**Definition of a vector**

Let $A$ and $B$ be two different points of a plane.

**A directed line segment **$\overrightarrow{AB}$ is a line segment in which we know precisely which point is the initial point and which one is the terminal point.

Two directed line segments are equivalent if there exists a translation in which one translates into another.

**A vector** is the set of all directed line segments which are equivalent to each other.

Any directed line segment we will refer to as a vector, due to the simplicity of the speech. Vectors we will denote by $\overrightarrow{AB}, \overrightarrow{CD}, \ldots$ or simply as $\overrightarrow{a}, \overrightarrow{b}, \ldots$.

Vectors are uniquely determined by their magnitude, direction and orientation.

## Drawing vectors in two-dimensional Cartesian coordinates

Every vector is determined by two points. This makes drawing them in the coordinate plane quite simple. For example, if we are given two points with their coordinates $A=(1, 2)$ and $ B = (5, 6)$ and our task is to draw vector $\overrightarrow{AB}$, we must first draw a line segment $\overline{AB}$ and then simply draw an arrow which denotes our terminal point.

To every point in a plane there can be assigned a unique vector whose initial point is in the origin and the terminal point is in the given point. These vectors are very important in vector geometry and they are called **position **or** radius vectors**.

## The magnitude of vectors

**The magnitude of a vector** $\overrightarrow{AB}$ is the length of a line segment $\overline{AB}$.

Some additional names for a vectors magnitude such as: vector norm, vector modulus or absolute value of a vector.

*The magnitude of any vector* is determined by the placement of its initial and terminal point, and it is calculated exactly the same as the length of a line segment. The magnitude of a vector is always either a positive number or a zero. The magnitude of a vector is equal to zero iff the initial point is equal to the terminal point.

If we have two points with their coordinates $ A = (x_1, y_1)$ and $ B = (x_2, y_2)$, then the magnitude of a vector $\overrightarrow{AB}$ is:

$$| \overrightarrow{AB}| = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}.$$

If we have a position vector, whose initial point is point $O=(0,0)$ end terminal point is $ T = (x, y)$, then the magnitude of the vector is:

$$|\overrightarrow{OT}|= \sqrt{x^2 + y^2}.$$

* ***A zero vector** or** null vector **is a vector whose length is equal to $0$. A zero vector is denoted by $\overrightarrow{0}$.

**A unit vector **is a vector whose length is equal to $1$, however, it can follow any direction. For a vector $\overrightarrow{a}$ of length $|\overrightarrow{a}|$, a unit vector $\overrightarrow{a_0}$ is defined as

$$\overrightarrow{a_0} = \frac{\overrightarrow{a}}{|\overrightarrow{a}|}.$$

.

*Example 1.*

Draw the following vectors in a plane and calculate their magnitude:

1. $\overrightarrow{AB}$ where $ A = (1, 5)$, $ B = (4, 2)$,

2. $\overrightarrow{CD}$ where $ C = (0, 0)$, $ D = (5, – 5)$,

3. $\overrightarrow{EF}$ where $ E = (6, 7)$, $ F = (6, 7)$.

*Solution*:

1. $\overrightarrow{AB} = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2} = \sqrt{(4 – 1)^2 + (2 – 5)^2} = \sqrt{18}$

$\sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}$.

2. $\overrightarrow{CD} = \sqrt{x^2 + y^2} = \sqrt{5^2 + (-5)^2} = \sqrt{50} = \sqrt{2 \cdot 25} = 5\sqrt{2}$.

3. $\overrightarrow{EF} = 0$, because $E=F$.

## The direction of vectors

The direction of a vector is the measure of the angle it encloses with the $y$ – axis. It can be observed as the slope of the line it lies on, this is because its slope is calculated in the same way we calculate the slope of a line. We will mark the direction of a vector with $\varphi$.

If we have a vector $\overrightarrow{PQ}$, $P = (x_1, y_1)$, $ Q = (x_2, y_2)$, then

$$ tan (\varphi) = \frac{y_2 – y_1}{x_2 – x_1}.$$

*Example 2*.

What is the direction of a vector a whose initial point is $ A = (1, 3)$, and terminal point $ B = (3, 5)$?

*Solution*:

By the formula above, we have: $ tan (\varphi) = \frac{y_2 – y_1}{x_2 – x_1} = \frac{5 – 3}{3 – 1} = 1 \Leftrightarrow \varphi = 45^{\circ}$.

## The orientation of vectors

**Vector orientation** is exclusively related to collinear vectors. Collinear vectors are vectors that lie on the same line or on a parallel line to the same. For example, if we have two vectors $\overrightarrow{AB}$ and $\overrightarrow{CD}$, then their orientation can only be equal or contrary.

How can we thus determine the orientation assuming that these two vectors are collinear? First we must take one vector and translate it in a way that its initial point falls into the initial point of the other vector. This means that we translate the vector $\overrightarrow{CD}$ so that $ C = A$. If the points $B$ and $D$ are on the same side of point $A$, then they are both of the same orientation, however, if they are on different sides of point $A$, then they are of contrary orientation.

Vectors with the same orientation:

Vectors with contrary orientation:

Note: Two vectors are equal if they have the same magnitude, direction and orientation. If two collinear vectors are of equal length, although different orientation they are called **contrary vectors**. If $\overrightarrow{a}$ is a vector we are observing, then its contrary vector is denoted by $\overrightarrow{- a}$.

## Vector addition

By adding vectors $\overrightarrow{AB}$ and $\overrightarrow{CD}$ we will get a new vector $\overrightarrow{AD}$ whose initial point is the same as the first addend, and the terminal point of the new vector is the same as the second addend. By applying this rule a triangle accrues. This is why this method of addition is called the **triangle rule**. When we have two vectors that we must to add together, first we translate one vector onto the other one, in a way that the terminal point of the first is the initial point of the second. Then, all that remains is to complete the triangle and mark the orientation of our new vector.

Addition with null vector:

$\overrightarrow{a}$ + $\overrightarrow{0}$ = $\overrightarrow{a}$

$\overrightarrow{0}$ + $\overrightarrow{a}$ = $\overrightarrow{a}$.

**The parallelogram law of vector addition**

If we have two vectors that have the same initial point, then we can use **the parallelogram law of vector** **addition**. We simply consider these two vectors as adjacent sides of a parallelogram, their sum will the diagonal of the parallelogram. The initial point of the sum of these two vectors will be their initial point.

Step by step:

**Properties of addition**

For every two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is valid:

$$\overrightarrow{a} + \overrightarrow{b} = \overrightarrow{b} + \overrightarrow{a}.$$

This means that the addition of vectors is commutative.

For every three vectors $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ is valid:

$$(\overrightarrow{a} + \overrightarrow{b}) + \overrightarrow{c} = \overrightarrow{a} + (\overrightarrow{b} + \overrightarrow{c}).$$

Therefore, addition of vectors is associative.

**Addition of more than two vectors**

Addition of $n$ inter-related vectors $\overrightarrow{A_1 A_2}, \overrightarrow{A_2 A_3}, \overrightarrow{A_3 A_4}, … , \overrightarrow{A_{n-1}A_n}$ is equal to the vector $\overrightarrow{A_1 A_n}$

Note: As we already know, $\overrightarrow{-a}$ is a vector whose orientation is contrary from orientation of vector $\overrightarrow{a}$, however, their magnitude and direction are equal. Using this statement we know how to subtract vectors. If we have two vectors, $\overrightarrow{a}$ and $\overrightarrow{b}$, and we need to subtract $\overrightarrow{b}$ from $\overrightarrow{a}$ $ (\overrightarrow{a} – \overrightarrow{b})$, we simply change the orientation of vector $\overrightarrow{b}$ and add them as such $\overrightarrow{a} + (\overrightarrow{-b})$.

## The multiplication of vectors and real numbers

The product of real number $ k \not = 0$ and vector $\overrightarrow{a}$ is a vector, which we denote as $ k \overrightarrow{a}$, with rules:

- Vectors $\overrightarrow{a}$ and $ k \vec{a}$ are collinear vectors of the same orientation if $ k > 0$ and of contrary orientation if $ k < 0$
- Magnitude of vector $ k \overrightarrow{a}$ is equal to $\mid k \mid \cdot\mid \overrightarrow{a} \mid$
- The product of a vector and zero is a null-vector
- The product of a vector and $1$ is the same vector

For example, if we have a vector $\overrightarrow{AB}$, where $ A = (2, 4)$, $ B = (5, 6)$, calculate $2\overrightarrow{AB}$ and $-2\overrightarrow{AB}$, the product of a vector and a real number will always be situated on the line on which the observed vector lies. This is why the first thing we must do is to draw a line on which the vector $\overrightarrow{AB}$ lies.

Now we need to calculate $ 2\overrightarrow{AB}$. It’s magnitude will be twice that of a vector $\overrightarrow{AB}$. The vector will be situated on the same line and will have the same orientation because $2 > 0$. Now we have all the required data:

Now we need to calculate $-2 \cdot\vec{AB}$. Since $-2 < 0$ this vector will have a contrary orientation to the vector $\overrightarrow{AB}$, double magnitude and the same direction.

**Properties of multiplying a vector with a real number**

It is valid for every two real numbers $k$ and $l$, and for every two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$:

$ k (\overrightarrow{a} + \overrightarrow{b}) = k \overrightarrow{a} + k \overrightarrow{b}$

$ (k + l) \overrightarrow{a} = k \overrightarrow{a} + l \overrightarrow{a}$

$(kl) \overrightarrow{a} = k (l \overrightarrow{a})$

## Linear combination – linear dependence and independence

If $\overrightarrow{a}$ and $\overrightarrow{b}$ are vectors and $\alpha$ and $\beta$ real numbers, then vector

$\overrightarrow{c} = \alpha \overrightarrow{a} + \beta \overrightarrow{b}$

is called the linear combination of vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ with coefficients $\alpha$ and $\beta$.

If $\overrightarrow{a}$ and $\overrightarrow{b}$ are two collinear and not null-vectors with the same orientation where $\overrightarrow{a}$ is $k$ times longer than $\overrightarrow{b}$, then we can write:

This means that one vector can be represented using the other and that $\overrightarrow{a}$ and $\overrightarrow{b}$ are **linearly dependent**.

The vectors $\overrightarrow{a_1}, \overrightarrow{a_2} … , \overrightarrow{a_n}$ are said to be linearly dependent if there exist real numbers $\alpha_1, \alpha_2, … , \alpha_n$ such that:

$\overrightarrow{0} = \alpha_1 \overrightarrow{a_1} + \alpha_2 \overrightarrow{a_2}, … , \alpha_n \overrightarrow{a_n}$.

If linear combination $\alpha_1 \overrightarrow{a_1} + \alpha_2 \overrightarrow{a_2}, … , \alpha_n \overrightarrow{a_n}$ is equal to $\overrightarrow{0}$ only when $\alpha_1, \alpha_2, … , \alpha_n$ are all equal to zero, then it is said that vectors $\overrightarrow{a_1}, \overrightarrow{a_2} … , \overrightarrow{a_n}$ are **linearly independent**.

Every two collinear vectors in a plane are linearly dependent and every two non-collinear vectors are linearly independent.

Every vector in a plain can be presented in a unique way as a linear combination of two non-collinear vectors.

## Vectors in a coordinate plane

Let $E$ be the unit point on the $x$-axis, $F$ the unit point on $y$-axis and point $O$ the origin. Then radius vector $\overrightarrow{OE}$ is equal to unit vector $\overrightarrow{i}$ and radius vector $\overrightarrow{OF}$ equal to unit vector $\overrightarrow{j}$.

Using these two vectors we can present any vector in a plane in a unique way.

If point $P$ has coordinates $ (x, y)$ then the radius vector $\overrightarrow{OP}$ has presentation $\overrightarrow{OP} = x \overrightarrow{i} + y \overrightarrow{j}$.

If $ A_1 = (x_1, y_1)$ and $ A_2 = (x_2, y_2)$ are two points of a plane then: $\overrightarrow{A_1A_2} = (x_2 – x_1) \overrightarrow{i} + (y_2 – y_1) \overrightarrow{j}$.

For example, if we have two points $ A = (1, 3)$ and $ B = (2 , 5)$ then the vector $\overrightarrow{AB}$ is equal to:

$\overrightarrow{AB} = (x_2 – x_1 ) \vec{i} + (y_2 – y_1) \overrightarrow{j} = (2 – 4) \overrightarrow{i} + (5 – 3) \overrightarrow{j} = \overrightarrow{i} + 2 \overrightarrow{j}$.

## Scalar or dot product

Let’s say there are two vectors in a plain, both different from $\overrightarrow{0}$. If they don’t have the same initial point, we simply translate one vector onto the other in a way that they do have the same initial point.

The angle between those two vectors is the smaller angle of the two angles enclosed by the half lines those two vectors are lying on.

If $\overrightarrow{a}$ and $\overrightarrow{b}$ are two vectors different from $\overrightarrow{0}$, the product

$$\overrightarrow{a} \cdot \overrightarrow{b} = |\overrightarrow{a}| \cdot |\overrightarrow{b}| cos \angle (\overrightarrow{a}, \overrightarrow{b})$$

is called the **scalar or dot product** of vectors $\overrightarrow{a}$ and $\overrightarrow{b}$.

**Properties of scalar product**

1. $ \cdot \overrightarrow{a} \cdot \overrightarrow{b} = \overrightarrow{b} \cdot \overrightarrow{a}$, for every two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$

2. $ (\overrightarrow{a} + \overrightarrow{b}) \cdot \overrightarrow{c} = \overrightarrow{a} \cdot \overrightarrow{c} + \overrightarrow{b} \cdot \overrightarrow{c}$, for every three vectors $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$

3. $(\alpha \cdot \overrightarrow{a}) \cdot \overrightarrow{b} = \overrightarrow{a}\cdot (\alpha \cdot \overrightarrow{b}) = \alpha \cdot (\overrightarrow{a} \cdot \overrightarrow{b})$ , for every two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ and $\forall \alpha \in \mathbb{R}$

4. $ \overrightarrow{a} \cdot \overrightarrow{a} = |\overrightarrow{a}| ^2$, for every vector $\overrightarrow{a}$

5. $\overrightarrow{a} \cdot \overrightarrow{b} = 0$ , if $\cos \angle(\overrightarrow{a}, \overrightarrow{b}) = 0$

For unit vectors $\overrightarrow{i}$ and $\overrightarrow{j}$ is valid:

$$\overrightarrow{i} \cdot \overrightarrow{j} = \overrightarrow{j} \cdot \overrightarrow{i} = 0.$$

## Worksheets

**Naming vertexes and vectors** (419.4 KiB, 573 hits)

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