5/12) ≈ 22.6° south of east.
8. The net of the two given forces is ⟨12, 8⟩ N. Its opposite, the equilibrant, is ⟨−12, −8⟩ N.