Vectors, basic ideas | Grade 10

Suppose a friend texts you: "The canteen is 200 m from the gate." Can you find it? Not really: you also need to know which way to walk. A number such as 200 m only gives a size. A size together with a direction is the idea behind a vector. Forces, velocities and displacements all work this way, which is why vectors are the language of physics, navigation and computer graphics.

From a segment to a vector

A line segment $AB$ has two endpoints and a length. If we decide which endpoint is the start and which is the end, the segment gets a direction. We draw an arrow from the start to the end.

The vector

\overrightarrow{AB}

with its initial point A and terminal point B

Two things describe a vector:

its direction, given by the line $AB$ and the arrow;
its magnitude (length) $|\overrightarrow{AB}|$ , which is the length of the segment $AB$ . So $|\overrightarrow{AB}| = AB$ .

A vector with magnitude $1$ is called a unit vector.

The zero vector

If the initial and terminal points are the same point, we get the vector $\overrightarrow{AA}$ . It has no length and no particular direction. All such vectors are regarded as one vector, the zero vector, written $\vec{0}$ . So $|\vec{0}| = 0$ , and $\overrightarrow{AA} = \overrightarrow{BB} = \vec{0}$ for any points $A$ and $B$ .

Same direction, opposite direction, collinear

The line of action (or support) of a nonzero vector $\overrightarrow{AB}$ is the line $AB$ . Two nonzero vectors are collinear if their lines of action are parallel or coincide. Collinear vectors point either the same way or in opposite ways.

Same direction and opposite directions

In the figure, $\vec{a}$ and $\vec{b}$ have the same direction, while $\vec{c}$ has the opposite direction to both. All three are collinear, even though they do not lie on one line, because their lines of action are parallel.

Equal vectors and opposite vectors

Notice that an equal vector can be drawn anywhere in the plane: we may slide a vector without turning it and without changing its length, and it stays the same vector. This is what lets us add forces drawn from different points later on.

In the parallelogram

ABCD

\overrightarrow{AB} = \overrightarrow{DC}

A handy fact: for four points $A, B, C, D$ not on one line, $ABCD$ is a parallelogram if and only if $\overrightarrow{AB} = \overrightarrow{DC}$ .

Example Vectors in a rectangle

Rectangle $ABCD$ has $AB = 6$ and $AD = 8$ . Let $O$ be the intersection of the diagonals.

a) Find $|\overrightarrow{AC}|$ . b) Name a vector equal to $\overrightarrow{AO}$ , and one opposite to $\overrightarrow{AB}$ . c) Is $\overrightarrow{AB} = \overrightarrow{CD}$ ?

Solution. a) $|\overrightarrow{AC}| = AC$ . In the right triangle $ABC$ (right angle at $B$ , with $BC = AD = 8$ ), Pythagoras gives $AC = \sqrt{6^2 + 8^2} = \sqrt{100} = 10$ . So $|\overrightarrow{AC}| = 10$ .

b) The diagonals of a rectangle bisect each other, so $O$ is the midpoint of $AC$ . Then $\overrightarrow{AO}$ and $\overrightarrow{OC}$ have the same direction and the same length, so $\overrightarrow{AO} = \overrightarrow{OC}$ . The opposite vector of $\overrightarrow{AB}$ is $\overrightarrow{BA}$ . Another one is $\overrightarrow{CD}$ (see part c).

c) $\overrightarrow{AB}$ and $\overrightarrow{CD}$ lie on parallel lines and have the same length $6$ , but $\overrightarrow{AB}$ points from $A$ to $B$ while $\overrightarrow{CD}$ points the other way (from $C$ to $D$ , and $\overrightarrow{DC} = \overrightarrow{AB}$ ). So they have opposite directions: $\overrightarrow{AB} \neq \overrightarrow{CD}$ , and in fact $\overrightarrow{CD} = -\overrightarrow{AB}$ .

Regular hexagon with centre O

Example A regular hexagon

$ABCDEF$ is a regular hexagon with centre $O$ (see the figure). List all vectors equal to $\overrightarrow{AB}$ whose endpoints are among the seven points $A, B, C, D, E, F, O$ , other than $\overrightarrow{AB}$ itself.

Solution. A regular hexagon is made of six equilateral triangles meeting at $O$ , so $OA = OB = \dots = AB$ . We look for segments parallel to $AB$ , of the same length, pointing the same way.

$ABCO$ is a rhombus, so $OC \parallel AB$ , $OC = AB$ , and both point the same way. Hence $\overrightarrow{OC} = \overrightarrow{AB}$ .
$ABOF$ is a rhombus as well, so $\overrightarrow{FO} = \overrightarrow{AB}$ .
The diagonals $AD$ and $BE$ of the quadrilateral $ABDE$ are diameters, so they bisect each other at $O$ and have equal length. Thus $ABDE$ is a rectangle, with $AB$ and $ED$ as opposite sides, and $\overrightarrow{ED} = \overrightarrow{AB}$ .

So the vectors are $\overrightarrow{OC}$ , $\overrightarrow{FO}$ and $\overrightarrow{ED}$ .

Key terms

English	Tiếng Việt
vector	vectơ
initial point	điểm đầu
terminal point	điểm cuối
direction	hướng
magnitude (length)	độ dài (môđun) của vectơ
zero vector	vectơ không
collinear vectors	các vectơ cùng phương
same direction	cùng hướng
equal vectors	hai vectơ bằng nhau
opposite vector	vectơ đối

Summary

A vector $\overrightarrow{AB}$ is a segment with a direction from the initial point $A$ to the terminal point $B$ ; its magnitude is $|\overrightarrow{AB}| = AB$ .
The zero vector $\vec{0}$ has magnitude $0$ and no definite direction.
Collinear vectors have parallel (or identical) lines of action; they point in the same or in opposite directions.
$\vec{a} = \vec{b}$ means same direction and same magnitude, so a vector can be moved freely without changing it.
The opposite vector $-\vec{a}$ has the same magnitude and the opposite direction; $\overrightarrow{BA} = -\overrightarrow{AB}$ .
$ABCD$ is a parallelogram if and only if $\overrightarrow{AB} = \overrightarrow{DC}$ .

Practice

Which of these are vectors and which are only numbers (scalars)? (a) the mass of a bag, 5 kg; (b) a wind of 30 km/h blowing from the west; (c) the temperature, 25 °C; (d) a force of 20 N pushing downwards; (e) a walk of 3 km due east.
$ABCD$ is a square of side $3$ . Find $|\overrightarrow{AB}|$ and $|\overrightarrow{AC}|$ . Is $\overrightarrow{AB} = \overrightarrow{DC}$ ? Is $\overrightarrow{AB} = \overrightarrow{BC}$ ?
$ABCD$ is a parallelogram with centre $O$ . Name a vector equal to $\overrightarrow{AO}$ , a vector equal to $\overrightarrow{BO}$ , and two vectors opposite to $\overrightarrow{AB}$ .
True or false? Explain briefly. (a) Equal vectors are collinear. (b) Collinear vectors are equal. (c) $\vec{0}$ is collinear with every vector. (d) $|\overrightarrow{AB}| = |\overrightarrow{BA}|$ . (e) If $|\vec{a}| = |\vec{b}|$ then $\vec{a} = \vec{b}$ .
Boat M moves at $15$ km/h towards the east and boat N at $15$ km/h towards the west. How are their velocity vectors related?
$ABCDEF$ is a regular hexagon with centre $O$ . List all vectors equal to $\overrightarrow{OA}$ whose endpoints are among $A, B, C, D, E, F, O$ .
A parallelogram has vertices $A, B, C, D$ . How many nonzero vectors have both endpoints among these four points? How many different vectors are these (counting equal vectors once)?
Points $A, B, C, D$ are not on one line and $\overrightarrow{AB} = \overrightarrow{CD}$ . Show that $ABDC$ is a parallelogram.
A point $A$ is marked on the plane and a vector $\vec{u}$ with $|\vec{u}| = 3$ is given. (a) Describe all points $B$ with $|\overrightarrow{AB}| = 3$ . (b) How many points $B$ satisfy $\overrightarrow{AB} = \vec{u}$ ?

Show solutions

A vector needs a magnitude and a direction. (b) wind, (d) force, (e) walk are vectors. (a) mass and (c) temperature are scalars.
$|\overrightarrow{AB}| = AB = 3$ . By Pythagoras $|\overrightarrow{AC}| = \sqrt{3^2+3^2} = 3\sqrt{2}$ . $ABCD$ is a parallelogram, so $\overrightarrow{AB} = \overrightarrow{DC}$ (yes). $\overrightarrow{AB}$ and $\overrightarrow{BC}$ have the same length but perpendicular directions, so they are not equal (no).
The diagonals bisect each other, so $O$ is the midpoint of $AC$ and of $BD$ . Hence $\overrightarrow{AO} = \overrightarrow{OC}$ and $\overrightarrow{BO} = \overrightarrow{OD}$ . Opposite to $\overrightarrow{AB}$ : $\overrightarrow{BA}$ and $\overrightarrow{CD}$ (because $\overrightarrow{AB} = \overrightarrow{DC}$ ).
(a) True: equal vectors have the same direction, so their lines of action are parallel or identical. (b) False: $\vec{a}$ and $-\vec{a}$ are collinear but not equal. (c) True, by convention. (d) True: both equal the length of segment $AB$ . (e) False: the directions may differ.
Same magnitude ( $15$ ) and opposite directions, so the two velocity vectors are opposite vectors.
$\overrightarrow{OA}$ goes from the centre to vertex $A$ . As in the example, the rhombi and parallelograms of the hexagon give three vectors of the same length, parallel to $OA$ and pointing the same way: $\overrightarrow{DO}$ (from $D$ through the centre to $O$ ), $\overrightarrow{CB}$ and $\overrightarrow{EF}$ (both parallel to the diagonal $AD$ , pointing towards the side of $A$ ). So the vectors are $\overrightarrow{DO}$ , $\overrightarrow{CB}$ , $\overrightarrow{EF}$ .
Each ordered pair of distinct points gives a nonzero vector: $4 \cdot 3 = 12$ . In parallelogram $ABCD$ : $\overrightarrow{AB} = \overrightarrow{DC}$ , $\overrightarrow{BA} = \overrightarrow{CD}$ , $\overrightarrow{AD} = \overrightarrow{BC}$ , $\overrightarrow{DA} = \overrightarrow{CB}$ , and $\overrightarrow{AC}$ , $\overrightarrow{CA}$ , $\overrightarrow{BD}$ , $\overrightarrow{DB}$ are all different. So there are $4 + 4 = 8$ different vectors.
Since $\overrightarrow{AB} = \overrightarrow{CD}$ , the segments $AB$ and $CD$ are parallel and have the same length, and the points are not on one line. Now apply the fact from the lesson to the quadrilateral $ABDC$ (vertices in this order): it says that a quadrilateral $PQRS$ is a parallelogram when $\overrightarrow{PQ} = \overrightarrow{SR}$ . With $P = A$ , $Q = B$ , $R = D$ , $S = C$ , the condition reads $\overrightarrow{AB} = \overrightarrow{CD}$ , which is true. So $ABDC$ is a parallelogram.
(a) These points form the circle with centre $A$ and radius $3$ . (b) Exactly one: $B$ is the point obtained by drawing a copy of $\vec{u}$ starting at $A$ .