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Module 6 · L18 of 20 ~35 min ⚡ +95 XP available

Applications of Vectors

A ship steaming north-east at 20 knots, a plane buffeted by a crosswind, a beam under two tension forces — all of these are vector problems in disguise. In this lesson you'll translate real-world scenarios into vectors, resolve forces and velocities, and use dot products and projections to extract distances and angles.

Today's hook — A plane's airspeed vector is $(300, 0)$ km/h (due east) and the wind vector is $(0, 50)$ km/h (due north). Before reading on — what is the plane's actual ground speed and direction? Make a rough estimate.

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Recall — your gut answer first

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A plane's airspeed vector is $(300, 0)$ km/h and the wind vector is $(0, 50)$ km/h. Without calculating — estimate the ground speed and the direction the plane actually travels. Write your reasoning below.

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Four application types to master

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Vector applications all reduce to the same four toolkit items:

1. Resultant (addition): Combine forces or velocities — $\mathbf{R} = \mathbf{F}_1 + \mathbf{F}_2$. Speed = $|\mathbf{R}|$.

2. Component resolution: Split a vector at angle $\theta$ into horizontal and vertical parts — $(F\cos\theta,\, F\sin\theta)$.

3. Relative velocity: Velocity of $A$ relative to $B$ is $\mathbf{v}_{A/B} = \mathbf{v}_A - \mathbf{v}_B$.

4. Dot product applications: Angle between vectors, work done ($W = \mathbf{F}\cdot\mathbf{d}$), projection onto a direction.

$|\mathbf{R}| = \sqrt{R_x^2 + R_y^2}$

North/East convention

In navigation, north = positive $y$, east = positive $x$. A bearing of $\theta°$ from north gives direction vector $(\sin\theta, \cos\theta)$ — note the swap from the standard angle.

Equilibrium

A particle is in equilibrium when the net force is zero: $\mathbf{F}_1 + \mathbf{F}_2 + \cdots = \mathbf{0}$. Equate $x$ and $y$ components to zero separately.

Work done formula

Work $W = \mathbf{F} \cdot \mathbf{d} = |\mathbf{F}||\mathbf{d}|\cos\theta$. Only the component of force in the direction of displacement does work. Unit: joules (N·m).

What you'll master

Know

Key facts

Resultant force/velocity: add component vectors
Relative velocity: $\mathbf{v}_{A/B} = \mathbf{v}_A - \mathbf{v}_B$
Work: $W = \mathbf{F} \cdot \mathbf{d}$; equilibrium: net force $= \mathbf{0}$

Understand

Concepts

Why vector addition gives the resultant effect of multiple forces or velocities
How the dot product measures the "overlap" between force and displacement
The difference between speed (scalar) and velocity (vector)

Can do

Skills

Resolve forces and velocities into components and find resultants
Solve navigation problems using bearing-to-component conversion
Calculate work done and find angles using the dot product

Key terms

Resultant vectorThe single vector equal to the sum of two or more vectors. Represents the combined effect of all components.

Component resolutionSplitting a vector into perpendicular parts, usually horizontal ($x$) and vertical ($y$), using trigonometry.

Relative velocityThe velocity of one object as observed from another. $\mathbf{v}_{A/B} = \mathbf{v}_A - \mathbf{v}_B$.

EquilibriumA state in which the net force on an object is zero: $\sum \mathbf{F} = \mathbf{0}$, implying no acceleration.

Work$W = \mathbf{F} \cdot \mathbf{d}$. The scalar quantity measuring energy transferred by a force through a displacement. Measured in joules.

BearingA direction measured clockwise from north (0°) to 360°. A bearing of $\theta°$ corresponds to vector $(\sin\theta, \cos\theta)$ (unit vector).

Forces and resultants

core concept

When several forces act on a particle simultaneously, the resultant is their vector sum. To find it:

Resolve each force into $x$ and $y$ components (using magnitude and angle).
Sum all $x$-components: $R_x = \sum F_{ix}$.
Sum all $y$-components: $R_y = \sum F_{iy}$.
Find magnitude: $|\mathbf{R}| = \sqrt{R_x^2 + R_y^2}$.
Find direction: $\theta = \tan^{-1}\!\left(\dfrac{R_y}{R_x}\right)$ (adjust for quadrant).

Equilibrium condition: For a particle in equilibrium, $R_x = 0$ and $R_y = 0$ simultaneously. This gives a system of two equations to solve for unknown magnitudes or angles.

Answer to the hook. Ground velocity $= (300,0)+(0,50)=(300,50)$ km/h. Speed $= \sqrt{300^2+50^2}=\sqrt{92500}\approx 304.1$ km/h. Direction: $\tan^{-1}(50/300)\approx 9.5°$ north of east — or a bearing of about $80.5°$.

Resultant: $\vec{F}_{\text{net}}=\sum\vec{F}_i$ component-wise. Bearing $\theta^\circ$ to components: $\begin{pmatrix}\sin\theta\\\cos\theta\end{pmatrix}\times\text{speed}$.

Pause — copy the resultant force method: express each force as a component vector, sum all $x$-components and all $y$-components separately into your book.

Quick check: Two forces act on a particle: $\mathbf{F}_1 = (3,4)$ N and $\mathbf{F}_2 = (-1,2)$ N. What is the magnitude of the resultant force?

Velocities and navigation

core concept

We just saw that when multiple forces act on a particle, the resultant is their vector sum, computed component by component. That raises a question: for navigation problems (ship/plane with velocity plus current/wind), how do you convert a bearing angle to a component vector, and what is the resultant velocity? This card answers it → bearing $\theta^\circ$ gives direction vector $\begin{pmatrix}\sin\theta\\\cos\theta\end{pmatrix}$; multiply by speed; add all force/velocity vectors component-wise.

In navigation problems, velocities add as vectors:

$$\mathbf{v}_{\text{ground}} = \mathbf{v}_{\text{self}} + \mathbf{v}_{\text{current/wind}}$$

Bearing to component conversion: A bearing of $\theta°$ (measured clockwise from north) gives direction vector $(\sin\theta°, \cos\theta°)$. Multiply by the speed to get the velocity vector.

Example: A ship travels on bearing $060°$ at speed 20 knots. Its velocity vector is:

$20(\sin 60°, \cos 60°) = 20\!\left(\dfrac{\sqrt{3}}{2}, \dfrac{1}{2}\right) = (10\sqrt{3},\, 10)$ knots.

Relative velocity. If ship A has velocity $(3,4)$ and ship B has velocity $(1,-2)$, then the velocity of A relative to B is $(3,4)-(1,-2)=(2,6)$. This is what an observer on B would see A doing.

Bearing to component conversion: A bearing of $\theta°$ (measured clockwise from north) gives direction vector $(\sin\theta°, \cos\theta°)$. Multiply by the speed to get the velocity vector.

Pause — copy the bearing-to-component conversion: bearing $\theta^\circ\to\begin{pmatrix}\sin\theta\\\cos\theta\end{pmatrix}$; include a worked navigation example into your book.

Did you get this? True or false: A bearing of $090°$ (due east) corresponds to direction vector $(1, 0)$ when north is the positive $y$-axis.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · RESULTANT FORCE

Two forces act on a particle: $\mathbf{F}_1 = 10$ N at $30°$ above the positive $x$-axis, and $\mathbf{F}_2 = 8$ N at $120°$ above the positive $x$-axis. Find the magnitude and direction of the resultant.

Resolve: $\mathbf{F}_1 = (10\cos30°, 10\sin30°) = (5\sqrt{3},\, 5)$. $\mathbf{F}_2 = (8\cos120°, 8\sin120°) = (-4,\, 4\sqrt{3})$.

Convert each force to component form using $F\cos\theta$ and $F\sin\theta$.

PROBLEM 2 · NAVIGATION

A boat travels on bearing $030°$ at 12 km/h in still water. A current flows due east at 3 km/h. Find the boat's actual speed and bearing.

Boat velocity: $12(\sin30°, \cos30°) = (6,\, 6\sqrt{3})$. Current: $(3, 0)$. Ground velocity: $(6+3,\, 6\sqrt{3}) = (9,\, 6\sqrt{3})$.

Convert bearing to components and add the current vector.

PROBLEM 3 · WORK DONE

A force $\mathbf{F} = (4, 3)$ N acts on a particle that moves with displacement $\mathbf{d} = (6, -2)$ m. Find (a) the work done, and (b) the angle between $\mathbf{F}$ and $\mathbf{d}$.

(a) $W = \mathbf{F}\cdot\mathbf{d} = (4)(6)+(3)(-2) = 24-6 = 18$ J.

Work is the dot product of force and displacement — a scalar quantity.

Fill the gap: A particle is in equilibrium when the sum of all forces equals (the zero vector), which means both the $x$ and $y$ components of the net force equal .

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Adding magnitudes, not vectors

You cannot find resultant speed by adding the individual speeds: $|\mathbf{v}_1|+|\mathbf{v}_2| \neq |\mathbf{v}_1+\mathbf{v}_2|$ in general. Always add the component vectors first, then find the magnitude of the sum.

Trap 02

Mixing up bearing and standard angle

Standard angles ($\theta$) are measured anticlockwise from the positive $x$-axis (east). Bearings are measured clockwise from north. Bearing $030°$ is NOT the same as $30°$ in standard form — always convert explicitly.

Trap 03

Forgetting the angle quadrant

$\tan^{-1}$ only returns angles in $(-90°, 90°)$. If $R_x < 0$, the resultant points into the second or third quadrant — add $180°$ to the $\tan^{-1}$ result. Sketch the component diagram to confirm the correct quadrant.

Did you get this? True or false: If $|\mathbf{F}_1|=5$ N and $|\mathbf{F}_2|=3$ N, the magnitude of the resultant must equal 8 N.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

A boat heads due north at 10 km/h and a current flows due east at 4 km/h. Find the actual speed and bearing of the boat.

Forces $\mathbf{F}_1 = (5, 0)$ N and $\mathbf{F}_2 = (0, -12)$ N act on a particle. Find the resultant magnitude and the angle it makes with the positive $x$-axis.

Find the work done by $\mathbf{F} = (2, 5)$ N as a particle moves from $A(1,1)$ to $B(4,3)$.

A particle is in equilibrium under forces $\mathbf{F}_1 = (3, k)$ N and $\mathbf{F}_2 = (-3, 5)$ N. Find $k$.

Find the angle between $\mathbf{a} = (1, \sqrt{3})$ and $\mathbf{b} = (\sqrt{3}, 1)$ using the dot product formula.

Odd one out: Three of these statements are correct. Which one is FALSE?

Revisit your thinking

Earlier you estimated the plane's ground speed when airspeed was $(300,0)$ and wind was $(0,50)$.

The ground velocity is $(300,50)$. Speed $= \sqrt{300^2+50^2}=\sqrt{92500}\approx \mathbf{304.1}$ km/h — only slightly faster than airspeed because the wind is perpendicular. The direction is $\tan^{-1}(50/300)\approx 9.5°$ north of east, giving bearing $\approx 080.5°$. Key insight: perpendicular vectors add by Pythagoras; small cross-components barely change speed but noticeably shift direction.

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Multiple choice

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Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 32 marks

Q1. A plane flies on bearing $045°$ at 200 km/h in still air. A wind blows from the south (i.e. in the direction north) at 30 km/h. Find the actual ground speed and bearing of the plane. (2 marks)

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ApplyBand 43 marks

Q2. A force $\mathbf{F} = (6, 8)$ N acts on a particle that moves from $P(2, 3)$ to $Q(7, 6)$. Find (a) the work done, and (b) the angle between $\mathbf{F}$ and $\overrightarrow{PQ}$. (3 marks)

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AnalyseBand 53 marks

Q3. A particle is in equilibrium under three forces: $\mathbf{F}_1 = (3, 2)$ N, $\mathbf{F}_2 = (-1, k)$ N, and $\mathbf{F}_3 = (m, -5)$ N. Find $k$ and $m$. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:
1. Ground velocity $(4,10)$. Speed $=\sqrt{16+100}=\sqrt{116}=2\sqrt{29}\approx 10.77$ km/h. Bearing $=\tan^{-1}(4/10)\approx 21.8°$ east of north $\approx$ bearing $022°$.
2. $\mathbf{R}=(5,-12)$. $|\mathbf{R}|=\sqrt{25+144}=\sqrt{169}=13$ N. $\theta=\tan^{-1}(-12/5)\approx -67.4°$ (i.e. $67.4°$ below the positive $x$-axis).
3. $\mathbf{d}=(3,2)$. $W=(2)(3)+(5)(2)=6+10=16$ J.
4. $y$: $k+5=0\Rightarrow k=-5$.
5. $\mathbf{a}\cdot\mathbf{b}=\sqrt{3}+\sqrt{3}=2\sqrt{3}$. $|\mathbf{a}|=|\mathbf{b}|=2$. $\cos\theta=\tfrac{2\sqrt{3}}{4}=\tfrac{\sqrt{3}}{2}\Rightarrow\theta=30°$.

Q1 (2 marks): Plane: $200(\tfrac{\sqrt{2}}{2},\tfrac{\sqrt{2}}{2})=(100\sqrt{2},100\sqrt{2})$. Wind: $(0,30)$. Ground velocity: $(100\sqrt{2},100\sqrt{2}+30)$ [1]. Speed $=\sqrt{20000+(100\sqrt{2}+30)^2}$. $100\sqrt{2}\approx141.4$, so $R_y\approx171.4$. Speed $\approx\sqrt{20000+29378}\approx\sqrt{49378}\approx222.2$ km/h. Bearing: $\tan^{-1}(141.4/171.4)\approx\tan^{-1}(0.825)\approx39.6°$, bearing $\approx040°$ [1].

Q2 (3 marks): $\overrightarrow{PQ}=(5,3)$ [1]. $W=(6)(5)+(8)(3)=30+24=54$ J [1]. $|\mathbf{F}|=\sqrt{36+64}=10$, $|\overrightarrow{PQ}|=\sqrt{25+9}=\sqrt{34}$. $\cos\theta=\dfrac{54}{10\sqrt{34}}\Rightarrow\theta=\cos^{-1}\!\left(\dfrac{54}{10\sqrt{34}}\right)\approx\cos^{-1}(0.926)\approx22.2°$ [1].

Q3 (3 marks): $x$: $3-1+m=0\Rightarrow m=-2$ [1]. $y$: $2+k-5=0\Rightarrow k=3$ [1]. Verify: $\mathbf{F}_1+\mathbf{F}_2+\mathbf{F}_3=(3-1-2,2+3-5)=(0,0)$ ✓ [1].

Boss battle · The Vector Master

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Science Jump · platform challenge

Climb platforms by answering vector application questions. Lighter alternative to the boss.

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Module overview · Extension 1 · Checkpoint 4