Vector

Vectors are a part of mathematics used all through science to solve problems where both the size (magnitude) and the direction of a quantity matter. The idea grew in the 1800s. In 1835 the Italian Giusto Bellavitis treated parallel segments of the same length and direction as one quantity, what we now call equal vectors, and the Irish mathematician William Rowan Hamilton (1805–1865) gave us the words vector (Latin for “carrier”) and scalar. (His story, and the bridge in Dublin where he carved a formula, is in Grade 10’s unit on vectors.)

Now think of Rima, who is trying to hit a round mark on a wall with a ball. She throws many times and misses. Then she uses a catapult (a slingshot), and the ball flies fast and almost straight, at 15 m/s, and hits the mark.

• On path A the ball went (very nearly) in a straight line, east, at 15 m/s, and reached the mark.
• On path B it also went at 15 m/s, but along a curve, and missed.

On path A the ball had a fixed size of speed and a fixed direction. On path B the size was the same but the direction kept changing. A quantity like path A’s, with a definite magnitude and a definite direction, is a vector. A quantity that has a magnitude but no direction is a scalar.

Physical quantities

Which of weight, force, distance, knowledge, density, area and feelings can be measured? Weight, force, distance, density and area can; knowledge and feelings can’t, not with a number and a unit. Quantities that can be measured are called physical quantities, and they are of two kinds: scalars and vectors.

i. Scalar

Water boils at about 100 ∘°C at sea level. The number 100 and the unit ∘°C say everything about that temperature: there is no direction to give. So temperature is a scalar.

Physical quantities that have only magnitude are called scalar quantities. Examples: distance, speed, time, temperature, mass, work, power, energy.

ii. Vector

A girl stands at O and walks 30 m in a straight line. Where does she reach?

It depends on which way she goes. Walking 30 m east she reaches B, west she reaches C, north-east she reaches A. Each time she moves 30 m from O in a definite direction to a definite point. This move is her displacement: its magnitude is 30 m every time, but its direction, and so where she ends, differs. Displacement has both magnitude and direction, so it is a vector.

Physical quantities that have both magnitude and direction are called vector quantities. Examples: displacement, velocity, acceleration, force, weight (a force, always pointing down).

Types of vector

On a grid, a vector is drawn as an arrow: its length is the magnitude and the way it points is the direction. In the figure below, the displacement from O to R(3, 3) is the vector OR: from O it goes 3 units along the X-axis (its x-component) and 3 units up (its y-component). There are two ways to write those components.

i. Column vector

If OR is called a, then with x-component 3 and y-component 3,
$$\vec{a} = \begin{pmatrix} 3 \\ 3 \end{pmatrix}$$
When the components are written one above the other in brackets, x on top, the vector is a column vector. Another example: $\vec{b} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}$.

ii. Row vector

The same vector can be written in a row, a = (3, 3): the x-component first, then the y-component, separated by a comma. This is a row vector. For example, b = (3, −2). Column and row are two ways of writing the same vector; choose whichever is handier.

iii. Position vector

Where does OA start? At the origin. If OA moves O(0, 0) to A(x, y), it tells you exactly where A is, so OA is called the position vector of A. A position vector always starts at the origin O(0, 0), and its components are the coordinates of its end point. For example, the position vector of B(3, 2) is OB = (3, 2).

So a vector starting from the origin is the position vector of its end point (its terminal point).

iv. Zero or null vector

Two people pull the two ends of a rope with equal forces in opposite directions. The rope doesn’t move: the two forces cancel, and the total force on it is the zero vector.

If a vector moves the point A(x, y) to A(x, y) itself, how big is the move? Nothing at all. A vector of magnitude 0 is a zero vector (or null vector): 0 = (0, 0). It is the only vector with no direction.

v. Equal vectors

AB moves A to B by a certain distance in a certain direction, and CD moves C to D by the same distance in the same direction. Vectors with equal magnitude and the same direction are called equal vectors. They can sit in different places: what matters is the move, not where it starts.

In components: a = (x₁, y₁) and b = (x₂, y₂) are equal exactly when x₁ = x₂ and y₁ = y₂.

vi. Negative vector

AB moves A to B; BA moves B back to A, by the same distance but the opposite way. Vectors with equal magnitude but opposite directions are called negative vectors of each other:
$$\overrightarrow{AB} = -\overrightarrow{BA}$$
For example, a = (6, 2) and b = (−6, −2) are negative vectors of each other: every component has changed sign.

vii. Like and unlike vectors

In the figure, which vectors have the same direction? Which have opposite directions? Which have neither?

• a and b have the same direction, though not the same length. Vectors with the same direction are like vectors. For example, a = (6, 4) and b = (3, 2) are like vectors, because (6, 4) = 2(3, 2): one is a positive number times the other.
• c and d have opposite directions. Vectors with opposite directions are unlike vectors. For example, c = (−6, −4) and d = (3, 2) are unlike, because (−6, −4) = −2(3, 2): a negative number times the other.
• Like and unlike vectors lie along parallel lines (or the same line), so both are called parallel (or collinear) vectors.
• e and f have neither the same nor opposite directions, so they are neither like nor unlike.

Scalar	Vector
A physical quantity that has only magnitude.	A physical quantity that has both magnitude and direction.
It is given by a number and a unit.	It is given by a number, a unit and a direction.
Scalars are added like ordinary numbers: 3 kg + 4 kg = 7 kg.	Vectors are added by the triangle law: 3 km east + 4 km north is 5 km, not 7 km.
Examples: temperature, speed, distance, mass, time.	Examples: velocity, acceleration, displacement, force, weight.

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In Figure 4.10, PT ∥ QR, QS ∥ RT and PQ ∥ SR, so PQRS and QRTS are parallelograms.

If a point moves from O(0, 0) to A(2, 3) (books say OA displaces O to A), the move is shown by the directed line segment OA: a segment with an arrowhead. O is the initial point (the tail) and A the terminal point (the head). So OA is a vector. OA means the move from O to A, while AO means the move from A back to O: they are different vectors. In the same way OC and AB are vectors.

In print, vectors are often written as bold letters, a; by hand, with an arrow over the letter, a, read “vector a”. A vector named by its ends gets an arrow over both letters, OA.

If a = OA goes from O to A(x, y), its move across is x and its move up is y, so
$$\vec{a} = (x, y) = \begin{pmatrix} x \\ y \end{pmatrix}$$
x is its x-component and y its y-component.

A vector between any two points

Join P(x₁, y₁) to Q(x₂, y₂) to get the vector PQ, called b here. Drop the perpendiculars PM and QN to the X-axis and draw PR across to QN. Then

• the move across, the horizontal displacement, is PR = MN = x₂ − x₁, the x-component;
• the move up, the vertical displacement, is QR = y₂ − y₁, the y-component.

$$\overrightarrow{PQ} = (x_2 - x_1, y_2 - y_1) = \begin{pmatrix} x_2 - x_1 \\ y_2 - y_1 \end{pmatrix}$$
Remember it as end minus start: the head’s coordinates minus the tail’s.

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Find AB and BA for each pair of points. Which pairs of vectors are equal?

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AB moves the point A to B. Draw AB and write it as a column vector.

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If AB moves a point from A to B and BC moves it on from B to C, the two moves together take it from A to C:
$$\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$$
In the same way AC + CB = AB and CA + AB = CB. This rule is the triangle law of vector addition: draw the second vector starting where the first ends (head to tail); their sum is the arrow from the first tail to the last head. The two vectors and their sum make a triangle.

The sum is called the resultant. Notice that the end letter of one vector is the start letter of the next, and the sum keeps the first start and the last end: $\overrightarrow{A\mathbf{B}} + \overrightarrow{\mathbf{B}C} = \overrightarrow{AC}$.

The polygon law

The triangle law can be used again and again. In the pentagon ABCDE,

$$\begin{aligned} &\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD} + \overrightarrow{DE} \\ &= \overrightarrow{AC} + \overrightarrow{CD} + \overrightarrow{DE} \quad [\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}] \\ &= \overrightarrow{AD} + \overrightarrow{DE} \quad [\overrightarrow{AC} + \overrightarrow{CD} = \overrightarrow{AD}] \\ &= \overrightarrow{AE} \quad [\overrightarrow{AD} + \overrightarrow{DE} = \overrightarrow{AE}] \end{aligned}$$
This extended form of the triangle law is the polygon law of vector addition: any number of vectors joined head to tail add up to the single arrow from the first tail to the last head. And if the path comes back to where it started, the sum is the zero vector.

a. Addition of vectors

If a = (3, 2) and b = (6, 4), what is a + b? And if $\vec{c} = \begin{pmatrix} -3 \\ 4 \end{pmatrix}$ and $\vec{d} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$, what is c + d?

Head to tail, the move across of the sum is the two moves across together, and the same for the moves up:
$$\vec{a} + \vec{b} = (3 + 6, 2 + 4) = (9, 6)$$
$$\vec{c} + \vec{d} = \begin{pmatrix} -3 + 2 \\ 4 + 3 \end{pmatrix} = \begin{pmatrix} -1 \\ 7 \end{pmatrix}$$
So, to add vectors, add the x-components together and the y-components together: for a = (x₁, y₁) and b = (x₂, y₂),
$$\vec{a} + \vec{b} = (x_1 + x_2, y_1 + y_2)$$

b. Subtraction of vectors

Is a − b = a + (−b)? Yes: taking a vector away is the same as adding its negative, as with numbers (7 − 3 = 7 + (−3)). Subtracting b means going along b the wrong way.

If $\vec{a} = \begin{pmatrix} x_1 \\ y_1 \end{pmatrix}$ and $\vec{b} = \begin{pmatrix} x_2 \\ y_2 \end{pmatrix}$, then $-\vec{b} = \begin{pmatrix} -x_2 \\ -y_2 \end{pmatrix}$, and
$$\vec{a} - \vec{b} = \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} + \begin{pmatrix} -x_2 \\ -y_2 \end{pmatrix} = \begin{pmatrix} x_1 - x_2 \\ y_1 - y_2 \end{pmatrix}$$
So, to subtract vectors, subtract x-component from x-component and y-component from y-component.

This is also where “end minus start” comes from. For any points A and B, OA + AB = OB by the triangle law, so
$$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$$
the position vector of the end minus the position vector of the start.

Take a = (3, 2) and b = (6, 4). Their magnitudes are $\sqrt{9 + 4} = \sqrt{13}$ and $\sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}$, and both point at $\tan^{-1}\frac{2}{3} = 33.69^\circ$ to the X-axis (you will see how in 4.7). Same direction, twice the length: b = 2a.

If a = (x, y) and k is any number (a scalar), then
$$k\vec{a} = k(x, y) = (kx, ky) \qquad |k\vec{a}| = |k| \, |\vec{a}|$$

• If k is positive, ka has the same direction as a: they are like vectors.
• If k is negative, ka has the opposite direction: unlike vectors.
• Either way, a and ka lie on parallel lines. So two vectors are parallel when one is a number times the other, b = ka. This is how parallel lines are proved with vectors.

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From Figure 4.30, find:

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In the regular hexagon ABCDEF (Figure 4.31), AB = a and BC = b. Express these in terms of a and b.

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If a = (4, −2), find:

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Prove that each pair of vectors is parallel.

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If $\vec{a} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ and $\vec{b} = \begin{pmatrix} -3 \\ 4 \end{pmatrix}$, find:

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Magnitude of a vector

How long are OM and AB in Figure 4.36, and OA and PQ in the drawings below? Each is the distance between two points, found by the distance formula. The length of the arrow of a vector is called its magnitude. The magnitude of a is written |a|, read “the modulus (or absolute value) of vector a”.

• $OM = \sqrt{(3 - 0)^2 + (0 - 0)^2} = \sqrt{9} = 3$ units.
• $|\overrightarrow{AB}| = \sqrt{(4 - 2)^2 + (3 - 1)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$ units.
• For A(x, y), $|\overrightarrow{OA}| = \sqrt{(x - 0)^2 + (y - 0)^2} = \sqrt{x^2 + y^2}$ units (Figure 4.37).
• For P(x₁, y₁) and Q(x₂, y₂), $|\overrightarrow{PQ}| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ units (Figure 4.12).

The two components are the legs of a right-angled triangle and the arrow its hypotenuse, so this is Pythagoras’ rule. In general,
$$|\vec{a}| = \sqrt{(\text{x-component})^2 + (\text{y-component})^2}$$
For example, if a = (3, 4), then $|\vec{a}| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$ units.

A length is never negative, so a magnitude is never negative. It is 0 only for the zero vector; every other vector has a positive magnitude. And | − a| = |a|: turning a vector round doesn’t change its length.

Direction of a vector

What angle does OR, with R(3, 3), make with the positive X-axis? Call it θ (Figure 4.3). In the right-angled triangle ORM,
$$\tan\theta = \frac{RM}{OM} = \frac{3}{3} = 1 = \tan 45^\circ, \quad \text{so } \theta = 45^\circ$$
OR makes 45∘° with the positive X-axis. This angle is called the direction of OR.

In the same way, if OA with A(x, y) makes θ with the positive X-axis, $\tan\theta = \dfrac{AM}{OM} = \dfrac{y}{x}$, so $\theta = \tan^{-1}\left(\dfrac{y}{x}\right)$. And for PQ in Figure 4.12, $\tan\theta = \dfrac{y_2 - y_1}{x_2 - x_1}$.

The angle a vector makes with the positive X-axis, measured anticlockwise, is its direction:
$$\tan\theta = \frac{\text{y-component}}{\text{x-component}}$$
For example, for a = (3, 2), $\theta = \tan^{-1}\left(\frac{2}{3}\right) = 33.69^\circ$.

Which quadrant?

$\tan\theta = \frac{y}{x}$ can’t tell (3, 2) from (−3, −2): both give $\frac{2}{3}$, yet they point opposite ways. So first find the acute angle α with the X-axis, $\alpha = \tan^{-1}\left|\frac{y}{x}\right|$, then turn it into θ by the signs of the components, as in Unit 2’s trigonometric ratios of any angle:

x-component	y-component	Quadrant	Direction θ
+	+	I	α
−	+	II	180∘° − α
−	−	III	180∘° + α
+	−	IV	360∘° − α

On the axes: (+, 0) is 0∘°, (0, +) is 90∘°, (−, 0) is 180∘°, (0, −) is 270∘°.

Unit vector

Find the magnitudes of a = (1, 0) and b = (0, 1):
$$|\vec{a}| = \sqrt{1^2 + 0^2} = 1, \qquad |\vec{b}| = \sqrt{0^2 + 1^2} = 1$$
Both are 1 unit long. A vector whose magnitude is 1 unit is called a unit vector.

Divide any vector (other than 0) by its own magnitude and you get the unit vector in its direction, written $\hat{a}$ (“a hat”):
$$\hat{a} = \frac{\vec{a}}{|\vec{a}|}$$
For example, if a = (3, 4), then |a| = 5 and $\hat{a} = \frac{(3, 4)}{5} = \left(\frac{3}{5}, \frac{4}{5}\right)$. Check: $\left(\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2 = \frac{9 + 16}{25} = 1$. ✓ The unit vector keeps the direction and throws the length away; multiply it by any length to get a vector of that length in the same direction.

Unit vectors along the axes

The unit vector along the X-axis is i = (1, 0), and along the Y-axis j = (0, 1).

Any vector is so many i steps across and so many j steps up:
$$\vec{a} = (x, y) = x\vec{i} + y\vec{j}$$
For example, (5, −3) = 5i − 3j.

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Find the magnitude and direction of each vector.

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PQ moves the point P to Q. Write PQ as a column vector, and find its magnitude and direction.

Question 4

AB moves A to B and CD moves C to D. Prove that |AB| = |CD|.

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Find the unit vector in the direction of:

Question 7

AB moves A to B. Write AB as xi + yj and find the unit vector in its direction.

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Are these pairs of vectors equal or unequal?

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Jamuna walked 3 km east and then 4 km north. With east along the X-axis and north along the Y-axis, her two moves are (3, 0) and (0, 4), and by the triangle law her whole displacement is
$$(3, 0) + (0, 4) = (3, 4)$$

• How far: $|(3, 4)| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$ km. Her brother walks 5 km, not 7.
• Which way: first quadrant, $\theta = \tan^{-1}\frac{4}{3} = 53.13^\circ$ from east, anticlockwise towards north. As a bearing, clockwise from north, that is 90∘° − 53.13∘° = 36.87∘°: between north and north-east (45∘°), a little nearer north-east.
• His first step: the unit vector $\left(\frac{3}{5}, \frac{4}{5}\right)$: for every 5 m he walks, 3 m are eastward and 4 m northward.

Grade 9’s final exam follows CDC’s Specification Grid and Model Question for Optional Mathematics, Grade 9 (2082). There Vector has 7 marks of the 75:

Where	Questions	Marks	Levels
Group A, multiple choice	1	1	understanding
Vector’s own question	1 (three parts)	3	knowledge 1, application 1, higher ability 1
A question joining areas	parts of 1	3	application 2, higher ability 1

In the grid’s model paper the multiple-choice question asks where the vector a = 3i + 10j starts if it ends at (3, 10): at the origin, (0, 0), since its components are its end point’s coordinates. Vector’s own question draws a vector m on a graph and asks for it in column form, its magnitude, and a vector n in the direction of m with magnitude 2. And a question joining areas puts two friends at P(2, 3) and Q(7, 3) and asks whether the distance between them is the magnitude of PQ (yes: a vector’s magnitude is the length of its arrow), and then for PQ itself.