For those of you that have studied physics, you have learned that mass, length, and time is sometimes measured by their magnitude. To describe a force, displacement or velocity, you discovered that you needed a direction in which it acts as well as the magnitude. Putting these two concepts together, you realized that a vector fit the job perfectly. So let us describe a vector in mathematical terms.

FACT:

A vector in the plane is a directed line segment.

FACT:

Two vectors are equal or the same if they have the same length and direction. (see figure 1)

(NOTE: All vectors in this set of supplemental notes will be in bold face.)

AB = CD

 Suppose we want to multiply a vector by a positive number. What we are really doing is multiplying its length by the number.

 FACT:

If c is a nonzero real number and v a vector, the direction of cv agrees with that of v if c is positive and it is opposite to v if c is negative. c is called a scalar and cv is a scalar multiple of v. 

 FACT:

If we multiply a vector v by 0, it produces the zero vector 0. The length of the 0 is 0, and the 0 has no direction.

 FACT:

 Two vectors are said to be parallel if they are nonzero scalar multiples of one another, or, equivalently, if the line segments representing them are parallel.

 FACT:

Whenever a vector v can be written as a sum v = v ₁ + v ₂ of two non-parallel vectors, the vectors v ₁ and v ₂ are said to be components of v.

 The most common algebra of vectors is based on representing each vector in terms of components parallel to the coordinate axes. These basic component vectors have length of one. The basic component vector in the x-direction is i, and the basic component vector in the y-direction is j. i runs from (0, 0) to (1, 0), and j runs from (0, 0) to (0, 1).

The vector ai is a vector of length | a | parallel to the x-axis. If a > 0, it has the same direction as i. If a < 0, then it is opposite in direction from i.

The vector bj is a vector of length | b | parallel to the y-axis. If b > 0, it has the same direction as j. If b < 0, then it has opposite direction from j.

 FACT:

 A vector v can be written in terms of its i and j components. (i.e. v = ai + bj)

DEFINITION:

If v = ai + bj, the vectors ai and bj are the vector components of v in the direction of i and j. The numbers a and b are the scalar components of v in the direction of i and j.

 FACT:

Two vectors are equal if their components are equal. (i.e. ai + bj = ci +dj if and only if a = c and b = d.)

 Let v ₁ = ai + bj and v ₂ = ci + dj, then v ₁ + v ₂ = (a + c)i + (b + d)j and v ₁ - v ₂ = (a - c)i + (b - d)j.

EXAMPLE 1:

Let A = 2i - 3j and B = i + 4j, find A + B and A - B.

SOLUTION:

A + B = (2 + 1)i + (-3 + 4)j = 3i + j

A - B = (2 - 1)i + (-3 - 4)j = i - 7j

EXAMPLE 2:

Let A = i - 3j, B = 2i + j, and C = -4i + 2j find A - B + C.

SOLUTION:

First, find A - B.

A - B = (1 - 2)i + (-3 - 1)j = -i - 4j

Now add C to the result of A - B.

A - B + C = (-4 -1)i + (2 - 4)j = -5i - 2j

EXAMPLE 3:

Let A = 2i + 3j and B = i - 4j, find 3A - 2B.

SOLUTION:

First, let us find 3A. To find 3A, I will multiply the components of A by 3.

3A = 6i + 9j

Now find 2B the same way.

2B = 2i - 8j

Now subtract the results.

3A - 2B = (6 - 2)i + (9 - (-8))j = 4i + 17j

Suppose we are given two points and we want to find the vector that starts at one point and goes to the other. How can we do it? Here is how.

Let P₁ (x ₁, y ₁) and P₂ (x ₂, y ₂) be the two points. I want to find the vector P₁P₂. First of all, order does matter. In fact P₁P₂ ¹ P₂P₁, the go in opposite directions of each other. To find the vector, I will find the difference between the x-values. This will be the scalar component of i. Then I will find the difference between the y-values. This will be the scalar component of j.

P₁P₂ = (x ₂ - x ₁)i + (y ₂ - y ₁)j

EXAMPLE 4:

Find P₁P₂ and P₂P₁ if P₁ (3, 4) and P₂ (2, -5).

SOLUTION:

P₁P₂ = (2 - 3)i + (-5 - 4)j = -i - 9j

P₂P₁ = (3 - 2)i + (4 - (-5))j = i + 9j

Notice that P₁P₂ = - P₂P₁. This implies that they are going in opposite directions.

EXAMPLE 5:

Find P₃P₄ if P₃ is the point (1, 3) and P₄ is the midpoint of the line segment P₁P₂ joining P₁ (2, -1) and P₂ (-4, 3).

SOLUTION:

First of all, we need to find the midpoint of the line segment P₁P₂. This will be the point P₄.

Now to find the vector P₃P₄.

P₃P₄ = (-1 - 1)i + (1 - 3)j = -2i - 2j

EXAMPLE 6:

Find the vector from the point A to the origin where AB = 4i - 2j and B is the point (-2, 5).

SOLUTION:

What we want to find is the vector from A to the origin, O, but we are not given the point A. We are given the vector AB and the point B. From this, we should be able to determine A.

Let A be the point (x, y), then the vector AB has the form (-2 - x)i + (5 - y)j. Since we are given the vector AB, set the two equal to each other and solve for x and y.

(-2 - x)i + (5 - y)j = 4i - 2j

Since the two vectors are equal to each other, then the components are equal to each other.

-2 - x = 4 ® -x = 6 ® x = -6

5 - y = -2 ® -y = -7 ® y = 7

Therefore, the point A is (-6, 7). Now to find the vector AO.

AO = (0 - (-6))i + (0 - 7)j = 6i - 7j

Recall that in the beginning of these notes, I mentioned that in physics we need to measure a quantity, and we call this measurement the magnitude. Another word for magnitude is length, so when we say that we want to find the magnitude of a vector, we really are finding the length of the vector.

FACT:

The magnitude or length of v = ai + bj is

EXAMPLE 7:

Find the magnitude of the vector v = 3i + 4j.

SOLUTION:

Suppose that a scalar c ¹ 0 is multiplied to v = ai + bj, the resulting vector cv = c (ai + bj) = (ca)i + (cb)j. This is called scalar multiplication.

How does a scalar multiple affect the length of a vector? Consider the vector v = ai + bj and the scalar c.

cv = (ca)i + (cb)j

Now find the length of cv.

FACT:

If c is a scalar and v is a vector, then | cv | = | c || v |.

FACT:

The zero vector is the vector 0 = 0i + 0j and | 0 | = 0.

FACT:

Any vector whose length is 1 is called a unit vector.

We have already worked with two unit vectors, and they are the vectors i and j There is an infinite number of unit vectors. Consider the vector u = (cos q )i + (sin q )j. u is a unit vector because cos ² q + sin ² q = 1. So given any q, we can find the related unit vector u.

EXAMPLE 8:

Determine the vector formed by rotating i p / 4 radians in the positive direction.

SOLUTION:

Using the unit vector u = (cos q )i + (sin q )j and the fact that q = p / 4, plug q into the formula.

What if we are not given q, but we are given some ordinary vector v. How do we make v into a unit vector? Now recall that a vector has two attributes assigned to it. One is the length, and the other is the direction. We know how to find the length of a vector, but how do we find the direction? Furthermore, what connection does the direction of a vector have with a unit vector?

Consider a vector v ¹ 0, then

So v/ | v | is a unit vector in the direction of v.

FACT:

If v ¹ 0, then

1.

v/| v | is a unit vector in the direction of v;

2.

The equation

expresses v in terms of its length and direction.

 EXAMPLE 9:

Express v = 4i - 2j in terms of its length and direction. 

SOLUTION:

First, I will find the length of v.

Now, I will find the direction.

Now to express v in terms of its length and direction.

 FACT:

A vector is parallel to a line if the segments that represent the vector are parallel to the line. 

 FACT:

The slope of a vector that is not vertical is the slope shared by the lines parallel to the vector.

 FACT:

If a ¹ 0, the vector v = ai + bj has a well-defined slope which is b/a.

 FACT:

A vector is tangent or normal (perpendicular) to a curve at a point if it is parallel or normal to the line that is tangent to the curve at that point.

The question that you should be asking is why would we want to find the vectors that are tangent and normal to the curve? What is the practical application of it? Later on in this course, we will talk about the curvature of a curve, and the normal and tangent vectors will play an important role in finding it.

 EXAMPLE 10:

Find the tangent and normal vectors to the curve x ² + y ² = 25 at the point (3, 4). 

SOLUTION:

First of all, there will be two tangent vectors and two normal vectors to any point on the curve.

Now let us find dy/dx. This will give us the slope of the tangent line to the curve at the given point. I will have to use implicit differentiation to find dy/dx.

Now to find the slope at the point (3, 4).

Now to write the equation of the tangent vector.

v_T = 4i - 3j

Here are the unit tangent vectors to the curve x ² + y ² = 25 at the point (3, 4).

The perpendicular slope is 4/3, so v_N = 3i + 4j.

Here are the unit normal vectors to the curve x ² + y ² = 25 at the point (3, 4).

EXAMPLE 11: 

Find the tangent and normal vectors to the curve 3x ² - 4xy + 2y ² - 2 = 0 at the point (0, 1). 

 SOLUTION:

Now to determine the tangent vector to the curve.

v_T = i + j

Here are the unit tangent vectors to the curve at the given point.

The perpendicular slope is -1 and v_N = - i + j.

Here are the unit normal vectors to the curve at the given point.