Sum and difference of two vectors | Grade 10

You walk 300 m east to the school gate and then 400 m north to the library. Where do you end up compared with where you started? Not 700 m away: you are only 500 m from your start, in a direction between east and north. Two displacements combined into one single displacement. That is exactly what adding vectors means, and it is how physicists add forces and how pilots plan for the wind.

Adding vectors

The triangle rule

Take two vectors $\vec{a}$ and $\vec{b}$ . Choose a point $A$ , draw $\overrightarrow{AB} = \vec{a}$ , and then from $B$ draw $\overrightarrow{BC} = \vec{b}$ . The vector from the start to the end, $\overrightarrow{AC}$ , is the sum.

The triangle rule:

\vec{a} + \vec{b}

Notice that the rule works for any three points, even when they lie on one line, and the point $B$ in the middle cancels: " $B$ followed by $B$ " disappears.

The parallelogram rule

If two vectors start from the same point $O$ , say $\overrightarrow{OA} = \vec{a}$ and $\overrightarrow{OD} = \vec{b}$ , complete the parallelogram $OACD$ . Its diagonal from $O$ is the sum.

This follows from the triangle rule: in a parallelogram $\overrightarrow{AC} = \overrightarrow{OD}$ , so $\overrightarrow{OA} + \overrightarrow{OD} = \overrightarrow{OA} + \overrightarrow{AC} = \overrightarrow{OC}$ .

The parallelogram rule

Properties of addition

For all vectors $\vec{a}$ , $\vec{b}$ , $\vec{c}$ :

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

The commutative law is the parallelogram rule in disguise, since $\overrightarrow{OA} + \overrightarrow{OD}$ and $\overrightarrow{OD} + \overrightarrow{OA}$ are both the diagonal $\overrightarrow{OC}$ . Because of associativity we can add many vectors without brackets, for example $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD} = \overrightarrow{AD}.$ The last rule comes from the triangle rule: $\overrightarrow{AB} + \overrightarrow{BA} = \overrightarrow{AA} = \vec{0}$ .

Example Closing a loop

For four points $A$ , $B$ , $C$ , $D$ , show that $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD} + \overrightarrow{DA} = \vec{0}$ , and simplify $\overrightarrow{AB} + \overrightarrow{CD} + \overrightarrow{BC} + \overrightarrow{DA}$ .

Solution.

By the triangle rule used twice, $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$ and then $\overrightarrow{AC} + \overrightarrow{CD} = \overrightarrow{AD}$ .
So $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD} + \overrightarrow{DA} = \overrightarrow{AD} + \overrightarrow{DA} = \overrightarrow{AA} = \vec{0}$ .
In the second sum the order is different. Addition is commutative, so we may reorder the terms: $\overrightarrow{AB} + \overrightarrow{CD} + \overrightarrow{BC} + \overrightarrow{DA} = \overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD} + \overrightarrow{DA} = \vec{0}$ .

A path that returns to its starting point has total displacement $\vec{0}$ .

Subtracting vectors

The most useful form of subtraction is the following. For any three points $O$ , $A$ , $B$ : $\overrightarrow{OA} - \overrightarrow{OB} = \overrightarrow{BA}.$ Indeed, $\overrightarrow{OA} - \overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{BO} = \overrightarrow{BO} + \overrightarrow{OA} = \overrightarrow{BA}$ . In words: the difference of two vectors with the same tail is the vector joining their heads, pointing to the head of the first one.

The difference

\vec{a} - \vec{b}

Example Subtracting in a parallelogram

$ABCD$ is a rectangle with $AB = 6$ and $AD = 8$ . Compute $|\overrightarrow{AB} + \overrightarrow{AD}|$ and $|\overrightarrow{AB} - \overrightarrow{AD}|$ .

Solution.

Since $ABCD$ is a parallelogram, the parallelogram rule gives $\overrightarrow{AB} + \overrightarrow{AD} = \overrightarrow{AC}$ .
The difference rule gives $\overrightarrow{AB} - \overrightarrow{AD} = \overrightarrow{DB}$ .
In a rectangle the diagonals are equal, and by Pythagoras $AC = DB = \sqrt{6^2 + 8^2} = 10$ .
Hence $|\overrightarrow{AB} + \overrightarrow{AD}| = 10$ and $|\overrightarrow{AB} - \overrightarrow{AD}| = 10$ .

The two vectors are different (they point in different directions) but have the same length.

Midpoints and centroids

Why does the first identity hold? If $M$ is the midpoint, then $\overrightarrow{MA}$ and $\overrightarrow{MB}$ have the same length and opposite directions, so they are opposite vectors and their sum is $\vec{0}$ . The converse is the same argument read backwards.

Example Proving the centroid identity

Let $G$ be the centroid of triangle $ABC$ and $M$ the midpoint of $BC$ . Prove that $\overrightarrow{GA} + \overrightarrow{GB} + \overrightarrow{GC} = \vec{0}$ .

Solution.

Since $M$ is the midpoint of $BC$ , $\overrightarrow{GB} + \overrightarrow{GC} = 2\overrightarrow{GM}$ . (Indeed $\overrightarrow{GB} = \overrightarrow{GM} + \overrightarrow{MB}$ and $\overrightarrow{GC} = \overrightarrow{GM} + \overrightarrow{MC}$ , and $\overrightarrow{MB} + \overrightarrow{MC} = \vec{0}$ .)
The centroid lies on the median $AM$ with $GA = 2\,GM$ , and $G$ lies between $A$ and $M$ . So $\overrightarrow{GA}$ and $\overrightarrow{GM}$ have opposite directions and $|\overrightarrow{GA}| = 2|\overrightarrow{GM}|$ . Hence $\overrightarrow{GA} = -2\overrightarrow{GM}$ .
Therefore $\overrightarrow{GA} + \overrightarrow{GB} + \overrightarrow{GC} = -2\overrightarrow{GM} + 2\overrightarrow{GM} = \vec{0}$ .

Forces and velocities

When two forces act on the same object, the single force that has the same effect is their resultant: the sum of the two force vectors. The same holds for velocities, for example a boat and a river current.

A boat crossing a river

Example The boat and the current

A boat points straight across a river and moves at $4$ km/h relative to the water. The current flows at $3$ km/h, perpendicular to the direction of the boat. Find the speed of the boat relative to the bank.

Solution.

Let $\vec{b}$ be the velocity of the boat in still water ( $|\vec{b}| = 4$ ) and $\vec{c}$ the velocity of the current ( $|\vec{c}| = 3$ ). The velocity relative to the bank is $\vec{v} = \vec{b} + \vec{c}$ .
The parallelogram of $\vec{b}$ and $\vec{c}$ is a rectangle because the vectors are perpendicular. Its diagonal is $\vec{v}$ .
By Pythagoras, $|\vec{v}| = \sqrt{4^2 + 3^2} = \sqrt{25} = 5$ .

The speed relative to the bank is $5$ km/h. This is the same $3$ - $4$ - $5$ triangle you know from geometry, and it also explains why the walk in the opening was $500$ m.

Key terms

English	Tiếng Việt
sum of vectors	tổng của hai vectơ
triangle rule	quy tắc ba điểm
parallelogram rule	quy tắc hình bình hành
commutative property	tính chất giao hoán
associative property	tính chất kết hợp
opposite vector	vectơ đối
difference of vectors	hiệu của hai vectơ
midpoint	trung điểm
centroid	trọng tâm
resultant force	hợp lực

Summary

Triangle rule: $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$ . Parallelogram rule: if $OACD$ is a parallelogram, then $\overrightarrow{OA} + \overrightarrow{OD} = \overrightarrow{OC}$ .
Addition is commutative and associative, $\vec{a} + \vec{0} = \vec{a}$ and $\vec{a} + (-\vec{a}) = \vec{0}$ .
Difference: $\vec{a} - \vec{b} = \vec{a} + (-\vec{b})$ , and $\overrightarrow{OA} - \overrightarrow{OB} = \overrightarrow{BA}$ .
$M$ is the midpoint of $AB$ iff $\overrightarrow{MA} + \overrightarrow{MB} = \vec{0}$ ; $G$ is the centroid of $ABC$ iff $\overrightarrow{GA} + \overrightarrow{GB} + \overrightarrow{GC} = \vec{0}$ .
Forces and velocities add as vectors, so the resultant is a diagonal of the parallelogram.

Practice

Simplify (a) $\overrightarrow{MN} + \overrightarrow{NP} + \overrightarrow{PQ}$ ; (b) $\overrightarrow{AB} - \overrightarrow{AC}$ ; (c) $\overrightarrow{AB} + \overrightarrow{BC} - \overrightarrow{AD}$ .
$ABCD$ is a square of side $4$ . Find $|\overrightarrow{AB} + \overrightarrow{AD}|$ , $|\overrightarrow{AB} - \overrightarrow{AD}|$ and $|\overrightarrow{AB} + \overrightarrow{CD}|$ .
For any four points $A$ , $B$ , $C$ , $D$ prove that $\overrightarrow{AB} + \overrightarrow{CD} = \overrightarrow{AD} + \overrightarrow{CB}$ .
$ABCD$ is a parallelogram with centre $O$ . Prove that $\overrightarrow{OA} + \overrightarrow{OC} = \vec{0}$ and $\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} = \vec{0}$ .
$I$ is the midpoint of side $AB$ of triangle $ABC$ . Prove that $\overrightarrow{CA} + \overrightarrow{CB} = 2\overrightarrow{CI}$ .
A boat moves at $6$ km/h in still water, pointing straight across a river whose current flows at $8$ km/h perpendicular to the boat. What is its speed relative to the bank?
Two forces of $20$ N act on the same point and make an angle of $120^\circ$ . Find the magnitude of their resultant.
$ABC$ is an equilateral triangle of side $6$ . Find $|\overrightarrow{AB} + \overrightarrow{AC}|$ and $|\overrightarrow{AB} - \overrightarrow{AC}|$ .
(Challenge) $G$ is the centroid of triangle $ABC$ and $M$ is any point. Prove that $\overrightarrow{MA} + \overrightarrow{MB} + \overrightarrow{MC} = 3\overrightarrow{MG}$ .
(Challenge) $ABCD$ is any quadrilateral, $M$ is the midpoint of $AB$ and $N$ is the midpoint of $CD$ . Prove that $\overrightarrow{MN} = \dfrac{1}{2}(\overrightarrow{AD} + \overrightarrow{BC})$ .

Show solutions

(a) By the triangle rule, $\overrightarrow{MN} + \overrightarrow{NP} + \overrightarrow{PQ} = \overrightarrow{MP} + \overrightarrow{PQ} = \overrightarrow{MQ}$ . (b) $\overrightarrow{AB} - \overrightarrow{AC} = \overrightarrow{CB}$ (same tail $A$ , the difference joins the heads and points to $B$ ). (c) $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$ , and $\overrightarrow{AC} - \overrightarrow{AD} = \overrightarrow{DC}$ .
$\overrightarrow{AB} + \overrightarrow{AD} = \overrightarrow{AC}$ , so its length is the diagonal $4\sqrt{2}$ . $\overrightarrow{AB} - \overrightarrow{AD} = \overrightarrow{DB}$ , with length $4\sqrt{2}$ as well. In a square $\overrightarrow{CD} = \overrightarrow{BA}$ , so $\overrightarrow{AB} + \overrightarrow{CD} = \overrightarrow{AB} + \overrightarrow{BA} = \vec{0}$ and its length is $0$ .
Insert the point $D$ and the point $B$ : $\overrightarrow{AB} + \overrightarrow{CD} = (\overrightarrow{AD} + \overrightarrow{DB}) + (\overrightarrow{CB} + \overrightarrow{BD}) = \overrightarrow{AD} + \overrightarrow{CB} + (\overrightarrow{DB} + \overrightarrow{BD}) = \overrightarrow{AD} + \overrightarrow{CB} + \vec{0}$ .
The diagonals of a parallelogram bisect each other, so $O$ is the midpoint of $AC$ and of $BD$ . Hence $\overrightarrow{OA} + \overrightarrow{OC} = \vec{0}$ and $\overrightarrow{OB} + \overrightarrow{OD} = \vec{0}$ . Adding the two equalities gives the second claim.
$\overrightarrow{CA} = \overrightarrow{CI} + \overrightarrow{IA}$ and $\overrightarrow{CB} = \overrightarrow{CI} + \overrightarrow{IB}$ . Adding, $\overrightarrow{CA} + \overrightarrow{CB} = 2\overrightarrow{CI} + (\overrightarrow{IA} + \overrightarrow{IB}) = 2\overrightarrow{CI}$ because $I$ is the midpoint of $AB$ .
The velocity relative to the bank is the diagonal of a rectangle with sides $6$ and $8$ : $\sqrt{6^2 + 8^2} = \sqrt{100} = 10$ km/h.
Draw $\overrightarrow{OA}$ and $\overrightarrow{OB}$ with $OA = OB = 20$ and $\widehat{AOB} = 120^\circ$ . The parallelogram $OACB$ is a rhombus and the resultant is $\overrightarrow{OC}$ . Angle $\widehat{OAC} = 180^\circ - 120^\circ = 60^\circ$ and $AO = AC$ , so triangle $OAC$ is equilateral and $OC = 20$ . The resultant has magnitude $20$ N.
$\overrightarrow{AB} + \overrightarrow{AC} = 2\overrightarrow{AM}$ where $M$ is the midpoint of $BC$ (exercise 5). The median of an equilateral triangle of side $6$ is $AM = 3\sqrt{3}$ , so the length is $6\sqrt{3}$ . Next, $\overrightarrow{AB} - \overrightarrow{AC} = \overrightarrow{CB}$ has length $BC = 6$ .
Write $\overrightarrow{MA} = \overrightarrow{MG} + \overrightarrow{GA}$ , and similarly for $B$ and $C$ . Adding: $\overrightarrow{MA} + \overrightarrow{MB} + \overrightarrow{MC} = 3\overrightarrow{MG} + (\overrightarrow{GA} + \overrightarrow{GB} + \overrightarrow{GC}) = 3\overrightarrow{MG}$ .
Going from $M$ to $N$ in two ways: $\overrightarrow{MN} = \overrightarrow{MA} + \overrightarrow{AD} + \overrightarrow{DN}$ and $\overrightarrow{MN} = \overrightarrow{MB} + \overrightarrow{BC} + \overrightarrow{CN}$ . Adding, $2\overrightarrow{MN} = (\overrightarrow{MA} + \overrightarrow{MB}) + (\overrightarrow{AD} + \overrightarrow{BC}) + (\overrightarrow{DN} + \overrightarrow{CN})$ . The first and third brackets are $\vec{0}$ because $M$ and $N$ are midpoints, so $\overrightarrow{MN} = \dfrac{1}{2}(\overrightarrow{AD} + \overrightarrow{BC})$ .