Adding and Subtracting Vectors

If you walk $3$ km north and then $4$ km east, you end up $5$ km from where you started, not $7$. Vectors don’t add like ordinary numbers, because their directions matter. On this page you’ll learn how to add and subtract vectors by drawing them, and how to calculate the size and direction of the result with the cosine law and sine law. All angles are in degrees.

Key ideas

Adding tip to tail (the triangle law)

To find $\vec{a} + \vec{b}$, draw $\vec{a}$, then draw $\vec{b}$ starting at the tip (head) of $\vec{a}$. The sum goes from the tail of $\vec{a}$ to the tip of $\vec{b}$. The sum of two or more vectors is called their resultant.

In terms of points, this says that going from $A$ to $B$ and then from $B$ to $C$ has the same effect as going straight from $A$ to $C$:

$$\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$$

To add more than two vectors, keep going: place each new vector at the tip of the last one, and the resultant runs from the very first tail to the very last tip.

The parallelogram law

You can also draw $\vec{a}$ and $\vec{b}$ from the same starting point and complete the parallelogram. The diagonal from that starting point is $\vec{a} + \vec{b}$. It’s the same answer as tip to tail, because the opposite side of the parallelogram is an equal copy of $\vec{b}$.

Two ways to draw the same sum $\vec{a} + \vec{b}$.

The zero vector and subtraction

The zero vector $\vec{0}$ has magnitude $0$ and no particular direction. Adding a vector and its opposite gets you back where you started:

$$\vec{a} + (-\vec{a}) = \vec{0}$$

To subtract, add the opposite:

$$\vec{a} - \vec{b} = \vec{a} + (-\vec{b})$$

There’s a handy shortcut: if $\vec{a}$ and $\vec{b}$ are drawn from the same point, then $\vec{a} - \vec{b}$ goes from the tip of $\vec{b}$ to the tip of $\vec{a}$. (Check: $\vec{b} + (\vec{a} - \vec{b}) = \vec{a}$, which is tip to tail.) So the two diagonals of the parallelogram are $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$.

With a common tail, $\vec{a} - \vec{b}$ points from the tip of $\vec{b}$ to the tip of $\vec{a}$.

A useful special case: for any points $O$, $A$ and $B$,

$$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$$

Properties of addition

These all follow from tip-to-tail diagrams, and they mean you can rearrange and regroup vector sums just like sums of numbers.

Property	Statement
Commutative	$\vec{a} + \vec{b} = \vec{b} + \vec{a}$
Associative	$(\vec{a} + \vec{b}) + \vec{c} = \vec{a} + (\vec{b} + \vec{c})$
Zero vector (identity)	$\vec{a} + \vec{0} = \vec{a}$
Opposite (inverse)	$\vec{a} + (-\vec{a}) = \vec{0}$

The parallelogram shows the commutative property: going along $\vec{a}$ then $\vec{b}$ (bottom then right side) reaches the same corner as $\vec{b}$ then $\vec{a}$ (left then top side). Subtraction, like ordinary subtraction, is not commutative: $\vec{b} - \vec{a} = -(\vec{a} - \vec{b})$.

Finding the magnitude and direction of a resultant

1. Sketch the vectors tip to tail and draw the resultant.
2. Find the angle inside the triangle between the two given vectors. If the angle between $\vec{a}$ and $\vec{b}$ (drawn tail to tail) is $\phi$, the angle inside the tip-to-tail triangle is $180^\circ - \phi$.
3. Use the cosine law to find the magnitude of the resultant.
4. Use the sine law to find an angle in the triangle, then turn it into a direction (for example, a bearing).

If the two vectors are perpendicular, the triangle has a right angle, and the Pythagorean theorem and $\tan$ are all you need.

The length of a resultant is never more than the sum of the lengths: $\lvert\vec{a} + \vec{b}\rvert \le \lvert\vec{a}\rvert + \lvert\vec{b}\rvert$ (the triangle inequality). It’s equal only when $\vec{a}$ and $\vec{b}$ point the same way.

Worked examples

Example 1: Simplifying with a parallelogram

$ABCD$ is a parallelogram (vertices in order). Write each expression as a single vector.

• (a) $\overrightarrow{AB} + \overrightarrow{BC}$
• (b) $\overrightarrow{AB} + \overrightarrow{AD}$
• (c) $\overrightarrow{AB} - \overrightarrow{AD}$
• (d) $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD}$

Solution.

(a) Tip to tail: $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$.

(b) In a parallelogram $\overrightarrow{AD} = \overrightarrow{BC}$, so $\overrightarrow{AB} + \overrightarrow{AD} = \overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$. (This is the parallelogram law: $\overrightarrow{AC}$ is the diagonal from $A$.)

(c) Add the opposite: $-\overrightarrow{AD} = \overrightarrow{DA}$. Then

$$\overrightarrow{AB} - \overrightarrow{AD} = \overrightarrow{AB} + \overrightarrow{DA} = \overrightarrow{DA} + \overrightarrow{AB} = \overrightarrow{DB}$$

That’s the other diagonal, from the tip of $\overrightarrow{AD}$ to the tip of $\overrightarrow{AB}$, as the shortcut says.

(d) Chain the tips and tails: $A \to B \to C \to D$, so the sum is $\overrightarrow{AD}$.

Example 2: Perpendicular vectors

A delivery robot moves $5$ m east and then $12$ m north. Find the resultant displacement.

Solution. Tip to tail, the two legs form a right angle, so

$$\lvert\vec{R}\rvert = \sqrt{5^2 + 12^2} = \sqrt{169} = 13 \text{ m}$$

At the start, let $\theta$ be the angle between north and $\vec{R}$. The east leg is opposite $\theta$ and the north leg is adjacent:

$$\tan\theta = \frac{5}{12} \quad\Rightarrow\quad \theta \approx 22.6^\circ$$

The resultant is $13$ m at N $22.6^\circ$ E (a true bearing of $022.6^\circ$), to one decimal place.

Example 3: Using the cosine law and sine law

A boat sails $8$ km on a bearing of $040^\circ$, then $5$ km on a bearing of $110^\circ$. Find its resultant displacement, to two decimal places for the distance and one decimal place for the bearing.

Solution. Let $O$ be the start, $P$ the turning point and $Q$ the finish. Sketch the trip tip to tail.

Angle inside the triangle at $P$. The boat turns from $040^\circ$ to $110^\circ$, a turn of $70^\circ$. The angle between the two vectors drawn tail to tail is $70^\circ$, so the angle inside the triangle is $\angle OPQ = 180^\circ - 70^\circ = 110^\circ$.

(Another way to see it: at $P$, the direction back to $O$ is the bearing $040^\circ + 180^\circ = 220^\circ$, and the direction to $Q$ is $110^\circ$. The difference is $220^\circ - 110^\circ = 110^\circ$.)

Magnitude, by the cosine law.

$$\begin{aligned} \lvert\vec{R}\rvert^2 &= 8^2 + 5^2 - 2(8)(5)\cos 110^\circ \\ &= 89 - 80\cos 110^\circ \\ &\approx 116.362 \end{aligned}$$

So $\lvert\vec{R}\rvert \approx 10.787$, about $10.79$ km.

Direction, by the sine law. Let $\alpha = \angle POQ$, the angle at the start between the first leg and the resultant. It’s opposite the $5$ km side:

$$\frac{\sin\alpha}{5} = \frac{\sin 110^\circ}{10.787} \quad\Rightarrow\quad \sin\alpha \approx 0.4356 \quad\Rightarrow\quad \alpha \approx 25.8^\circ$$

($\alpha$ must be acute, because it’s opposite the shortest side, so there’s no ambiguous case here.) The resultant is turned $25.8^\circ$ clockwise from the first leg, toward the second leg, so its bearing is $40^\circ + 25.8^\circ = 65.8^\circ$.

The boat ends up about $10.79$ km from its start on a bearing of $065.8^\circ$.

Example 4: A sum and a difference

$\lvert\vec{u}\rvert = 10$ and $\lvert\vec{v}\rvert = 6$, and the angle between $\vec{u}$ and $\vec{v}$ (tail to tail) is $60^\circ$. Find $\lvert\vec{u} + \vec{v}\rvert$ and $\lvert\vec{u} - \vec{v}\rvert$.

Solution. Use the parallelogram: $\vec{u} + \vec{v}$ and $\vec{u} - \vec{v}$ are its two diagonals.

Difference. $\vec{u} - \vec{v}$ joins the tips of $\vec{u}$ and $\vec{v}$ drawn from the same point, so it’s the third side of a triangle with sides $10$ and $6$ and the $60^\circ$ angle between them:

$$\lvert\vec{u} - \vec{v}\rvert^2 = 10^2 + 6^2 - 2(10)(6)\cos 60^\circ = 136 - 60 = 76$$

So $\lvert\vec{u} - \vec{v}\rvert = \sqrt{76} = 2\sqrt{19} \approx 8.72$.

Sum. Tip to tail, the angle inside the triangle is $180^\circ - 60^\circ = 120^\circ$:

$$\lvert\vec{u} + \vec{v}\rvert^2 = 10^2 + 6^2 - 2(10)(6)\cos 120^\circ = 136 + 60 = 196$$

So $\lvert\vec{u} + \vec{v}\rvert = 14$.

Check: both answers are between $10 - 6 = 4$ and $10 + 6 = 16$, as the triangle inequality requires. The sum is longer than the difference because the angle between the vectors is less than $90^\circ$.

Common mistakes

Adding the magnitudes. $\lvert\vec{a} + \vec{b}\rvert$ is usually not $\lvert\vec{a}\rvert + \lvert\vec{b}\rvert$. That only happens when the vectors point the same way. Always sketch and use the triangle.

Using the angle between the vectors instead of the angle inside the triangle. If $\vec{a}$ and $\vec{b}$ make an angle $\phi$ tail to tail, the tip-to-tail triangle has $180^\circ - \phi$ between them. In Example 3, using $70^\circ$ instead of $110^\circ$ in the cosine law would give about $7.85$ km instead of $10.79$ km.

Drawing vectors tail to tail and joining the tips for a sum. Joining the tips gives the difference $\vec{a} - \vec{b}$, not the sum. For a sum, go tip to tail (or use the diagonal of the parallelogram that starts at the common tail).

Getting the direction of a difference backwards. With a common tail, $\vec{a} - \vec{b}$ points toward the tip of $\vec{a}$ (the first vector). Check with $\vec{b} + (\vec{a} - \vec{b}) = \vec{a}$.

Stopping at an angle inside the triangle. The sine law gives an angle in the triangle, not a bearing. Use your sketch to decide whether to add it to or subtract it from a known direction, and say what it’s measured from.

Rounding too early. Keep the unrounded magnitude from the cosine law in your calculator when you use the sine law, and round only the final answers.

Practice

1. (Warm-up) Write each expression as a single vector.

• (a) $\overrightarrow{PQ} + \overrightarrow{QR}$
• (b) $\overrightarrow{PQ} + \overrightarrow{QR} + \overrightarrow{RP}$
• (c) $\overrightarrow{PQ} - \overrightarrow{RQ}$

Solution

(a) $\overrightarrow{PR}$ (tip to tail).

(b) $\overrightarrow{PR} + \overrightarrow{RP} = \overrightarrow{PP} = \vec{0}$: the trip ends where it started.

(c) $-\overrightarrow{RQ} = \overrightarrow{QR}$, so $\overrightarrow{PQ} - \overrightarrow{RQ} = \overrightarrow{PQ} + \overrightarrow{QR} = \overrightarrow{PR}$.

2. (Warm-up) $\vec{a}$ is $40$ km/h north and $\vec{b}$ is $15$ km/h south. Find $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$.

Solution

The vectors are parallel, so the triangle flattens into a line.

$\vec{a} + \vec{b}$: $40$ north and $15$ south partly cancel, leaving $25$ km/h north.

$\vec{a} - \vec{b} = \vec{a} + (-\vec{b})$, and $-\vec{b}$ is $15$ km/h north, so $\vec{a} - \vec{b}$ is $55$ km/h north.

3. (Core) A kayak’s displacement is $9$ m east followed by $12$ m south. Find the resultant displacement, with the direction as a quadrant bearing and a true bearing to one decimal place.

Solution

Magnitude: $\sqrt{9^2 + 12^2} = \sqrt{225} = 15$ m.

At the start, the angle $\theta$ between south and the resultant has the $9$ m east leg opposite it and the $12$ m south leg adjacent:

$$\tan\theta = \frac{9}{12} \quad\Rightarrow\quad \theta \approx 36.9^\circ$$

The resultant is $15$ m at S $36.9^\circ$ E, a true bearing of $180^\circ - 36.9^\circ = 143.1^\circ$.

4. (Core) $\lvert\vec{a}\rvert = 7$, $\lvert\vec{b}\rvert = 4$, and the angle between $\vec{a}$ and $\vec{b}$ is $50^\circ$. Find $\lvert\vec{a} + \vec{b}\rvert$ and the angle between $\vec{a} + \vec{b}$ and $\vec{a}$, each to one decimal place.

Solution

Tip to tail, the angle inside the triangle is $180^\circ - 50^\circ = 130^\circ$.

$$\lvert\vec{a} + \vec{b}\rvert^2 = 7^2 + 4^2 - 2(7)(4)\cos 130^\circ = 65 - 56\cos 130^\circ \approx 100.996$$

So $\lvert\vec{a} + \vec{b}\rvert \approx 10.05$, or $10.0$ to one decimal place.

The angle $\alpha$ between $\vec{a}$ and the resultant is opposite the side of length $4$:

$$\sin\alpha = \frac{4\sin 130^\circ}{10.050} \approx 0.3049 \quad\Rightarrow\quad \alpha \approx 17.8^\circ$$

5. (Core) For the same $\vec{a}$ and $\vec{b}$ as in question 4, find $\lvert\vec{a} - \vec{b}\rvert$ to two decimal places.

Solution

$\vec{a} - \vec{b}$ joins the tips of $\vec{a}$ and $\vec{b}$ drawn from the same point, so the included angle is $50^\circ$:

$$\lvert\vec{a} - \vec{b}\rvert^2 = 7^2 + 4^2 - 2(7)(4)\cos 50^\circ = 65 - 56\cos 50^\circ \approx 29.004$$

So $\lvert\vec{a} - \vec{b}\rvert \approx 5.39$.

6. (Core) Simplify $\overrightarrow{AB} + \overrightarrow{CD} + \overrightarrow{BC} + \overrightarrow{DE}$, and state which properties of addition you used.

Solution

Rearrange (commutative property) and regroup (associative property) so tips meet tails:

$$\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD} + \overrightarrow{DE} = \overrightarrow{AC} + \overrightarrow{CD} + \overrightarrow{DE} = \overrightarrow{AD} + \overrightarrow{DE} = \overrightarrow{AE}$$

7. (Core) A ferry travels $15$ km on a bearing of $070^\circ$, then $20$ km on a bearing of $160^\circ$. Find its resultant displacement, with the bearing to one decimal place.

Solution

The turn is $160^\circ - 70^\circ = 90^\circ$, so the triangle has a right angle at the turning point:

$$\lvert\vec{R}\rvert = \sqrt{15^2 + 20^2} = \sqrt{625} = 25 \text{ km}$$

At the start, the angle $\alpha$ between the first leg and the resultant satisfies

$$\tan\alpha = \frac{20}{15} \quad\Rightarrow\quad \alpha \approx 53.1^\circ$$

The resultant is turned clockwise from $070^\circ$ toward the second leg, so its bearing is $70^\circ + 53.1^\circ = 123.1^\circ$. The ferry ends up $25$ km away on a bearing of $123.1^\circ$.

8. (Challenge) $\lvert\vec{a}\rvert = 5$, $\lvert\vec{b}\rvert = 8$ and $\lvert\vec{a} + \vec{b}\rvert = 11$. Find the angle between $\vec{a}$ and $\vec{b}$, to one decimal place.

Solution

The tip-to-tail triangle has sides $5$, $8$ and $11$. Let $C$ be the angle inside the triangle between $\vec{a}$ and $\vec{b}$; it’s opposite the resultant:

$$\cos C = \frac{5^2 + 8^2 - 11^2}{2(5)(8)} = \frac{-32}{80} = -0.4 \quad\Rightarrow\quad C \approx 113.6^\circ$$

The angle between the vectors (tail to tail) is $180^\circ - 113.6^\circ = 66.4^\circ$.

9. (Challenge) $\lvert\vec{a}\rvert = 6$ and $\lvert\vec{b}\rvert = 10$, but their directions are unknown.

• (a) What are the largest and smallest possible values of $\lvert\vec{a} + \vec{b}\rvert$? When does each happen?
• (b) Explain why $\lvert\vec{a} + \vec{b}\rvert \le \lvert\vec{a}\rvert + \lvert\vec{b}\rvert$ for any two vectors.

Solution

(a) The largest value is $6 + 10 = 16$, when $\vec{a}$ and $\vec{b}$ point the same way. The smallest is $10 - 6 = 4$, when they point in opposite directions (the resultant then points the same way as $\vec{b}$).

(b) If $\vec{a}$ and $\vec{b}$ aren’t parallel, then $\vec{a}$, $\vec{b}$ and $\vec{a} + \vec{b}$ form a triangle, and any side of a triangle is shorter than the sum of the other two. If they’re parallel, the “triangle” is flat, and the resultant’s length is either the sum (same direction) or the difference (opposite directions), which is never more than the sum.