## Summary

IIntroduction to vectorsACoordinate plane and complex plane reviewBDefinition of vectorsCRepresentation of vectorsIIProperties and operations with vectorsAOperations withs vectorsBColinear and equal vectorsIIIDot products and orthogonal vectorsADot productsBOrthogonal vectors## Introduction to vectors

### Coordinate plane and complex plane review

#### Coordinate plane

The coordinate plane is the plane with two axes intersecting at a right angle: the x -axis and the y -axis. Points in the coordinate plane are labeled by their x -coordinate and their y -coordinate.

#### Complex plane

The complex plane is the plane with a real axis and an imaginary axis. A complex number of the form a+bi is plotted on the imaginary plane via its real component and imaginary component.

### Definition of vectors

#### Vector

A vector is the information of a magnitude and a direction. A vector is pictured as an arrow.

- Its direction is the direction the arrow points.
- Its magnitude is the length of the arrow.

The following graphic contains several vectors.

#### Head and base of a vector

- The endpoint of the vector with the arrow is called the head of the vector.
- The opposite endpoint is called the base of the vector.

Two vectors are said to be equal or equivalent if they have the same magnitude and direction. They do not have to have the same base and head.

The following graphic contains several vectors which are all equivalent to one another.

Let P and Q be points in a plane. The vector whose base is P and whose head is Q is denoted \vec{PQ}.

### Representation of vectors

#### Algebraic representation of vectors

Let a,b be real numbers. We let \langle a,b \rangle be the vector in the xy -plane whose base is \left(0{,}0\right) and whose head is \left(a,b\right).

In the complex plane, we can let \langle a,b\rangle be the vector whose base is at 0 and whose head is at a+bi.

The notations \left(a,b\right) and \langle a,b\rangle represent two different mathematical objects. The notation \left(a,b\right) is a point and \langle a,b\rangle is the vector starting from the point \left(0{,}0\right) and ending at the point \left(a,b\right).

#### Vector connecting two points in the Cartesian plane

Let P=\left(a,b\right) and Q=\left(c,d\right) be points in the Cartesian plane. Then:

** \vec{PQ}=\langle c-a,d-b \rangle **

#### Magnitude of a vector

Let a,b be real numbers. The magnitude of the vector \langle a, b\rangle is:

** ||\langle a, b\rangle||=\sqrt{a^2+b^2} **

Consider the following vector:

\langle 3,-4\rangle

The magnitude of the vector is:

||\langle 3,-4\rangle||=\sqrt{3^2+\left(-4\right)^2}=\sqrt{25}=5

The formula for the magnitude of a vector is derived from the Pythagorean Theorem.

In the above graphic, a right triangle is formed from the vector \langle a,b \rangle. The length of the base is a and the length of the side is b. According to the Pythagorean Theorem, the length of the vector is \sqrt{a^2+b^2}.

#### Unit vector

A unit vector is a vector of magnitude 1.

Consider the following vector:

\left\langle \dfrac{\sqrt{3}}{2},1/2\right\rangle

Then:

\left|\left|\left\langle \dfrac{\sqrt{3}}{2},1/2\right\rangle\right|\right|=\sqrt{\left(\dfrac{\sqrt{3}}{2}\right)^2+\left(\dfrac{1}{2}\right)^2}=\sqrt{1}=1

Therefore, the vector \left\langle \dfrac{\sqrt{3}}{2},1/2\right\rangle is a unit vector.

#### Scalar multiplication

Let a,b be real numbers and consider the vector \langle a,b \rangle. If \alpha is a real number, then:

** \alpha\langle a,b\rangle=\langle \alpha a, \alpha b\rangle **

Consider the following vector:

\langle 2, -3\rangle

Then:

4\langle 2,-3\rangle=\langle 8,-12\rangle

The process of scalar multiplication changes the size of the vector. In the following graphic, the vector \langle 2, -3\rangle is in black and the scaled vector 4\langle 2,-3\rangle =\langle 8,-12\rangle is in red.

Every vector can be written as the scalar multiple of a unit vector. Let \vec{v} be a nonzero vector. Then \vec{u}=\dfrac{1}{||\vec{v}||}.\overrightarrow{v} is a unit vector and:

** \vec{v}=||\vec{v}||\vec{u} **

Consider the vector \langle 2{,}3\rangle. Then:

||\langle 2{,}3\rangle ||=\sqrt{4+9}=\sqrt{13}

Therefore the following vector is a unit vector:

\dfrac{1}{\sqrt{13}}\langle 2{,}3\rangle=\left\langle \dfrac{2}{\sqrt{13}},\dfrac{3}{\sqrt{13}}\right\rangle

And:

\langle 2{,}3\rangle=\sqrt{13}\left\langle \dfrac{2}{\sqrt{13}},\dfrac{3}{\sqrt{13}}\right\rangle

## Properties and operations with vectors

### Operations withs vectors

#### Adding vectors

Suppose \langle a,b \rangle and \langle c, d \rangle are vectors. Then:

** \langle a,b \rangle +\langle c,d \rangle =\langle a+c, b+d \rangle **

Consider the sum of the following two vectors:

\langle 2, 3\rangle +\langle 1{,}9\rangle=\langle 2+1, 3+9 \rangle=\langle 3, 12 \rangle

#### Parallelogram law part I

If two vectors \vec{u} and \vec{v} are represented by the adjacent sides of a parallelogram, then the sum \vec{u}+\vec{v} is the vector which is the diagonal of the parallelogram and whose base is the base of \vec{u} and \vec{v}.

#### Subtracting vectors

Suppose \langle a,b \rangle and \langle c, d \rangle are vectors. Then:

** \langle a,b \rangle -\langle c,d \rangle =\langle a-c, b-d \rangle **

Consider the difference of the following two vectors:

\langle 2, 3\rangle -\langle 1{,}9\rangle=\langle 2-1, 3-9 \rangle=\langle 1, -6 \rangle

#### Parallelogram law part II

If two vectors \vec{u} and \vec{v} are represented by the adjacent sides of a parallelogram, then the difference \vec{u}-\vec{v} is the vector which is the diagonal of the parallelogram connecting the head of \vec{v} with the head of of \vec{u}.

### Colinear and equal vectors

#### Colinear vectors

Two vectors are colinear if they have the same direction.

The following graphic contains several vectors which are all colinear to one another.

Two vectors \vec{v} and \vec{u} are colinear if and only if there is a real number \alpha such that:

** \vec{v}=\alpha\vec{u} **

Equivalently, the vectors \langle a,b\rangle and \langle c,d \rangle are colinear if and only if there is some number \alpha such that:

** \langle a,b\rangle =\langle \alpha c, \alpha d\rangle **

The vectors \langle 2, -3\rangle and \langle 8,-12\rangle are colinear because:

\langle 2, -3\rangle =\dfrac{1}{4}\langle 8, -12\rangle

#### Criterion for vectors to be colinear

Two vectors \langle a,b \rangle and \langle c,d \rangle are colinear if and only if:

** \dfrac{1}{\sqrt{a^2+b^2}}\langle a,b\rangle =\dfrac{1}{\sqrt{c^2+d^2}}\langle c,d\rangle **

or

** \dfrac{1}{\sqrt{a^2+b^2}}\left \lt a,b\right \gt =\dfrac{-1}{\sqrt{c^2+d^2}}\left \lt c,d\right \gt **

Consider the following vectors:

\langle 4{,}0\rangle and \langle 8{,}0\rangle

Then:

\dfrac{1}{\sqrt{4^2+0^2}}\langle 4{,}0\rangle=\langle 1{,}0\rangle

And:

\dfrac{1}{\sqrt{8^2+0^2}}\langle 8{,}0\rangle=\langle 1{,}0\rangle

Therefore, these two vectors are colinear.

Two vectors are equal or opposite each other if and only if they are colinear and have the same magnitude.

## Dot products and orthogonal vectors

### Dot products

#### Dot product

Let \langle a, b \rangle and \langle c,d \rangle be vectors. The dot product of these two vectors is:

** \langle a, b \rangle\cdot \langle c,d \rangle=ac+bd **

Consider the following vectors:

\langle 2, 3\rangle and \langle 5{,}5\rangle

The dot product of these two vectors is:

\langle 2, 3\rangle\cdot \langle 5{,}5\rangle =2\left(5\right)+3\left(5\right)=10+15=25

### Orthogonal vectors

#### Orthogonal vectors

Suppose \vec{v} and \vec{u} are vectors. \vec{v} is orthogonal to \vec{u} if the angle obtained by moving the base of \vec{v} to the base of \vec{u} is 90^\circ.

The following graphic contains a pair of orthogonal vectors.

#### Criterion for orthogonality

Let \langle a, b \rangle and \langle c, d \rangle be vectors. Then \langle a, b \rangle is orthogonal to \langle c, d \rangle if and only if:

** \langle a, b \rangle\cdot \langle c,d \rangle=0 **

Consider the vectors:

\langle 2, 1 \rangle and \langle -3, 6 \rangle

The dot product of these two vectors is:

\langle 2, 1 \rangle\cdot \langle -3, 6 \rangle=2\left(-3\right)+1\left(6\right)=-6+6=0

Therefore, the vectors \langle 2, 1 \rangle and \langle -3, 6 \rangle are orthogonal.