To begin our discussion, we use the word vector to show a displacement. What separates a vector, as a quantity, is that a vector is identified with length (magnitude) and direction. Calculus uses vectors for many other purposes such as velocity or force. In -space, a displacement from a point to is shown by a vector which has a magnitude and directed as an arrow with initial point and terminal We use bold fonts to show vectors in this document and a sharp set of brackets to provide the coordinates of a vector such as .

Using this description, the magnitude of a vector must represent the length of the displacement that is the distance between initial and terminal point of the given vector. We show this by . Any vector with length equal to is said to be a unit vector.

In -space, -space or more general -space for any natural number using terminal points and initial points for displacement, one can construct such arrows. One needs to review the concept of vectors in general, not only in the sense of displacement. This shall be done in a linear algebra course. In the way that we construct vectors, one can start with a different initial and terminal points and get the same displacement. This is why, vectors with the same magnitude and same direction are considered as one vector (or better, a class of vectors). Vector and below are said to be equivalent to each other.

In -space, addition and multiplication by real numbers of vectors are coordinate-wise. In fact, for any two vectors and we define

(1)

to be the resultant vector, or sum of the two vectors and obtained by placing the initial point of at the terminal point of and connecting the initial of with terminal point of .

Suppose is a real number and given and ,

(2)

(3)

Any linear algebra book (or these days any calculus book) provides a list of vector properties. Suppose and are vectors in -space and and are real numbers. Here is the zero vector. We assign the explanation for each of these as an exercise.

1.

2.

3.

4.

5.

6.

7.

8.

We use and to show the basic vectors , and with respect to order.

The notion represents a point in -space as explained in previous section and is a vector. The vector can be drawn in such a way that it has its initial point at the origin with terminal point

Given vectors and we draw the vector and find its magnitude to be

(4)

Try to do this similar exercise. Given and , find and using a geometric argument.

Given a number and a unit vector we can construct the vector with length in direction of . Here, our goal is to find two vectors parallel to the vector and length Let us begin with making a unit vector

(5)

Now a new vector has length and the desired direction. Observe that the vector with the same magnitude and opposite direction also satisfies the given condition that is .

We want to show that is a unit vector. For this, all we have to do is to show the magnitude of this vector is equal to Indeed,

(6)

In general, the vector for any is a unit vector.

Given any real number and a vector in -space, show

(7)

We start by writing and so

(8)

Given any vector we want to show is a unit vector. We evaluate its magnitude

(9)

The vector makes the angle with -axis and has length We want to write in terms of and First we can get a direction by defining the unit vector . Next if we multiply this vector by we get the desired vector . Note that the length of this vector is Now we write this vector as a linear combination of the basic vectors in -space

(10)

Given two vectors and in -space, we define the dot product of and to be

(11)

Note that the result of the dot product of two vectors is a number. For example, dot product of and is

(12)

A set of properties of dot product is below. Suppose and are vectors and is a real number. Here is the zero vector unless said otherwise.

1. .

To see this, write Now using definition of dot product

(13)

and we are done.

2.

3.

4. Here, the first zero is the zero vector.

5.

6. The following statements are equivalent to each other

a) Here, is the number zero.

b) Here, is the vector zero.

We show this part:

a) implies b) Suppose we know Then so This means

(14)

The only way that sum of non-negative numbers is equal to zero is if all of them are zero. This means all coordinates of are equal to zero.

b) implies a) Suppose Then all coordinates of are zero that is . Use definition of dot product to get

(15)

And we are done.

Consider the two vectors shown below.

A difference vector is shown in red. The law of cosines provides us with the fact that

(16)

On the left of this equality we have

(17)

Setting this equal to the right side of equality (6) one gets

(18)

(19)

that is

(20)

One can use this relation to notice, saying that two vectors are perpendicular to each other is exactly saying that their dot product is equal to zero. Therefore, we have a tool to identify if two vectors are perpendicular to each other.

Consider the following two vectors and ; more, consider the vector shadow of on if light is going downward perpendicular to

Observe that to find the coordinate of this shadow vector, we need its magnitude. This is because, we can multiply this magnitude to and get the shadow vector. We call this shadow vector, vector projection of onto and show it by . Now, if is the angle between and

(21)

and so

(22)

Suppose and are any two vectors. Then it is a fact that for the vector we have

(23)

this means

(24)

(25)

Simplify to get

(26)

In fact using the same process with the vector instead, provides

(27)

From these two inequality, we get the well-known Cauchy-Schwarz Inequality

(28)

Suppose and are two given vectors. Then

(29)

This provides the well-known Triangle Inequality which implies on your way back home from school, you should not go anywhere else.

(30)

Find a nonzero vector that is perpendicular to . Our job is to find a vector such that

(31)

We cannot choose all entries zero by the condition of this problem. We cannot choose 2 entries zero, since we will be forced to choose the last one zero as well. But we can choose one of them zero, say the last one. Now, choose the first entry any number other than zero, say We need to choose the second entry such that

(32)

Solve and get So

(33)

Given and find and We can use the formula that we built here

(34)

(35)

Find such that a triangle with vertices , and is a right triangle. We choose to be the vertex where the angle of the triangle is Then the vectors and must be perpendicular to each other. Vector is and vector is . Therefore

(36)

(37)

See the second part of vector fundamentals.