Vector

A vector is a quantity with both a magnitude (a size) and a direction: a displacement of 5 km to the north, a force of 20 N pulling down, the wind at 10 m/s from the west. A scalar has a size only: a mass of 3 kg, a temperature of 25 ∘°C, a time of 10 s. The words are Hamilton’s (below); vector is Latin for “carrier”, since a vector carries a point A to a point B.

What you need from Grade 9

This unit builds on these, so keep them handy.

• A vector from A to B is written AB, and a vector with no named ends a. In the plane it is a column of two numbers, the move along the X-axis on top: $\vec{a} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$, also written (3, 4).
• $\vec{i} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\vec{j} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ are the unit vectors along the axes, each of length 1, so $\begin{pmatrix} x \\ y \end{pmatrix} = x\vec{i} + y\vec{j}$.
• The magnitude of $\vec{a} = \begin{pmatrix} x \\ y \end{pmatrix}$ is its length, $|\vec{a}| = \sqrt{x^2 + y^2}$ (Pythagoras). For (3, 4) it is 5.
• The position vector of a point A is OA, the arrow from the origin O to A. Its column is A’s coordinates.
• End minus start: AB = OB − OA. For A(x₁, y₁) and B(x₂, y₂), $\overrightarrow{AB} = \begin{pmatrix} x_2 - x_1 \\ y_2 - y_1 \end{pmatrix}$.
• ka is a stretched k times (turned round if k is negative). Two vectors are parallel when one is a number times the other: EF = k BC.
• BA = −AB: the same length, the opposite way. And AA = 0, the zero vector.

10.2.1 Introduction to scalar product

So the product of two vectors, the force and the displacement, has given a single number, the work. The work done by a force is the part of the force along the displacement times the distance moved:

$$W = \vec{F} \cdot \vec{d} = |\vec{F}|\,|\vec{d}|\cos\theta$$

Work is a scalar, measured in joules (J).

Power is the work done each second: the part of the force along the motion times the velocity, P = F · v = |F| |v| cos θ, a scalar measured in watts (W).

Both activities multiply two vectors and get a scalar, in the same way. That gives the definition.

If θ is the angle between two vectors a and b ($0° \le \theta \le 180°$), their scalar product is

$$\vec{a} \cdot \vec{b} = |\vec{a}|\,|\vec{b}|\cos\theta$$

It is called the scalar product because the answer is a scalar, and the dot product because of the dot. Read a · b as “a dot b”. Turned round, the same equation gives the angle between two vectors:

$$\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|\,|\vec{b}|}$$

The angle between two vectors is measured with both of them starting from the same point, tail to tail.

10.2.2 Conditions for parallel and perpendicular vectors

a. Parallel vectors. When two vectors are parallel, the angle between them is 0∘° (the same direction) or 180∘° (opposite directions):

$$\theta = 0°:\ \ \vec{a} \cdot \vec{b} = |\vec{a}|\,|\vec{b}|\cos 0° = |\vec{a}|\,|\vec{b}| \qquad \theta = 180°:\ \ \vec{a} \cdot \vec{b} = |\vec{a}|\,|\vec{b}|\cos 180° = -|\vec{a}|\,|\vec{b}|$$

So if a and b are parallel, a · b = ±|a| |b|. In coordinates it is quicker to check that one vector is a number times the other: $\begin{pmatrix} x_1 \\ y_1 \end{pmatrix}$ and $\begin{pmatrix} x_2 \\ y_2 \end{pmatrix}$ are parallel when x₁/x₂ = y₁/y₂, that is, when x₁y₂ = x₂y₁.

b. Perpendicular vectors. When two vectors are perpendicular, the angle between them is 90∘°:

$$\vec{a} \cdot \vec{b} = |\vec{a}|\,|\vec{b}|\cos 90° = |\vec{a}|\,|\vec{b}| \times 0 = 0$$

So if a and b are perpendicular, a · b = 0, and the other way round: if two vectors, neither of them zero, have a scalar product of 0, they are perpendicular. This is the test used most in this unit.

What the sign says, for any two vectors that aren’t zero:

a · b	cos θ	the angle between them
positive	positive	acute (less than 90∘°)
zero	zero	a right angle (perpendicular)
negative	negative	obtuse (more than 90∘°)

10.2.3 Scalar product in terms of coordinates

Let $\vec{a} = \begin{pmatrix} x_1 \\ y_1 \end{pmatrix} = x_1\vec{i} + y_1\vec{j}$ and $\vec{b} = \begin{pmatrix} x_2 \\ y_2 \end{pmatrix} = x_2\vec{i} + y_2\vec{j}$. First, the unit vectors with each other: i and i point the same way and are 1 long, so $\vec{i} \cdot \vec{i} = 1 \times 1 \times \cos 0° = 1$, and likewise j · j = 1; i and j are perpendicular, so i · j = j · i = 0. Now multiply out, as with brackets:

$$\begin{aligned} \vec{a} \cdot \vec{b} &= (x_1\vec{i} + y_1\vec{j}) \cdot (x_2\vec{i} + y_2\vec{j}) \\ &= x_1x_2\,(\vec{i} \cdot \vec{i}) + x_1y_2\,(\vec{i} \cdot \vec{j}) + y_1x_2\,(\vec{j} \cdot \vec{i}) + y_1y_2\,(\vec{j} \cdot \vec{j}) \\ &= x_1x_2 + y_1y_2 \end{aligned}$$

For a = (x₁, y₁) and b = (x₂, y₂), a · b = x₁x₂ + y₁y₂: multiply the tops, multiply the bottoms, add. So the angle between two vectors comes from their coordinates alone:

$$\cos\theta = \frac{x_1x_2 + y_1y_2}{\sqrt{x_1^2 + y_1^2}\,\sqrt{x_2^2 + y_2^2}}$$

10.2.4 Geometrical meaning of scalar product

Let PA = a and PB = b, with the angle θ between them. Drop a perpendicular from B to the line PA, meeting it at M, as in Figure 10.3. In the right-angled triangle PMB,

$$\cos\theta = \frac{PM}{PB} = \frac{PM}{|\vec{b}|}, \qquad \text{so} \qquad PM = |\vec{b}|\cos\theta$$

PM is the projection of b on a: the shadow b casts on a’s line when light falls straight down onto it. So

$$\vec{a} \cdot \vec{b} = |\vec{a}|\,(|\vec{b}|\cos\theta) = (\text{magnitude of } \vec{a}) \times (\text{projection of } \vec{b} \text{ on } \vec{a})$$

and in the same way, a · b = |b| (|a| cos θ), the magnitude of b times the projection of a on b.

The scalar product is the length of one vector times the projection of the other onto it. When the angle is obtuse, the projection falls behind P and counts as negative.

10.2.5 Properties of scalar product

• a.Commutative: a · b = b · a (the angle between them is the same either way).
• b.Distributive: a · (b + c) = a · b + a · c, so brackets multiply out as in algebra.
• c.A scalar comes out: (ka) · b = a · (kb) = k(a · b).
• d.The greatest (maximum) value: in the same direction ($\theta = 0°$), a · b = |a| |b|. Since cos θ is never more than 1, no angle gives more.
• e.The least (minimum) value: in opposite directions ($\theta = 180°$), a · b = −|a| |b|.

Two more follow, and the proofs of 10.4 use both:

• A vector dotted with itself ($\theta = 0°$) gives the square of its length: $\vec{a} \cdot \vec{a} = |\vec{a}|^2$, also written $(\vec{a})^2 = |\vec{a}|^2$. In coordinates, $x^2 + y^2$.
• A difference of squares: by b and a, $(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = \vec{a} \cdot \vec{a} - \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} - \vec{b} \cdot \vec{b} = |\vec{a}|^2 - |\vec{b}|^2$.

10.2.6 Triangle law of vector addition

PQ is the displacement from P to Q, and QR the displacement from Q on to R. Together they make the displacement from P to R:

$$\overrightarrow{PQ} + \overrightarrow{QR} = \overrightarrow{PR}$$

This is the triangle law of vector addition: if two vectors are drawn as two sides of a triangle, head to tail (in order), their sum is the third side, drawn from the start of the first to the end of the second. The middle letters meet: PQ + QR, Q to Q.

Worked examples

Back to the sack

The opening asked how much of a pull really moves a load. Pull with a force F along a rope at an angle θ to the ground, over a displacement d along it: the work is F · d = |F| |d| cos θ. Only the fraction cos θ of your pull moves the sack along. With 150 N of pull over 20 m:

angle of the rope	cos θ	work done moving the sack along
0∘° (level)	1	3,000 J
30∘°	0.866	2,598 J
40∘°	0.766	2,298 J
60∘°	0.5	1,500 J
90∘° (straight up)	0	0 J

At 30∘° you lose only about 13 %; at 60∘°, half. (A level pull isn’t always the easiest, though: a slight upward tug lifts some of the sack’s weight off the ground and so lessens the rubbing. The dot product says how much of the pull goes along the path; how much the ground holds back is a question for physics.)

Question 1

Questions 2 and 3

Question 4

Find the scalar product of these vectors.

Question 5

Find the angle between the vectors of Question 4.

Question 6

Are these vectors perpendicular or parallel?

Question 7

If these vectors are perpendicular, find m.

Question 8

If these vectors are parallel, find m.

Question 9

$\vec{a} = \begin{pmatrix} x_1 \\ y_1 \end{pmatrix}$ and $\vec{b} = \begin{pmatrix} x_2 \\ y_2 \end{pmatrix}$ are two vectors.

Question 10

Four points A(−2, 1), B(−1, −3), C(3, −2) and D(2, 2) are given.

Questions 11 and 12

10.3.1 Mid-point formula

In any triangle, the median from a vertex is half the sum of the vectors along the two sides from that vertex: in △ABC, with D the midpoint of BC,

$$\overrightarrow{AD} = \frac{1}{2}\left(\overrightarrow{AB} + \overrightarrow{AC}\right)$$

Proof. Let D, E and F be the midpoints of BC, CA and AB, so AD is a median. In △ABD, by the triangle law,
$$\overrightarrow{AD} = \overrightarrow{AB} + \overrightarrow{BD} \quad \text{(i)}$$
and in △ACD,
$$\overrightarrow{AD} = \overrightarrow{AC} + \overrightarrow{CD} \quad \text{(ii)}$$
Adding (i) and (ii):
$$2\overrightarrow{AD} = (\overrightarrow{AB} + \overrightarrow{AC}) + (\overrightarrow{BD} + \overrightarrow{CD})$$
D is the midpoint of BC, so BD and CD are equal in length and point in opposite directions: BD + CD = 0. So
$$2\overrightarrow{AD} = \overrightarrow{AB} + \overrightarrow{AC}, \qquad \overrightarrow{AD} = \frac{1}{2}\left(\overrightarrow{AB} + \overrightarrow{AC}\right)$$
In the same way, for the other two medians, $\overrightarrow{BE} = \frac{1}{2}(\overrightarrow{BA} + \overrightarrow{BC})$ and $\overrightarrow{CF} = \frac{1}{2}(\overrightarrow{CA} + \overrightarrow{CB})$.

The position vector of a midpoint. Put the vertex at the origin: if OA = a and OB = b, the midpoint M of AB is the end of the median from O in △OAB, so

$$\overrightarrow{OM} = \frac{\vec{a} + \vec{b}}{2}$$

Another way to see it: $\overrightarrow{OM} = \overrightarrow{OA} + \frac{1}{2}\overrightarrow{AB} = \vec{a} + \frac{1}{2}(\vec{b} - \vec{a}) = \frac{1}{2}(\vec{a} + \vec{b})$. With coordinates A(x₁, y₁) and B(x₂, y₂), this is the midpoint formula you know:

$$\overrightarrow{OM} = \begin{pmatrix} \dfrac{x_1 + x_2}{2} \\[2mm] \dfrac{y_1 + y_2}{2} \end{pmatrix}$$

10.3.2 Internal division theorem

A point P on the segment AB divides it internally in the ratio m:n when AP:PB = m:n: cut AB into m + n equal parts, and P is m parts from A and n parts from B.

If the position vectors of A and B are a and b, and P divides AB internally in the ratio m:n, then the position vector of P is

$$\vec{p} = \frac{m\vec{b} + n\vec{a}}{m + n}$$

Proof. Let O be the origin, with OA = a, OB = b and OP = p. P lies between A and B, so AP and PB point the same way along the line, and their lengths are in the ratio m:n:
$$\frac{AP}{PB} = \frac{m}{n}, \qquad \text{so} \qquad n\,\overrightarrow{AP} = m\,\overrightarrow{PB} \quad \text{(i)}$$
End minus start: AP = p − a and PB = b − p. Put these in (i):
$$\begin{aligned} n(\vec{p} - \vec{a}) &= m(\vec{b} - \vec{p}) \\ n\vec{p} - n\vec{a} &= m\vec{b} - m\vec{p} \\ (m + n)\vec{p} &= m\vec{b} + n\vec{a} \\ \vec{p} &= \frac{m\vec{b} + n\vec{a}}{m + n} \end{aligned}$$
With m = n (a ratio of 1 : 1), P is the midpoint, and the formula gives $\frac{1}{2}(\vec{a} + \vec{b})$, as in 10.3.1.

10.3.3 External division theorem

A point P on the line AB but outside the segment divides AB externally in the ratio m:n when AP:BP = m:n. With m bigger than n, P lies beyond B; with m smaller than n, beyond A.

If the position vectors of A and B are a and b, and P divides AB externally in the ratio m:n, then

$$\overrightarrow{OP} = \vec{p} = \frac{m\vec{b} - n\vec{a}}{m - n}$$

Proof. Let OA = a, OB = b and OP = p. A, B and P lie on one line, with P outside AB, so AP and BP point the same way, with lengths in the ratio m:n:
$$\frac{AP}{BP} = \frac{m}{n}, \qquad \text{so} \qquad n\,\overrightarrow{AP} = m\,\overrightarrow{BP} \quad \text{(i)}$$
AP = p − a and BP = p − b. Put these in (i):
$$\begin{aligned} n(\vec{p} - \vec{a}) &= m(\vec{p} - \vec{b}) \\ n\vec{p} - n\vec{a} &= m\vec{p} - m\vec{b} \\ m\vec{b} - n\vec{a} &= (m - n)\vec{p} \\ \vec{p} &= \frac{m\vec{b} - n\vec{a}}{m - n} \end{aligned}$$
The formula breaks down when m = n: the denominator is 0. No point outside AB is equally far from A and B.

You may have met some of these theorems in geometry, proved with congruent or isosceles triangles. Vectors prove them another way: write each side as a sum of other sides (the triangle law), and use a scalar product of 0 for a right angle. The proofs are short, and they work without any figure-by-figure argument about which triangles match.

Theorem 1: the mid-point theorem

The line segment joining the midpoints of two sides of a triangle is parallel to the third side and half of it.

Given: in △ABC, E and F are the midpoints of AB and AC.

To prove: $\overrightarrow{EF} = \frac{1}{2}\overrightarrow{BC}$, so $EF \parallel BC$ and $EF = \frac{1}{2}BC$.

Proof. In △AEF, by the triangle law,
$$\overrightarrow{EF} = \overrightarrow{EA} + \overrightarrow{AF} \quad \text{(i)}$$
E and F are midpoints, so $\overrightarrow{EA} = \frac{1}{2}\overrightarrow{BA}$ and $\overrightarrow{AF} = \frac{1}{2}\overrightarrow{AC}$. From (i),
$$\overrightarrow{EF} = \frac{1}{2}\overrightarrow{BA} + \frac{1}{2}\overrightarrow{AC} = \frac{1}{2}\left(\overrightarrow{BA} + \overrightarrow{AC}\right) = \frac{1}{2}\overrightarrow{BC}$$
since BA + AC = BC in △ABC. EF is a number ($k = \frac{1}{2}$) times BC, so the two are parallel, and EF is half of BC. Hence proved.

Theorem 2: the midpoint of the hypotenuse

In a right-angled triangle, the midpoint of the hypotenuse is equally far from (equidistant from) all three vertices.

Given: △ABC with $\angle ABC = 90°$, so AC is the hypotenuse, and D the midpoint of AC.

To prove: AD = BD = CD.

Proof. The angle at B is 90∘°, so BA and BC are perpendicular:
$$\overrightarrow{BA} \cdot \overrightarrow{BC} = 0 \quad \text{(i)}$$
By the triangle law, in △ABD, BA = BD + DA, and in △BCD, BC = BD + DC. D is the midpoint of AC, so DC = −DA. Put these in (i):
$$\begin{aligned} (\overrightarrow{BD} + \overrightarrow{DA}) \cdot (\overrightarrow{BD} - \overrightarrow{DA}) &= 0 \\ |\overrightarrow{BD}|^2 - |\overrightarrow{DA}|^2 &= 0 \qquad [(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = |\vec{a}|^2 - |\vec{b}|^2] \\ BD^2 &= DA^2 \end{aligned}$$
So BD = DA, lengths being positive. And DA = DC, since D is the midpoint. Hence AD = BD = CD.
This means the circle with the hypotenuse as its diameter passes through the right angle’s vertex: the reverse of Theorem 4.

Theorem 3: the midpoints of a quadrilateral

The quadrilateral formed by joining the midpoints of the sides of any quadrilateral, in order, is a parallelogram.

Given: a quadrilateral ABCD, with F, G, H and E the midpoints of AB, BC, CD and DA, joined in order to make EFGH.

To prove: EFGH is a parallelogram (EF = HG, or EH = FG).

Construction: join B and D.

Proof. In △ABD, F and E are the midpoints of AB and AD. By the mid-point theorem (Theorem 1),
$$\overrightarrow{EF} = \frac{1}{2}\overrightarrow{DB} \quad \text{(i)}$$
In △CBD, G and H are the midpoints of CB and CD, so in the same way
$$\overrightarrow{HG} = \frac{1}{2}\overrightarrow{DB} \quad \text{(ii)}$$
From (i) and (ii), EF = HG: the sides EF and HG are equal in length and parallel. A quadrilateral with one pair of opposite sides equal and parallel is a parallelogram. (Joining A and C gives $\overrightarrow{EH} = \overrightarrow{FG} = \frac{1}{2}\overrightarrow{AC}$ in the same way.) Hence EFGH is a parallelogram.

Theorem 4: the angle in a semicircle

The angle that a diameter of a circle makes (subtends) at any point on the circle (its circumference) is a right angle.

Given: O is the center of a circle, AB a diameter and C any other point on the circle.

To prove: $\angle ACB = 90°$.

Proof. Every point on the circle is a radius r from the center, so |OA| = |OB| = |OC| = r. O is the midpoint of AB, so OB = −OA. End minus start:
$$\begin{aligned} \overrightarrow{AC} \cdot \overrightarrow{BC} &= (\overrightarrow{OC} - \overrightarrow{OA}) \cdot (\overrightarrow{OC} - \overrightarrow{OB}) \\ &= (\overrightarrow{OC} - \overrightarrow{OA}) \cdot (\overrightarrow{OC} + \overrightarrow{OA}) \\ &= |\overrightarrow{OC}|^2 - |\overrightarrow{OA}|^2 \\ &= r^2 - r^2 = 0 \end{aligned}$$
AC · BC = 0, so AC and BC are perpendicular, and $\angle ACB = 90°$. Hence proved.

Worked examples

Question 1

Questions 2 to 4

Question 5

In △ABC, F and E are the midpoints of AB and AC.