Applications of Vectors

Pilots, sailors, and engineers use vector addition every day. A plane flying into a crosswind doesn’t travel the way its nose points, and a bridge cable has to balance the forces pulling on it. On this page you’ll solve these problems with a vector diagram and the cosine law and sine law, using the tip-to-tail methods from adding and subtracting vectors. All angles are in degrees.

Key ideas

Resultant and equilibrant forces

When several forces act on the same object, their vector sum is the resultant force: the single force that would have the same effect.

The equilibrant is the force that exactly balances the resultant, keeping the object still (or moving at a steady velocity). It has the same magnitude as the resultant and the opposite direction:

$$\vec{E} = -\vec{R}$$

An object is in equilibrium when all the forces on it add to $\vec{0}$. Drawn tip to tail, forces in equilibrium form a closed shape: the last arrow ends where the first one began. Three forces in equilibrium form a triangle, which you can solve with the sine and cosine laws.

Forces are measured in newtons (N). The weight of a mass of $m$ kilograms is about $9.8m$ N, pointing straight down.

Navigation: air speed and ground speed

A plane moves through the air, and the air itself moves (the wind). Its velocity relative to the ground is the sum:

$$\vec{v}_{\text{ground}} = \vec{v}_{\text{air}} + \vec{w}$$

Term	Meaning
heading	the direction the plane’s nose points (the direction of $\vec{v}_{\text{air}}$)
air speed	the plane’s speed relative to the air, $\lvert\vec{v}_{\text{air}}\rvert$
track (course)	the direction the plane actually travels over the ground
ground speed	the plane’s speed relative to the ground, $\lvert\vec{v}_{\text{ground}}\rvert$

The same idea works for boats and swimmers: the velocity relative to the shore is the velocity relative to the water plus the velocity of the current.

Careful with wind directions. Winds are named for where they come from. A “west wind” or a “wind from the west” blows toward the east. Currents are described by where they flow to.

A strategy for vector word problems

1. Draw a clear sketch with a north line at each vertex where you need to measure a bearing.
2. Draw the given vectors tip to tail. In navigation, draw $\vec{v}_{\text{air}}$ first, then $\vec{w}$ at its tip; the ground velocity joins the start to the end.
3. Work out the angle inside the triangle from the bearings.
4. Use the cosine law for a missing side and the sine law for a missing angle. If there’s a right angle, use the Pythagorean theorem and basic trig instead.
5. Answer the question asked, with a direction (bearing or angle) and units.

You can also solve all of these problems by breaking each vector into horizontal and vertical components and adding the components. That method is on the next page, Cartesian vectors, and it’s a great way to check your answers.

Worked examples

Example 1: Crossing a river

A boat that moves at $4$ m/s in still water points straight across a river, while the current flows downstream at $3$ m/s. The river is $120$ m wide.

• (a) Find the boat’s resultant velocity.
• (b) How long does it take to cross, and how far downstream does it land?

Solution.

(a) The boat’s velocity relative to the water (straight across) and the current (downstream) are at right angles, so

$$\lvert\vec{v}\rvert = \sqrt{4^2 + 3^2} = 5 \text{ m/s}$$

Let $\theta$ be the angle between the resultant and the straight-across direction: $\tan\theta = \dfrac{3}{4}$, so $\theta \approx 36.9^\circ$. The boat moves at $5$ m/s, angled about $36.9^\circ$ downstream from straight across.

(b) The current doesn’t help or hinder the crossing itself: only the $4$ m/s straight-across part does. So the crossing time is

$$t = \frac{120 \text{ m}}{4 \text{ m/s}} = 30 \text{ s}$$

In that time, the current carries the boat $3 \times 30 = 90$ m downstream. (Check: the boat travels $5 \times 30 = 150$ m in total, and $\sqrt{120^2 + 90^2} = 150$. ✓)

Example 2: Resultant and equilibrant of two forces

Two forces act on a crate: $60$ N and $40$ N, with an angle of $70^\circ$ between them. Find the resultant and the equilibrant, giving the directions relative to the $60$ N force.

The resultant $\vec{R}$ is the diagonal of the parallelogram formed by the two forces.

Solution. In the parallelogram (or the tip-to-tail triangle), the angle opposite $\vec{R}$ is $180^\circ - 70^\circ = 110^\circ$.

Magnitude, by the cosine law.

$$\begin{aligned} \lvert\vec{R}\rvert^2 &= 60^2 + 40^2 - 2(60)(40)\cos 110^\circ \\ &\approx 6841.70 \end{aligned}$$

So $\lvert\vec{R}\rvert \approx 82.71$ N.

Direction, by the sine law. Let $\alpha$ be the angle between $\vec{R}$ and the $60$ N force. In the triangle, it’s opposite the $40$ N side:

$$\sin\alpha = \frac{40\sin 110^\circ}{82.715} \approx 0.4544 \quad\Rightarrow\quad \alpha \approx 27.0^\circ$$

The resultant is about $82.71$ N at $27.0^\circ$ from the $60$ N force, toward the $40$ N force.

Equilibrant. It has the same magnitude and the opposite direction: about $82.71$ N, pointing at $180^\circ - 27.0^\circ = 153.0^\circ$ from the $60$ N force, on the other side of the $60$ N force from the $40$ N one.

Example 3: A plane in a crosswind

A plane has an air speed of $450$ km/h on a heading of N $40^\circ$ E. A wind of $60$ km/h blows from the west. Find the plane’s ground speed and track, to one decimal place.

The wind arrow is drawn longer than scale so it’s easy to see.

Solution. A wind from the west blows toward the east (a bearing of $090^\circ$). Draw the air velocity, then the wind at its tip.

Angle inside the triangle. At the tip of the air velocity, the direction back to the start is $040^\circ + 180^\circ = 220^\circ$, and the wind points along $090^\circ$. The angle between them is $220^\circ - 90^\circ = 130^\circ$.

Ground speed, by the cosine law.

$$\begin{aligned} \lvert\vec{v}_{\text{ground}}\rvert^2 &= 450^2 + 60^2 - 2(450)(60)\cos 130^\circ \\ &\approx 240\,810.53 \end{aligned}$$

So the ground speed is about $490.7$ km/h.

Track, by the sine law. Let $\alpha$ be the angle at the start between the heading and the track. It’s opposite the wind side:

$$\sin\alpha = \frac{60\sin 130^\circ}{490.72} \approx 0.0937 \quad\Rightarrow\quad \alpha \approx 5.4^\circ$$

The wind pushes the plane toward the east, so the track is clockwise from the heading: $40^\circ + 5.4^\circ = 45.4^\circ$.

The plane travels at about $490.7$ km/h on a track of N $45.4^\circ$ E (a bearing of $045.4^\circ$). The tailwind part of the wind makes the ground speed a little more than the air speed.

Example 4: Choosing a heading

A pilot wants to fly due north to a town $1000$ km away. The plane’s air speed is $500$ km/h, and there’s a wind of $80$ km/h from the east. What heading should the pilot fly, what is the ground speed, and how long will the trip take?

Solution. The wind blows toward the west. To end up going due north, the pilot has to point the plane a little east of north, so that the eastward part of the air velocity cancels the wind.

Draw the triangle: the ground velocity (unknown length) points north, the wind ($80$ km/h) points west, and $\vec{v}_{\text{ground}} = \vec{v}_{\text{air}} + \vec{w}$. The wind is perpendicular to the track, so the triangle has a right angle, with the air velocity ($500$ km/h) as the hypotenuse.

Let $\theta$ be the angle between the heading and north. The wind side is opposite $\theta$:

$$\sin\theta = \frac{80}{500} = 0.16 \quad\Rightarrow\quad \theta \approx 9.2^\circ$$

The heading is N $9.2^\circ$ E (a bearing of about $009.2^\circ$). The ground speed is the remaining leg:

$$\lvert\vec{v}_{\text{ground}}\rvert = \sqrt{500^2 - 80^2} = \sqrt{243\,600} \approx 493.6 \text{ km/h}$$

The trip takes about $\dfrac{1000}{493.56} \approx 2.03$ h, which is about $2$ h $2$ min.

Common mistakes

Getting the wind direction backwards. A wind “from the east” blows west. Draw the arrow pointing the way the air moves, which is away from the direction named.

Confusing heading with track (or air speed with ground speed). The heading is where the plane points; the track is where it actually goes. The given “air speed” belongs on the heading vector, not on the resultant.

Using the angle between the bearings instead of the angle inside the triangle. In Example 3, the bearings differ by $50^\circ$, but the angle in the triangle is $130^\circ$. A sketch with a north line at each vertex makes this clear.

Forgetting that the equilibrant is opposite the resultant. The equilibrant has the same size as the resultant but points the other way. Give its direction explicitly; don’t just repeat the resultant.

Adding the angle from the sine law the wrong way. After finding the angle between the heading and the track, decide from the diagram whether the wind turns the track clockwise (add to the bearing) or counterclockwise (subtract).

Thinking a current slows down a crossing. If a boat points straight across, the current changes where it lands, not how long the crossing takes. If it points upstream to land straight across, then the crossing is slower, because only part of its velocity is going across (practice question 6).

Practice

1. (Warm-up) Two forces act along the same line: $120$ N and $80$ N.

• (a) Find the resultant if they act in the same direction.
• (b) Find the resultant and the equilibrant if they act in opposite directions.

Solution

(a) $120 + 80 = 200$ N, in the common direction.

(b) $120 - 80 = 40$ N, in the direction of the $120$ N force. The equilibrant is $40$ N in the direction of the $80$ N force.

2. (Warm-up) A swimmer can swim at $1.5$ m/s in still water, in a river whose current is $0.5$ m/s.

• (a) What is her speed relative to the shore when she swims downstream? Upstream?
• (b) In which direction does a “north wind” blow?

Solution

(a) Downstream the velocities point the same way: $1.5 + 0.5 = 2$ m/s. Upstream they’re opposite: $1.5 - 0.5 = 1$ m/s.

(b) Winds are named for where they come from, so a north wind blows toward the south.

3. (Core) Forces of $25$ N east and $60$ N north act on an object. Find the resultant and the equilibrant, with directions as quadrant bearings to one decimal place.

Solution

The forces are perpendicular, so $\lvert\vec{R}\rvert = \sqrt{25^2 + 60^2} = \sqrt{4225} = 65$ N.

The angle $\theta$ between north and $\vec{R}$ satisfies $\tan\theta = \dfrac{25}{60}$, so $\theta \approx 22.6^\circ$. The resultant is $65$ N at N $22.6^\circ$ E.

The equilibrant is $65$ N at S $22.6^\circ$ W.

4. (Core) Forces of $150$ N and $200$ N act on a point with $45^\circ$ between them. Find the magnitude of the resultant to the nearest newton, and the angle between the resultant and the $200$ N force to one decimal place.

Solution

The angle inside the triangle is $180^\circ - 45^\circ = 135^\circ$:

$$\lvert\vec{R}\rvert^2 = 150^2 + 200^2 - 2(150)(200)\cos 135^\circ \approx 104\,926.4$$

So $\lvert\vec{R}\rvert \approx 323.9$, which is $324$ N to the nearest newton.

The angle $\alpha$ between $\vec{R}$ and the $200$ N force is opposite the $150$ N side:

$$\sin\alpha = \frac{150\sin 135^\circ}{323.92} \approx 0.3274 \quad\Rightarrow\quad \alpha \approx 19.1^\circ$$

5. (Core) A plane has an air speed of $300$ km/h on a heading of $120^\circ$. A $50$ km/h wind blows from the south. Find the ground speed and the track, to one decimal place.

Solution

A wind from the south blows north (bearing $000^\circ$). At the tip of the air velocity, the direction back to the start is $120^\circ + 180^\circ = 300^\circ$; the wind points along $360^\circ$. The angle inside the triangle is $360^\circ - 300^\circ = 60^\circ$.

$$\lvert\vec{v}_{\text{ground}}\rvert^2 = 300^2 + 50^2 - 2(300)(50)\cos 60^\circ = 90\,000 + 2500 - 15\,000 = 77\,500$$

The ground speed is $\sqrt{77\,500} \approx 278.4$ km/h.

The angle $\alpha$ between the heading and the track is opposite the wind side:

$$\sin\alpha = \frac{50\sin 60^\circ}{278.39} \approx 0.1555 \quad\Rightarrow\quad \alpha \approx 8.9^\circ$$

The wind pushes the plane north, which turns the track counterclockwise from $120^\circ$: the track is $120^\circ - 8.9^\circ = 111.1^\circ$.

6. (Core) A river is $200$ m wide and its current flows at $1.2$ m/s. A boat moves at $2$ m/s in still water. The driver wants to land directly across from the starting point.

• (a) At what angle upstream from straight across should the boat point?
• (b) How long does the crossing take?

Solution

(a) The resultant must point straight across, so the upstream part of the boat’s velocity has to cancel the current. In the right triangle, the boat’s $2$ m/s is the hypotenuse and the current’s $1.2$ m/s is opposite the angle $\theta$ upstream:

$$\sin\theta = \frac{1.2}{2} = 0.6 \quad\Rightarrow\quad \theta \approx 36.9^\circ$$

(b) The speed straight across is $\sqrt{2^2 - 1.2^2} = \sqrt{2.56} = 1.6$ m/s, so the time is $\dfrac{200}{1.6} = 125$ s.

7. (Core) A $50$ kg crate sits on the floor, and two people pull on it: one with $300$ N east, the other with $200$ N at N $30^\circ$ E (both horizontal). Find the magnitude of the resultant horizontal force, to the nearest newton.

Solution

The angle between the forces (tail to tail) is $90^\circ - 30^\circ = 60^\circ$ (east is a bearing of $90^\circ$ and the second force is at $30^\circ$). The angle inside the tip-to-tail triangle is $180^\circ - 60^\circ = 120^\circ$:

$$\lvert\vec{R}\rvert^2 = 300^2 + 200^2 - 2(300)(200)\cos 120^\circ = 130\,000 + 60\,000 = 190\,000$$

So $\lvert\vec{R}\rvert = \sqrt{190\,000} \approx 436$ N. (The crate’s mass and weight don’t affect the horizontal resultant.)

8. (Challenge) A $100$ N sign hangs from two ropes. The left rope makes an angle of $40^\circ$ with the horizontal and the right rope makes $60^\circ$ with the horizontal. Find the tension (force) in each rope, to one decimal place.

Solution

The sign is in equilibrium, so the weight ($100$ N down) and the two tensions $\vec{T}_1$ (left rope) and $\vec{T}_2$ (right rope) form a closed triangle when drawn tip to tail.

Angles between the forces (tail to tail): $\vec{T}_1$ and $\vec{T}_2$ make $180^\circ - 40^\circ - 60^\circ = 80^\circ$; $\vec{T}_1$ and the weight make $90^\circ + 40^\circ = 130^\circ$; $\vec{T}_2$ and the weight make $90^\circ + 60^\circ = 150^\circ$. In the closed triangle, each interior angle is $180^\circ$ minus one of these, giving $100^\circ$, $50^\circ$ and $30^\circ$ (which add to $180^\circ$ ✓).

Each force is opposite the interior angle made by the other two: the weight is opposite $100^\circ$, $\vec{T}_1$ is opposite $30^\circ$ and $\vec{T}_2$ is opposite $50^\circ$. By the sine law,

$$\frac{\lvert\vec{T}_1\rvert}{\sin 30^\circ} = \frac{\lvert\vec{T}_2\rvert}{\sin 50^\circ} = \frac{100}{\sin 100^\circ}$$$$\lvert\vec{T}_1\rvert = \frac{100\sin 30^\circ}{\sin 100^\circ} \approx 50.8 \text{ N}, \qquad \lvert\vec{T}_2\rvert = \frac{100\sin 50^\circ}{\sin 100^\circ} \approx 77.8 \text{ N}$$

Check: the steeper rope carries more of the load, as you’d expect.

9. (Challenge) A pilot must fly to a city $600$ km away on a bearing of $070^\circ$. The plane’s air speed is $400$ km/h, and a $75$ km/h wind blows from a bearing of $340^\circ$. Find the heading the pilot should fly, the ground speed, and the flight time, to one decimal place where needed.

Solution

A wind from $340^\circ$ blows toward $340^\circ - 180^\circ = 160^\circ$. The track is $070^\circ$, and $160^\circ - 70^\circ = 90^\circ$, so the wind is perpendicular to the track: it pushes the plane to the right of its track.

In the triangle $\vec{v}_{\text{ground}} = \vec{v}_{\text{air}} + \vec{w}$, the right angle is between the track and the wind, so the air velocity ($400$ km/h) is the hypotenuse. Let $\theta$ be the angle between the heading and the track:

$$\sin\theta = \frac{75}{400} = 0.1875 \quad\Rightarrow\quad \theta \approx 10.8^\circ$$

To cancel a wind pushing right, the pilot points left of the track (counterclockwise): heading $70^\circ - 10.8^\circ = 59.2^\circ$, a bearing of $059.2^\circ$.

Ground speed: $\sqrt{400^2 - 75^2} = \sqrt{154\,375} \approx 392.9$ km/h.

Time: $\dfrac{600}{392.91} \approx 1.53$ h, which is about $1$ h $32$ min.