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Vectors

• A vector is a way to talk about magnitude (length/size) and direction (angle) at the same time.
• We can visualize a vector as a directed line segment-a length with direction.
• Vectors are normally denoted by an overhead arrow u  or put in bold face u. [And once in a while it's just assumed: u.] Vectors are usually named with lowercase letters, but just like variables, any symbol can be used in theory. Most often you will see u and v.
• If we know the vector's location in the plane, we can give the vector algebraically in component form. For example, u = 〈3,4〉. We normally use angle brackets 〈,  〉 to denote vectors, but we will often also see parentheses (,   ). [After we learn about unit vectors, we'll also talk about using i, j.]
• We denote the magnitude (length) of a vector u by  |u|. [Also sometimes written as ||u|| .] If we know the component form of a vector, we can figure out its length by the Pythagorean theorem. This gives the formula
 | →u |  = | 〈a, b 〉|   = √ a2 + b2 .
• If we know the component form of a vector, we can figure out its angle through trigonometry. We normally talk about direction as the counter-clockwise angle from the positive x-axis (just like the unit circle), but sometimes our reference location changes. In any case, always draw a picture before trying to figure out or use angles.
• If we know the magnitude and angle of a vector, we can figure out its component form. First draw a sketch to help you see it, then use trigonometry.
• If we have a vector u, we can scale it to a different length by multiplying by a scalar: a real number. Algebraically, the scalar just multiplies each component. In general, for a scalar k and vector u = 〈a, b 〉,
 k →u = k ·〈a, b 〉 = 〈ka, kb 〉.
• A unit vector is a vector with a length of 1. It still has a direction, but it must have a magnitude of exactly 1. We can create a unit vector out of any vector u by dividing u by its length, |u|.
Unit vector:   →u

 | →u |

Note that the above has the same direction as u, but is length 1.
• We can combine two (or more) vectors through addition (or subtraction). We simply add them component-wise: the horizontal components from each vector combine, as do the vertical components.
 →u + →v = 〈u1, u2 〉+ 〈v1, v2 〉 = 〈u1 + v1,   u2 + v2 〉
• We can see this idea of combining vectors geometrically as well. Adding vectors is done by placing the tail of one at the head of the other. Then we draw in a vector from the original start to the final end.
• We can put together the idea of combining vectors with the idea of unit vectors to get a new way to express a vector's components. We start by creating two standard unit vectors, one horizontal and one vertical:
 i = 〈1, 0 〉,                      j = 〈0, 1 〉.
With this, we can express any vector in terms of i and j:
 〈3, 4 〉 = 3 i + 4 j,               〈4.7, π〉 = 4.7 i + πj.
• We can also talk about a zero vector, denoted 0, that has 0's in all its components. It has no length (and so its direction does not matter).
• While vectors have many operations that are similar to things we are already used to, there is no analogue to multiplication. There is no good way to define multiplying two vectors, so we do not define vector multiplication. [Still, there is an operation somewhat similar to multiplication called the dot product. We'll explore that and what it means in the next lesson.]
• Motion in a medium is the combination of the object's motion vector relative to the medium and the medium's motion vector:
 →v + →v = →v .
• Vectors can work in arbitrarily high dimensions. While all of the above was only done in two dimensions, a vector can have any number of components. By the way we defined scalars and vector combination, everything we've discussed so far about vectors still works fine. They might get confusing to picture in higher dimensions, but everything still makes sense. [In three dimensions, we have another standard unit vector:
 k = 〈0, 0 , 1 〉           (and in three dimensions  i = 〈1,0,0〉,     j = 〈0,1,0 〉)
• For any n-dimensional vector
 →x = 〈x1,  x2, …,  xn 〉,
the length of the vector is simply the square root of the sums of each of its components squared:
 | →x | = √ x1  2 + x2  2 + …+ xn  2 .

Vectors

Given that u = 〈4,  5 〉,  v = 〈2,  −3 〉,  and w = 〈−2,  0 〉, find the below.
 →u + →v →u + →v + →w
• When adding vectors, we add component-wise, that is, we add the first components together to make a new first component, we add the second components together to make a new second component, and so on.
• For the first expression, we have
 →u + →v =     〈4,  5 〉+ 〈2,  −3 〉    =     〈4+2,    5 −3 〉    =     〈6,  2 〉
• For the second expression, we have
 →u + →v + →w =     〈4,  5 〉+ 〈2,  −3 〉+ 〈−2,  0 〉    =     〈4+2−2,    5 −3 +0 〉    =     〈4,  2 〉
u+v = 〈6,  2 〉,       u + v+w = 〈4,  2 〉
Given that u = 〈−7,  3 〉 and v = 〈5,  10 〉, find the below.
 8 →u − 1 2 →v
• The number in front of the vector is called a scalar. When a scalar is in front of a vector, it multiplies the vector. Applying a scalar to a vector is as simple as multiplying each of the components by the scalar (like distribution).
• For the first expression, we have
 8 →u =     8 ·〈−7,  3 〉    =     〈8 ·(−7),    8·3 〉    =     〈−56,  24 〉
• For the second expression, we have
 − 1 2 →v =     − 1 2 ·〈5,  10 〉    = ⎛⎝ − 1 2 ⎞⎠ ·5, ⎛⎝ − 1 2 ⎞⎠ ·10 = − 5 2 ,  − 5
8 u = 〈−56,  24 〉,        − [1/2] v = 〈 −[5/2],  − 5 〉
Given that u = 〈1,  2 〉,  v = 〈3,  −1 〉,  and w = 〈−2,  −2 〉, find the below.
 5 →u −3 →v 3( →u + 2 →v ) −5 →w
• Vector addition (or subtraction) is done component-wise: each portion of the vector only interacts with the equivalent portion in the other vector. Scalar multiplication (a normal, real number in front of the vector) is the same as distributing that number through multiplication.
• Start off by applying the scalars to their vectors, then add the results together. Here's the first expression:
 5 →u −3 →v =     5 ·〈1,  2 〉− 3 ·〈3,  −1 〉    =     〈5,  10 〉+ 〈−9,  3 〉    =     〈−4,  13 〉
• We work similarly with the second expression, but pay attention to the parentheses-they mean the same thing as we're used to. Expressions inside of a parentheses get priority, but they can also be distributed on to.
 3( →u + 2 →v ) −5 →w =     3 (  〈1,  2 〉+ 2·〈3,  −1 〉 ) −5 ·〈−2,  −2 〉    =     3(  〈1,  2 〉+ 〈6,  −2 〉 ) + 〈10,  10 〉

 = 3(  〈7,  0 〉 ) + 〈10,  10 〉    =     〈21,  0 〉+ 〈10,  10 〉    =     〈31,  10〉
[Alternatively, instead of figuring out the vector inside of the parentheses first, we could distribute right at the start:
 3( →u + 2 →v ) −5 →w =     3 →u + 6 →v − 5 →w =     3 〈1,  2 〉+ 6 〈3,  −1 〉−5 〈−2,  −2 〉    =     〈31,  10 〉
Both methods work fine, and they will give the same result. Use whichever you feel more comfortable with.]
5u −3v = 〈−4,  13 〉,        3(u + 2 v) −5 w = 〈31,  10 〉
Let u = 〈3,  4 〉 and v = 〈−10,  −2 〉. Show how we can geometrically find the resultant vector of the below expression.
 2 →u + →v
• To draw the first vector of u have the tail start at the origin and the head of the vector will fall on the point (3,  4), since the vector is 〈3,  4 〉.
• Next, we need to apply the scalar of 2 to u. The scalar scales the vector by that amount. In this case, we scale it by a factor of 2, so we will double the length of the vector. This means we will now draw from the origin out twice as far to (6,  8).
• Finally, we need to add the vector v. Adding a vector geometrically is simple: instead of having the second vector also "start" at the origin, it starts from where the first vector "ended" (the first vector's head). Thus, we will place v so that it's tail is at the point (6,  8), and it will travel 〈−10,  −2 〉 to have its head at the point (−4,  6). The resultant vector of the expression 2u + v has its tail at the origin and its head goes to that same final point.
• We can check our geometric work that we did graphically with algebra. If we did this algebraically, we would have
 2 →u + →v =     2 ·〈3,  4 〉+ 〈−10,  −2 〉    =     〈6,  8 〉+ 〈−10,  −2 〉    =     〈−4,  6 〉
This is exactly what we got from figuring it out geometrically, as you can see in the picture for the answer.
Find the magnitude and angle of the below vector.
 〈−2,  −5 〉
• Finding the magnitude of any vector is quite easy: just take the square root of the sum of each component squared and added together.
 | 〈−2,  −5 〉|     = √ (−2)2+(−5)2 = √ 29
• Finding the angle that the vector has is not much harder, but it is very important to make a diagram. If you don't make a diagram to help yourself see what's going on, you might make a mistake. We want to find the angle θ below:
• We can use trigonometry to solve for the angle that the vector makes, but it will not directly give us the value of θ. None of the inverse trigonometric functions can output an angle of that size. Instead, the safest way to do this problem is to ask ourselves, "What angle does the vector make with the horizontal axis?" Thus, we want to know the value of the green `?' symbol in the below diagram. To help us do this, we can treat the vector as a right triangle, and put down the length of its two legs.
• We can find the value of the `?' angle by simple right triangle trigonometry:
 tan (?) = 5 2 ⇒     ? = tan−1 ⎛⎝ 5 2 ⎞⎠
Plugging that in to a calculator, we get that    ?=68.199°. Remember though, our ultimate goal is to find θ. Notice that to make θ, it has to go over the entire top two quadrants (180°), then add on the angle of `?'. Therefore, the angle of θ is given by
 θ    =     ? + 180     =     68.199 + 180     =     248.199
Whenever you need to find the angle of a vector, make sure to draw a diagram first. It's extremely useful to have a visual reference so you can make sense of the angle you're looking for.
Magnitude: √{29},     Angle: 248.199°
The angle u has a magnitude of |u| = 19 and angle of θ = 147°. Give the component form of u. (Round your answer to three decimal places.)
• Finding the component form of a vector if we know the magnitude (length) and angle is quite easy: we simply use trigonometry. The cosine function allows us to find the x-value and the sine function allows us to find the y-value. If you have difficulty understanding how the below steps work, try drawing a sketch of the vector, then create a right triangle where the vector, the x portion, and the y portion make the three sides.
• The distance form the origin is the magnitude of the vector, and we know the angle that the vector is on. Thus, if we want to find the horizontal x component, we have
 cosθ = x r ⇒     cos147° = x 19 ⇒     19 ·cos147° = x
Plugging into a calculator, we get that x = −15.935. [If you get a different value on your calculator, make sure your calculator is set to degrees mode and not radians.]
• Working similarly to find the vertical y component, we have
 sinθ = y r ⇒     sin147° = y 19 ⇒     19 ·sin147° = y
Plugging into a calculator, we get that y = 10.348.
• Finally, because the vector u is comprised of both components, we have to put them together into a single vector:
 →u =     〈x,  y 〉    =     〈−15.935,   10.348 〉
u = 〈−15.935,   10.348 〉
Find the unit vector pointing in the same direction as the vector u = 〈36,  77 〉.
• A unit vector is a vector with magnitude 1. We can create a unit vector out of any vector (that is, create a vector that points in the exact same direction, but now has a length of 1). We do this by finding the magnitude of the vector, then dividing the vector by that.
Unit vector:  →u

 | →u |
Since [1/(|u|)] is just a scalar, this scales the vector to a length of 1, no matter what length u started at.
• Begin by finding |u|:
 | →u |     = √ 362 + 772 =     85
• Once you know the magnitude, divide the original vector by it:
Unit vector:  →u

 | →u |
=    〈36,  77 〉

85
=

36

85
,   77

85

Unit vector in same direction as u: 〈 [36/85],  [77/85] 〉
A large boat is being pulled along by two tugboats. Tugboat A is pulling along the boat with a force of 7000 N (N →newtons, the metric unit for force) and at an angle of 30° north of east. Tugboat B is pulling the boat with a force of 5000 N and at an angle of 50° south of east. What is the resultant force? (Give your answer in component form.) [The resultant force is the sum of the forces acting upon an object.]
• We can express the force exerted by each tugboat as a vector. Once we know the force vector of each tugboat, we can sum them together to find the combined effect of both pulls-the resultant force.
• Let us consider east and west as being the positive and negative x-directions respectively. Similarly, let us consider north and south as being the positive and negative y-directions respectively. With this set up, we can find the component form of each tugboat's force vector by using trigonometry. Tugboat A:
 cos(30°) = x 7000 ⇒     x = 6062.2 ⎢⎢ sin(30°) = y 7000 ⇒     y = 3500
Thus tugboat A has a force vector of FA=〈6062.2,   3500 〉. Tugboat B:
 cos(−50°) = x 5000 ⇒     x = 3213.9 ⎢⎢ sin(−50°) = y 5000 ⇒     y = −3830.2
Thus tugboat B has a force vector of FB = 〈3213.9,   −3830.2 〉.
• To find the resultant force, we simply add together the force vectors for each tugboat on its own:
 →F A + →F B =     〈6062.2,   3500 〉+ 〈3213.9,   −3830.2 〉    =     〈9276.1,   −330.2 〉
〈9276.1 N,   −330.2 N 〉
A crate weighing 700 N is suspended from a pair of cables A and B, as in the diagram. Using the diagram, figure out the magnitude of the tension in each cable.
• To do this problem, we must realize a deceptively simple fact: the crate is not moving. If the crate is not moving, that means the total force on the crate must come out to nothing. Since we can express force as a vector, that means the total force on the crate must be the zero vector:
 Total force on crate: 〈0,  0 〉
• The total force on the crate is the sum of all the other forces being applied to the crate. Let us call the force applied by cable A the vector A. Similarly, we can call the force applied by cable B the vector B. Notice that gravity is also pulling down on the crate (in the form of its weight); let's call that g for right now. Therefore, the sum of these three force vectors must result in the total force that we just talked about:
 →A + →B + →g = 〈0,  0 〉
• While we can't figure out the component form of A or B (yet), we can connect the magnitude (length/hypotenuse) of each vector to its components through trigonometry. For cable A, we can find the length of the horizontal component as
cos(52°) = xA

 | →A |
⇒     xA = |

A

| cos(52°)
Similarly, the length of the vertical component is
sin(52°) = yA

 | →A |
⇒     yA = |

A

| sin(52°)
From this we can create the component form of the vector A. However!, there is one slight issue. Notice that the cable will pull to the left on the crate. Thus, instead of a positive value for the x-component of A, it must be a negative value, since it pulls to the left. Put this all together:
 →A = 〈− | →A | cos(52°),     | →A | sin(52°) 〉

We can do a very similar thing to find B (although its x component will be positive, since it pulls the crate to the right):
 →B = 〈| →B | cos(39°),     | →B | sin(39°) 〉
• We can also figure out the component form for the force of gravity. Gravity only pulls down, so all of the weight is pulling in the y-direction, none in the x. Thus, for gravity, we have
 →g = 〈0,  −700 〉.
From earlier, we know
 →A + →B + →g = 〈0,  0 〉,
so we can plug in the vectors we now know:
 〈− | →A | cos(52°),   | →A | sin(52°) 〉+ 〈| →B | cos(39°),   | →B | sin(39°) 〉+ 〈0,  −700 〉 = 〈0,  0 〉
That means we know that the sum of each component type comes out to be 0.
• Doing this for the x-components first, we get
 − | →A | cos(52°) + | →B | cos(39°) + 0 = 0     ⇒     | →B | cos(39°) = | →A | cos(52°)
Doing this with the y-components next, we get
 | →A | sin(52°) + | →B | sin(39°) − 700 = 0     ⇒     | →A | sin(52°) + | →B | sin(39°) = 700
• We now have a system of equations that we can solve by substitution. Solve for one of the vector magnitudes in the first equation, then plug that in to the other one and solve:
 | →B | cos(39°) = | →A | cos(52°)     ⇒     | →B | = | →A | · cos(52°) cos(39°)
Plugging in:
 | →A | sin(52°) + ⎛⎝ | →A | · cos(52°) cos(39°) ⎞⎠ sin(39°) = 700

 | →A | · ⎛⎝ sin(52°) + cos(52°) ·sin(39°) cos(39°) ⎞⎠ = 700

|

A

| = 700

 sin(52°) + cos(52°) ·sin(39°) cos(39°)
=     544.09

Now that we know |A| = 544.09, we can plug in to find |B|:
 | →B | = (544.09) · cos(52°) cos(39°) =     431.03
Cable A's magnitude: 544.09 N,    Cable B's magnitude: 431.03 N
A boat is in the water and needs to go directly north at 10 m/s. However, the water is flowing from the NE to the SW at an angle of 20° south of west and a speed of 2 m/s. What speed and angle must the boat have relative to the water to achieve its goal velocity?
• We will approach this problem based on the idea of motion in a medium. The total motion of an object is affected by the motion of the medium and the object's motion relative to the medium it is in. Mathematically, using vectors, we have
 →v object + →v medium = →v total motion .
For this problem, the motion of the water combined with the motion of the ship relative to the water (what the problem asked for) must equal the ship's goal velocity.
• To continue, we need to put everything in vector component form so we can add things together. The goal velocity of the ship is going entirely to the north at 10 m/s, so we can write that as
 Total motion:    〈0,  10 〉
• Next, we figure out the velocity vector for the water. Using trig, we can find the length of the horizontal (xw) and vertical (yw) components:
 cos(20°) = xw2 ⇒     xw = 1.879 ⎢⎢ sin(20°) = yw2 ⇒     yw = 0.684
We can now put this together to find the velocity vector for the water. However!, we must pay attention to our diagram. The water's motion is moving to the left (west) and down (south). This means that the horizontal and negative components must be negative. While the lengths are the above, we have to pay attention and realize that those components will be negative because of their directions.
 Velocity of water:    〈−1.879,  −0.684 〉
• We don't know anything about the velocity of the boat relative to the water yet, so let's just call it v, and we can name its components as v = 〈xv, yv 〉. With this set up, we can now add together the velocity of the water and the velocity of the boat relative to the water and get the total velocity:
 〈−1.879,  −0.684 〉+ 〈xv, yv 〉 = 〈0,  10 〉
• Since vectors add component-wise, we don't have to worry about first components affecting second components and vice-versa. Thus we can break this into two separate equations:
 −1.879 + xv = 0 ⎢⎢ −0.684 + yv = 10
Solve for xv and yv:
 xv = 1.879 ⎢⎢ yv = 10.684
Therefore, the ship's velocity vector relative to the water must be
 →v = 〈1.879,   10.684 〉
• Finally, the problem asked for v's angle and speed. Speed just means the magnitude of the velocity vector, so we find that as
 | →v | = √ 1.8792 + 10.6842 =     10.848
To find the angle, we can just use simple trigonometry:
 tan(θ) = 10.684 1.879 ⇒     θ = tan−1 ⎛⎝ 10.684 1.879 ⎞⎠ =     80.025°
Ship's speed relative to water: 10.848 m/s,    Ship's angle of motion relative to water: 80.025° north of east

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Vectors

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Introduction 0:10
• Magnitude of the Force
• Direction of the Force
• Vector
• Idea of a Vector 1:30
• How Vectors are Denoted
• Component Form 3:20
• Angle Brackets and Parentheses
• Magnitude/Length 4:26
• Denoting the Magnitude of a Vector
• Direction/Angle 7:52
• Always Draw a Picture
• Component Form from Magnitude & Angle 10:10
• Scaling by Scalars 14:06
• Unit Vectors 16:26
• Combining Vectors - Algebraically 18:10
• Combining Vectors - Geometrically 19:54
• Resultant Vector
• Alternate Component Form: i, j 21:16
• The Zero Vector 23:18
• Properties of Vectors 24:20
• No Multiplication (Between Vectors) 28:30
• Dot Product
• Motion in a Medium 30:10
• Fish in an Aquarium Example
• More Than Two Dimensions 33:12
• More Than Two Dimensions - Magnitude 34:18
• Example 1 35:26
• Example 2 38:10
• Example 3 45:48
• Example 4 50:40
• Example 4, cont.
• Example 5 1:01:32

Transcription: Vectors

Hi--welcome back to Educator.com.0000

Today, we are going to talk about vectors; this starts an entirely new section for us.0002

We are getting into really new territory; we are going to talk about vectors, and then later on we will be talking about matrices.0007

But first, let's talk about vectors: when we talk about the force on an object, we need to know two things:0011

the magnitude of the force--how hard it is being pushed, and the direction--which way it is being pushed.0016

If we are going to talk about something being pushed, we can't just talk about how hard it is being pushed.0023

We also have to know which way it is being pushed; it is very different to push something this way than it is to push something this way.0027

But even if we know the direction, we also have to know how hard it is.0032

We are pushing really hard versus pushing very slightly--there are big differences here.0034

We need to know both magnitude and direction; we need to know how hard and which way.0039

Now, one number on its own won't be enough to get across both pieces of information.0044

So, this leads us to the idea of a vector--a way to be able to talk about magnitude and direction at the same time--0048

be able to talk about both of these things in one piece of information.0054

Vectors are massively useful: they are used throughout math; they are in the sciences everywhere, especially in physics;0058

they are in engineering, computer programming, business, medicine, and more fields.0065

Pretty much, if a field even vaguely uses math, it is going to use vectors; so this is a really useful thing that we are talking about in this lesson.0069

Now, I want to point out that this lesson will use some basic trigonometry to figure out angles.0076

So, make sure you have some familiarity with how trigonometry works, so that you can understand what is going on when we are figuring out angles.0080

All right, let's go on: the idea of a vector: a vector is just trying to get across the idea of length and direction.0087

Graphically, if we look at a picture of it, it is exactly that: it is a directed line segment, a length with direction.0095

It is both of these things at once.0102

So, in this case, we are starting at (0,0), and we have this length here; and the arrow at the end says that we are going in this direction.0104

So, we have a chunk of length, and we see which way it is pointed.0112

It is pointed in a very specific direction; it has some angle to it.0116

How do we call out a vector, if we want to talk about a vector?0120

Normally, we denote it with an overhead arrow like this: u with an arrow above it says that we are talking about the vector u.0123

Or we can put it in bold face with a bold u.0131

And once in a while, if we are talking about nothing but vectors, and there is nothing else showing up,0135

sometimes it will just be assumed; but we are pretty much never going to see that--not in this course.0140

Vectors are normally shown with lowercase letters; however, just like variables, you can use any symbol.0145

But for the most part, we will stick with lowercase letters: u and v are very common letters for talking about them.0150

One last thing: we use this u with the arrow on top, and that is how we will be denoting it in this course, although you might see bold face in textbooks.0157

When you are actually writing it by hand, I write it like this, where I don't really have a full arrow.0166

I have more of a harpoon, where it is just an arrow on one side--just this sort of arrow flange on one side.0171

That is how I write it; you could write it with an actual arrow on top--there is nothing wrong with that.0178

But I am lazy, like us all, and so I just tend to write it like that, because it is the fastest way that I know to write it.0182

But it still gets across the idea of a vector.0189

So, this is a perfectly fine way to write it by hand when you are working it out.0191

But when we have it actually written out in this lesson, we will have it like that with an arrow on top.0194

We can also give the vector algebraically by its location in the plane.0200

We have these nice rectangular coordinates; we can break that into components.0204

We call this the component form: we have a horizontal amount of 3 and a vertical amount of 4.0209

So, we say that its horizontal component is 3, and its vertical component is 4.0215

u, the vector u, equals (3,4); the first component is the horizontal one, and the second component is the vertical,0220

just like when we are talking about points in the plane.0227

We normally use angle brackets (these are angle brackets, because they are at an angle) to denote vectors.0230

But we will often also see parentheses; parentheses are very common, as well.0238

I personally actually tend to use parentheses more; but most precalculus/math analysis courses tend to use angle brackets.0241

So, I am teaching with angle brackets; but personally, when I am just doing math on my own, I often tend to use parentheses.0249

But either of them is just fine.0254

After we learn about unit vectors, we will see one more way to talk about the component form, using i and j.0256

But we will leave that until a little bit later, once we have actually talked about unit vectors.0261

If we know the component form of a vector, we can figure out its length by the Pythagorean theorem.0265

Remember: the Pythagorean theorem says that the hypotenuse to a triangle, squared, is equal to both of the legs, squared and then added together.0270

So, that means that we can take the square root of both sides of the equation.0284

We have that the hypotenuse is equal to the square root of each leg, squared and added together, the square root of (leg squared + leg squared).0287

In this case, we know that our vector is (3,4); so we have that one leg is 3, and one leg is 4.0295

So, we work this out; √(32 + 42) becomes √(9 + 16), or √25, so we get 5.0306

So, we see that the length of this vector is 5.0312

When we want to denote the magnitude of a vector, when we want to talk about the length of a vector,0315

the magnitude of a vector, we normally use the word "magnitude" to talk about length.0320

But in either case, with "length" or "magnitude," we are just talking about how long the vector is.0324

We normally use these vertical bars on either side of it, just like we do with absolute value.0329

And we will talk about that in just a moment--why it is like absolute value.0336

Also, sometimes you will see it written ||u||, vector u with double bars around it.0339

In either case, whether it is single bar or single bar, what we are talking about is how long that vector is, if we measure it from the origin out to its tip.0346

If u is equal to (a,b), then the length of u, the bars on either side of u (which would also be bars on either side of (a,b),0354

because that is just a vector, as well), will give us the square root of a2 + b2,0363

based on this exact same reasoning that the length of our hypotenuse is equal to the square root of each leg, squared and added together.0368

So in this case, a and b are just there and there on our vector.0376

Now, I would like to point out, really quickly, why in the world we are using vertical bars on either side,0381

just like we did with talking about absolute value.0385

Let's think about that for a bit: let's think about...when we work with absolute value, say we have 0 here, and here is +5, and here is -5.0389

Well, if we talk about the absolute value of 5, and we talk about the absolute value of -5,0398

in both cases, we are going to get the number 5 out of it.0403

We get 5 in either case, because what the absolute value is telling us is: it is saying how far you are from 0--how far you are from the origin.0408

So, the reason why the absolute value of 5 is 5 is because it takes 5 units of length to get from 0 to 5.0417

And that is the exact same reason why the absolute value of -5 is 5: because it takes 5 units of length to get from 0 to -5.0424

They go in different directions, but it is a question of how far away it is.0431

They are both 5 units away from the origin.0435

The same thing is going on when we are dealing with a vector.0437

When we talk about the size of a vector, what the length of a vector is, we are asking how far the vector goes out from the origin.0440

In both the case of the absolute value and the magnitude of a vector, what we are saying is, "How far are you from the origin?"0450

We put bars around the u because it is basically doing the same thing as bars around a number with absolute value.0457

Bars around a number are saying, "How far are you from 0?"; bars around a vector are saying, "How far are you from the origin?"0463

So, they are both a question of length, effectively.0468

All right, direction and angle: if we know the component form of a vector, we can figure out its angle for trigonometry.0472

So, once again, we have u = (3,4); so, tanθ = sideopposite/sideadjacent, when we have a right triangle.0477

And we have that here, because we know we are dealing with a nice rectangular coordinate system.0487

We have a 3 here and a 4 here; so our side opposite to our angle θ is this side here, and our side adjacent is this side here.0492

So, that gets us tanθ = 4/3; at this point, we can take the arctan of both sides0502

(the inverse tangent--arctan and tan-1 mean the same thing; I like arctan).0510

So, θ = arctan (tan-1) of 4/3; we plug that into a calculator or look it up,0514

and we see that that is approximately equal to 53.13 degrees; so we see that that is the angle in there.0520

Now, we normally talk about direction as the counterclockwise angle from the positive x-axis.0527

It is a little bit confusing, but it just means that we start over here at the positive x-axis,0532

and then we just keep turning counterclockwise until we get to whatever thing we are trying to measure out to,0536

at which point we stop, and that is the measure of the angle that we are going with.0542

It is just like we did with the unit circle: we started at the x-axis positive, and then we just kept spinning0546

until we got to whatever angle we were trying to get to.0552

But sometimes we won't be using the positive x-axis.0555

That is what we normally end up using, but sometimes the reference location--sometimes--will change,0558

and we won't be using this nice positive x-axis that we are used to using.0564

But maybe we will be talking about how far we are off of the vertical y-axis to the right,0567

or maybe some other thing, like we have this other angle here created,0574

and we are talking about how far we are off of this other thing, some angle, going clockwise...0579

So who knows how it is going to be done?0584

We have to pay attention to what the specific problem is and how it is set up.0585

It generally will be the positive x-axis, but it is not an absolute guarantee; you have to pay attention.0588

This means, in any case, whatever you are dealing with, that I recommend always drawing a picture,0593

because that is going to give you a way to be able to see what is going on.0599

Draw a picture before you try to figure out or use angles; it will help you get a sense of what is going on, and really help clarify things.0602

All right, now, what if we have the component from...0609

We talked about (previously) taking the magnitude and angle and getting component forms.0612

I'm sorry, we talked about the exact opposite of that.0619

So now, we are going to say, "What if we know the magnitude and angle, and we want to get the component form?"0621

So, if we are trying to find the component form from the magnitude and angle, we can figure that out.0626

The first thing first: always draw a sketch--it will help keep things clear and help you understand what is going on.0631

And then, from there, you just use trigonometry.0636

So, in this case, we have a length of 6; the length of our vector u is 6; our angle θ is 120 degrees.0638

We remember from trigonometry that the cosine of 120, the cosine of this angle here, is going to be equal to0647

the horizontal amount of change, up until it drops down on its crossover, divided by the length of the segment we are dealing with.0652

Cosine of 120 = x/6; multiply both sides by 6; 6cos120...cos120 is the same thing as -1/2, so 6 times -1/2...gets us -3.0662

So, -3 is our value for x.0673

A similar thing is going on for our y, the vertical component; sin(120), the sine of the angle we are dealing with,0676

is equal to the vertical component, divided by the length of the entire segment (6 in this case).0684

So, sin(120) = y/6; multiply both sides by 6; sin(120) is √3/2, so it simplifies to 3√3 = y.0689

At that point, we just take both of these pieces of information.0699

We slot them into our u, and we now have component form.0701

Our u is equal to (-3,3√3).0704

Now, I did tell us to do it with cosine 120; but most of us are probably used to dealing with trigonometry for degree angles under 90 degrees.0708

It is a little bit confusing--maybe just a bit--to be able to work with things over 90 degrees.0719

So, we also could have converted this to an angle of 120, so a total of 60 degrees over here, because it is 180 on the whole thing.0723

So, we could have figured out that inside of the triangle is 60 degrees.0732

Now, that is going to cause a little bit of difference here, because cos(120) brought that negative to the table,0735

because indeed, our x is going to the left, and remember: this is the negative direction for x.0740

Going this way gets us negative x values, when we go to the left, just like going down gets us negative y values.0745

If that is the case, we have cos(60), what we can figure out...0753

If you have things inside of a triangle, you are just going to figure out the length of each of those sides.0756

So, cosine of 60 equals x/6; so we will get 6 times...cos(60) is just 1/2, so we get that 3 = x.0760

But notice: what we are figuring out is...we are figuring out here to here as x, which is not the same thing as the x-value, as in the horizontal location.0772

What we are figuring out is just the length of the side.0780

We are figuring out how far it is from here to here; but we have to figure out the horizontal coordinates--not just the length, but the actual x-value.0783

If that is the case, what we are really figuring out is the absolute value of x--how long our x is.0792

And that means that, at the end, we have to look at this picture; and that is why we draw these sketches that are so handy.0798

We look at the picture, and we see that our x has a length of 3.0804

However, it is going to the left; so that means it has to be a negative thing.0809

By paying attention, we see that it is -3...not just length 3, but -3; but it is because we drew a sketch that we are able to see this.0816

You have this option: you can either just use the angle, and be able to be really good at trigonometry;0824

or you can make it in a slightly simpler form, where it is easier to understand.0828

But you have to be paying attention and realize that you have to set the sign at the end.0831

I have to pay attention, based on this sketch: is this length going to come out to be positive?0835

Is this length going to come out to be negative?0839

You really have to pay attention to that.0841

All right, we are ready to move on to a new idea.0843

Scaling by scalars: if we have a vector u, we can scale it to a different length by multiplying by a scalar, a real number.0845

A scalar is just some number; it is not a vector--it is just an actual, real number.0852

Algebraically, the scalar just ends up multiplying each component.0858

Let's say we start with this red vector on our picture, which is u = (3,-2).0862

We can scale this by some other thing, by just multiplying it by some number.0868

Like, for example: we could multiply u by 2, so we have 2; and now that is the blue vector that we see there.0872

Now notice that the blue vector is double the length of the red vector, because it is 2 times u.0877

So, it just takes that length, and it scales it by a factor of 2; it doubles that original length.0884

Algebraically, we just end up having this 2 multiply on the 3, and multiply on the -2; so we get (6,-4) as the components.0890

We can take this; we can try multiplying by something else--how about a negative number?0899

What would a negative get us? A negative ends up going in the opposite direction.0903

So, if we have a negative u, then the positive direction is the way it normally goes; negative will be the opposite way.0906

So, negative goes in the opposite direction; so now, we are going opposite the direction that u went, as we can see pictorially here.0913

And for how it is going to end up having it, it is just a negative now on each of the components.0919

So, negative cancels out there, and we have (-3,+2).0924

And that is what we see on our picture.0929

We could also have something that is not just a whole integer number, like, say, -3/2.0932

We end up having 3/2 times the length--1 and 1/2 the length of the original vector--0939

but then also, in the negative direction--opposite the direction of the first one.0946

Once again, numerically, it just ends up being -3/2 times the first number times the second number; so we get (-9/2,3).0950

That is what it ends up being.0959

So, algebraically it just multiplies each component; graphically it is a question of stretching and maybe also flipping.0960

In general, for any scalar k and some vector u that is (a,b), k times u...that is the same thing as k times (a,b),0966

so that is the same thing as just that k getting distributed to both the a and the b, so it gives (ka,kb).0976

Great; a unit vector is a vector with a length of 1; "unit vector" just means a length of 1.0983

It still has a direction; it can have any direction, but it has to have a magnitude of exactly 1.0990

Its magnitude--its length--how long it is...it is 1 in terms of length.0996

We create a unit vector out of any vector u by dividing u by its length.1000

Remember: if we divide by a number, that is the same thing as just using a scalar; we are multiplying by 1 over the number we are dividing by.1004

So, previously, we could scale; if we know its length is 10, and then we divide it by 10, we are going to get something with length 1 now.1011

We have some vector u that is some length; but then we divide it by that length, because the magnitude of u is just its length.1020

So, we have that; we divide by that; we have scaled it back to what it would be if it was just at a length of 1.1029

So, our unit vector is the original vector, divided by the length of the vector.1035

This is the same direction as u; so it will be in the same direction, but it is going to have a length of 1--it will just be length 1.1042

Now, the use of having a unit vector--the reason why it is so great to have a unit vector--1050

is that we can just take it later and multiply it by any scalar k that we want.1054

And we will know that we will have created a vector that is length k, because we started at length 1.1059

You scale that by k, and we are going to go with 1 times k; so we will just be at vector length k.1063

And we are going to be scaling in the direction we already started with, so we know that we will be in the unit vector's direction.1068

This gives us an ability to easily create vectors of any length we want in a known direction.1074

And this ability comes in really, really handy in a lot of situations, and that is why unit vectors are important.1079

All right, we can combine two or more vectors through addition or subtraction.1084

It is actually not that difficult; we just add them component-wise.1090

All of your first components go together; the second components go together, and so on and so forth, like that.1096

In this case, if we add (3,4) and (2,-5), then what we end up doing is: we have the 3 and the 2 getting combined.1102

And they become 3 + 2, because we are adding (3,4) + (2,-5).1110

The same basic thing: the 4 and the -5 get combined, and so 4 + -5 is 4 - 5.1116

We simplify that, and we get (5,-1); we are just taking the numbers at the beginning and adding them together,1124

and the numbers at the end and adding them together.1130

We are doing it component-wise: each component stays and only mixes with components that are the same type1132

(first components with first components, second components with second components, and so on).1137

So, in general, for some (u1,u2)...really, that is just saying the first component of u,1142

and the second component of u...and then (v1,v2), the first component of v1147

and the second component of v--if we add u and v, u + v, (u1,u2) + (v1,v2),1153

it is going to be the exact same thing: the u1 and the v1, the first components,1158

adding together, and then the second components, u2 and v2, adding together, as well.1163

It is a very similar thing if we end up subtracting: u - v is just going to be u1,1172

the first component of u, minus the first component of v, v1;1177

u2 - v2 (the second component of u minus the second component of v).1180

The components stay in their locations, but they do either addition or subtraction, depending on whether it is addition or subtraction; it makes sense.1184

We can also do this geometrically and get an understanding of what is going on,1191

because vectors are supposed to represent this thing--this directed line segment.1194

So, we can see this idea geometrically: you add vectors by placing the tail of one at the head of the other.1198

So, for example, in this one, you have u in red, (3,4); and then we have v in blue, (2,-5).1205

So, we put (3,4) out to here; that goes out to (3,4).1212

And then, this one goes over 2 and goes down 5; so (2,-5) is 2 to the right, down by 5.1218

We put those together, and we get (5,-1), which is exactly what we see here in the purple vector that is the combination of them.1229

So, u + v is the purple vector that we see there.1236

We can see this algebraically and geometrically, and the two ideas match up completely.1239

Not only that, but we get the same resultant vector.1245

This resultant, what we get when we put the two things together, is the same whether it is u + v or v + u.1248

It doesn't matter if we end up doing the blue one first, and then the red one.1254

We can add them geometrically in any order we please, and it still comes out to be the same thing.1260

We end up seeing this parallelogram; we can see this as another way of adding things--creating this parallelogram out of them, or just head-to-tail.1264

But we see geometrically what is going on.1271

All right, now that we have these two ideas, we can put together the idea of combining vectors1275

with the idea of unit vectors to get a new way to express a vector's components.1280

We start by creating two standard unit vectors, which is really just a fancy way of saying "things that make sense and are kind of fundamental."1284

One horizontal, one vertical: i = (1,0); it is a unit vector that is purely horizontal;1292

and j, which is (0,1), is a unit vector that is purely vertical.1302

This one is just one unit long that way; and this one is just one unit long that way--purely vertical.1308

And that is what we are seeing there.1317

One other thing: if we want to write this bold thing, we can't really write bold on our paper; that is very difficult.1318

You can end up writing it like an i, but instead of putting a dot on it, you put that little arrow on top (or in my case, the harpoon on top).1324

The same thing with the j: you can make the j and then put a little arrow on top.1331

And so, that is another way of talking about these units vectors, i and j, if you want to.1334

All right, with this, we can express any vector in terms of i and j.1339

If we have (3,4), well, that is the same thing as having 3 i's, three of these unit vectors that are horizontal,1343

plus four of these unit vectors that are vertical; so that is what we have there.1350

We can combine them; we can break them up into 3 horizontal motions plus four vertical motions.1354

And we get to the same thing as if we had just done (3,4), all at once.1359

3, our first component, matches up with the i's, because that is just our way of saying our horizontal, since our first component is our horizontal.1363

And then, our 4 matches up with the 4j, because that is just the same thing as saying 4 vertical, or 4 units' worth of vertical.1371

And this also works for numbers that aren't just whole integers, as well,1379

because 4.7 times i just scales i by a factor of 4.7, so it will be in the same place horizontally.1381

And π--we can also scale by a factor of π; as long as it is a real number, we can scale by it; so πj goes in there; great.1390

All right, we can also talk about a zero vector; we denote the zero vector with that same arrow top on top of a 0.1397

That has 0 in all of its components; so it will be nothing but 0's as the vector.1404

It has no length; it is 0--it just lives at the origin.1408

And so, since it has no length, its direction doesn't matter.1412

So, here is an example: (0,0)...the zero vector is (0,0).1414

It is just sitting at the origin; its distance from the origin is nothing, because it is currently at the origin.1421

Notice: for any vector u whatsoever, if we add u and the zero vector together, we will just end up being there.1426

We will not have moved anywhere, because head-to-tail we end up going someplace,1432

and then we don't move, because the zero vector doesn't move at all.1436

u - u: if we subtract u from itself, we will end up going out and then coming right back, so we will end up getting 0.1439

And then finally, if we take any vector and multiply it by a scalar of 0, we will have some length,1446

and then we bring that length to 0; so at a length of 0, we have the zero vector; great.1450

All right, at this point, we have talked about a lot of different ideas with vectors.1455

And we can turn this into a bunch of properties.1457

Don't worry too much about understanding all of these properties right away.1459

They will just make more sense, and you will be used to using them.1463

The beauty of all of these properties is that they are very much what we are already used to using with the real numbers.1465

So, vectors, as long as you remember to keep them in this form of components only interacting with other components,1470

are very similar to working with numbers in many of the ways we are already used to.1475

Let's talk through some of these properties.1478

u + v is the same thing as v + u; this is the idea of commutativity, that 5 + 8 is the same thing as 8 + 5.1481

We are used to that with the real numbers.1488

We also have associativity, (u + v) + w is the same thing as u + (v + w).1490

3 + 5 + 4 is the same thing, if you add the 3 and the 5 first, or if you add the 5 and the 4 first.1496

3 + 5, then add 4, or 3 + (5 + 4) (where you add the 3 second)--it doesn't matter which way you do it;1502

so once again, it is very similar to doing it with normal numbers.1509

k times l...if we have two scalars, k and l, times u, well, that is the same thing as k times an already-scaled lu.1513

So, we can either multiply our scalars together and then multiply the vector;1520

or we can have them multiply the vector, each one after another--the same thing, either way, which is pretty much what we would expect.1523

k times (u + v) is nice; it distributes: k times vector u plus vector v is k times vector u, plus k times vector v.1530

It also distributes in the other way: k + l times vector u: the vector can distribute out onto them, so we have k times vector u, plus l times vector u; great.1539

If we take u, and we add it to the zero vector, we end up just getting u; it has no effect.1548

u - u is going to get us back to the zero vector.1553

And then, a couple more: 0 times u is going to give us the zero vector; 1 times u has no effect--1557

we are scaling by just what we are already at; and then, -1 times u is going to flip us to the negative version;1563

it will just cause everything in there to become negative.1568

The only one that might be a little bit confusing is ku, the magnitude of a scaled u, k scaled on u, k scalar times vector u.1570

The magnitude of that is equal to the absolute value of k, times the magnitude of u.1581

Let's look at why that is the case.1588

A really quick, simple example: let's consider if we had (0,1).1590

We have this vector here that goes from here out to a length of 1.1596

So, we could scale it by k; and let's say that k is equal to -2.1603

So, we scale this; here is our u: u = (0,1), so we could scale it by k = -2; -2 times u will get us the same thing,1609

but now it is going to be flipped, and it will be twice the length.1625

We are going to go down two units now.1629

So, whereas the first u went up by 1 unit (it had a length of 1), this has a length of 2.1631

The fact that we are going down doesn't make it negative length; length always is positive.1640

So, it has a length of 2, which is why we have the magnitude of (0,-2): well, that ends up being equal to +2, as we can see from this diagram right here.1644

But this -2 times u--if we had separated this out into -2 times (0,1), what u started as,1658

well, we could break this, by this rule, into the absolute value of our scalar, times the length of our initial thing,1672

which would give us positive 2 times 1--the same thing.1682

So, what it is doing is saying that the reason why we have absolute value on the scalar here1685

is because it doesn't matter that we are flipping and pointing in a new direction;1690

ultimately, length is always going to come out to be positive.1694

So, we can't let a negative k cause our result to come out as a negative length, because that just doesn't make sense.1697

So, we have to figure out a way for it to always stay positive, and that is why we have this absolute value here.1703

All right, there is no multiplication between vectors.1708

Even with everything we have seen so far about vectors, there has been no mention of vector multiplication, other than scalars multiplying on vectors.1712

But other than that, we haven't talked about vector multiplication.1720

That is because there is no good way to define vector multiplication.1722

There is just no way to really do it that is going to make sense.1727

So, we could make up some numerical way to multiply vectors: some vector times some vector makes some other vector.1730

But it would probably be geometrically meaningless; we can't really come up with a good way that is going to have some deep geometric meaning.1736

And that is the problem here: while we could come up with something numerically,1743

we want all of this stuff to have a geometric connection, and all of this other stuff has.1746

It makes sense to combine vectors; we are doing one vector, and then we are doing another vector.1750

We are doing two pieces of motion.1754

Or we stretch them: we have some piece of motion, and then we just elongate it or shrink it or flip it.1756

Those things make sense geometrically; but what would it mean geometrically--1760

what would it mean as a picture--to multiply a line segment by another line segment?1763

It just doesn't really make sense; and because of that, we do not define vector multiplication.1767

There is just no vector multiplication, pretty much, to talk about.1772

Now, all that said, there is an operation similar to multiplication that is called the dot product.1775

That is different, though, because it will take two vectors and multiply them together (although it won't multiply them together)--1781

it will take two vectors, and this vector dot, this other vector, will give us a scalar.1786

It will give us a single real number.1790

Don't worry about that too much now; we are not going to talk about it in this lesson.1792

But we will explore it in the next lesson; so we will see it soon.1795

But for right now, there is just no way to multiply vectors.1799

And even later on, once you see something that is kind of close to it,1802

you will see that it is very different from actually multiplying vectors and getting a new vector out of it.1805

All right, motion in a medium: a really common use of vectors is to analyze motion--1810

the location of an object, the velocity of an object, the acceleration of an object.1816

However, what happens if something is moving relative to a medium, like water or air?1820

Say we have a boat in a river, and so the boat is moving up the river.1826

But at the same time, the water in the river is moving in another direction.1830

We have to do something to take this into account.1833

So, the object is moving relative to the medium, but the medium itself is also moving, like the boat in the water.1835

So, to understand this, let's consider a fish swimming in an aquarium.1843

I love this as an example.1847

First, we have that the aquarium is completely still; we have some table like this;1849

and imagine that my arm here is the aquarium, and so here is the table.1854

The fish is here, and the fish is swimming forward; the fish swims forward, and it gets to the other end of the aquarium.1858

That is how it starts in this picture here: the fish is the only thing moving.1863

But what happens if we don't just let the fish be the only thing moving?1867

If we grab the aquarium, and we actually slide it to the slide as the fish is swimming--1871

if we grab the aquarium, and we move it, we are going to see the aquarium move like this while the fish is moving like this,1877

because the fish is moving inside of the aquarium, but now the entire aquarium is also moving.1884

We see that the fish moves, but at the same time, the aquarium moves over.1889

So, it is very different from the world where the fish ended over here.1893

Now, the fish manages to move at the same time as the aquarium is moved.1896

So, we have to take both of these things into account.1900

We can see this pictorially here: our fish is moving to the side, but at the same time, the box is moving to the side, as well.1902

The fish manages to get over to the left, but the box is now way over to the right.1911

So, if we are going to talk about where the fish has gotten to, the velocity of the fish--1916

anything where we want to talk about the fish's motion and analyze the motion of the fish--1919

we have to take both of these things into account.1924

We have to take into account the aquarium's motion, but also the fish moving inside of the aquarium.1926

It is not enough to talk about just one of them; we have to combine these two ideas.1931

Motion in a medium is the combination of the object's motion vector relative to the medium,1935

the fish relative to being inside of an aquarium, and then the medium's motion vector--how the aquarium is moving.1940

How does the fish move in the aquarium? How does the aquarium move in the larger world?1948

That is what we mean by relative to the medium; it is just whatever the thing is inside of,1952

and then how the thing that you are inside of is moving.1956

So, we can break this down into the velocity of the object, plus the velocity of the medium, is equal to the velocity of the total motion.1960

The fish relative to the table is the addition of the velocity of the fish in the water, plus the velocity of the aquarium,1967

as vectors, because the velocity vector in one case is going to be positive;1976

in the other...for one of them, it will be positive; for the other one, it will be negative,1980

because they are going to be pointing in opposite directions.1983

All right, we can also talk about more than two dimensions.1986

At this point, we have only seen vectors in two dimensions; but we can expand this idea to any number of dimensions we want.1990

For example, a three-dimensional vector could be (5,-2,3)--no problem.1995

We will just keep putting in more components.1999

By the way we define scalars and vector combinations, everything we have discussed so far about vectors still works just fine.2001

It might get a little confusing to picture in our head in higher dimensions.2007

But everything still makes sense; it is hard to picture higher than three dimensions,2010

because we are used to living in a three-dimensional world, but it still makes sense in terms of the algebra of what is going on.2013

That is great; also, one little thing: if we are talking about three dimensions...you remember that i and j--2019

we could talk about unit vectors as an alternate way of talking about component form.2025

There is another standard unit vector; there is k, which is (0,0,1).2028

So, i is the first component; j is the second component; k is the third component.2033

With this, we can express three-dimensional vectors with (i,j,k), so (5,-2,3) would become 5i - 2j + 3k.2037

5 becomes 5i; -2 becomes -2j; 3 becomes 3k; great.2046

So, we can combine them in terms of standard unit vectors, as well.2051

We can also talk about the magnitude of something that is higher than two dimensions.2055

It might seem a little surprising at first, but it turns out that it is actually really easy to figure out the magnitude-- the length--of a vector in any dimension.2058

Consider the n-dimensional vector x, where it is x1, the first component of x,2065

comma, x2, the second component of x, all the way up until we get to xn, the nth component of x.2072

Now, it turns out that the magnitude of our vector x is just the square root of the sums of each of its components, squared.2078

Now, that seems a little bit confusing, but it makes sense.2085

The length of our vector x is equal to the square root of the first component squared, plus the second component squared,2087

plus...all the way up until the nth component, our last component, squared.2093

So, that seems really surprising, the very first time we see this.2098

We will actually explore why this is the case, and it will make sense why this has to always be the case,2100

if we expand the idea that we will see in Example 3 of higher dimensions.2104

Just think, "Oh, yes, that will just keep stair-stepping up, and that is why we see this square root of all of the components, squared and added together."2108

But you can also just memorize this formula, if you want to.2114

And it will be just fine, too, and work out.2118

All right, let's see some examples.2120

First, given that u = (1,3), v = (4,2), w = (-5,1), what are each of the following?2122

(1,3)--if we are talking about u + v, then that is the same thing as talking about (1,3) + (4,2).2129

So, remember: we add the components together, so it will be 1 + 4, because they are both the first components;2138

and then 3 + 2, because they are both of the second components.2143

1 + 4 gets us 5; 3 + 2 gets us 5 also, just by chance; so our answer here is (5,5).2146

Next, we have 6u: well, 6 times...our vector is (1,3), so the 6 distributes effectively.2154

It is not quite distribution--it is a little bit different--but it has the exact same effect and feel.2162

So, the 6 multiplies each of the components; so we have 6, comma, 6 times 3 is 18; so (6,18) comes out of that.2167

All right, we can probably start doing these scalars in our head; they are not too hard to do.2175

1/2 times v...well, v was (4,2), so 1/2 times (4,2) is going to produce 2 (1/2 of 4 is 2), and 1/2 of 2 is 1.2179

So, we have (2,1) for 1/2v; and then, minus...our w was (-5,1).2189

We can distribute this negative here; so this becomes positive; this will become positive; this will become negative.2195

We distribute that negative into there; and so, 2 + 5 is 7; 1 - 1 is 0.2200

There we go; and the last one: if we have 2u - 3v + w...2208

All right, 2 times u: 2 times (1,3) will become (2,6); minus...3 times (4,2) will become (12,6), plus w: w was (-5,1); great.2211

(2,6) -...so that would make that -12 + -12, comma, -6, plus (-5,1).2229

At this point, we could add them all together; we could add them one by one; it doesn't really matter how we approach this.2243

Let's just add the first two; so 2 and -12 becomes -10; 6 and -6 becomes 0; plus...bring down the rest of it...(-5,1)...2248

So, (-5,1) + (-10,0): -10 and -5 becomes -15; 0 and 1 becomes positive 1; and there it is.2259

So, that is the basics of vector addition and multiplication by scalars.2269

It is very similar to what we are used to doing with numbers normally.2273

It is just that everything stays inside of its slot; they only interact with other things from the same slot as them.2276

First slots interact; second slots interact; and if it is higher than two dimensions, third, fourth, fifth...whatever slots interact.2282

All right, the next example: If u = 4, v = 6, and the two vectors make angles of θu = 30 degrees, θv = 120 degrees,2289

to the positive x-axis, what is the component form of u + v? its length? its angle?2297

All right, let's do u in red; over here, we first draw a sketch to be able to figure this out.2302

So notice: if we are going to figure out u + v, if we are going to get the component form of u + v,2309

well, it is hard to add angles and lengths together.2313

We could draw...well, u is an angle of 30 degrees, so we will go out like this.2316

And then, v is 6 at 120 degrees, so we will be a little bit longer...6...120 degrees...like that.2322

And so, our final thing will end up being this; and we could measure what that is.2329

But we would have to have really, really precise stuff; we would have to have a really accurate ruler,2333

and be doing this with a protractor that was really good, and be really, really careful to get all this.2338

So, it is not the sort of thing where we could draw it out and get a very good answer.2342

So, our first step is to get a component format of each of them, because once we have a component form for u2345

and a component form for v, it is easy to add them together.2350

And we add them together, and then we can figure out the length and the angle.2353

So, our first step is to get a component form.2355

u = 4; our θu is 30 degrees; and remember, it was with the positive x-axis.2358

So, our angle is 30 degrees like this.2363

So, if we want to break down u into its x-component and its y-component, well, then, we know that cos(30)...2369

remember, its hypotenuse was 4, so cos(30) is going to be equal to the x-component of u,2378

the first component of u (here is ux); the side adjacent...cos(30) = side adjacent, ux,2388

over the hypotenuse, 4; so 4 times cos(30) = ux.2396

cos(30) is just the same thing as √3/2, so we have 4(√3/2); so we have 2√3 = ux.2403

A very similar thing is going on if we want to figure out what uy is.2413

It will be sin(30) = uy/4; multiply by 4 on both sides; 4sin(30) = uy.2417

4sin(30)...sin(30) is just 1/2, so 4(1/2)...so we have 2 = uy.2428

So, at this point, we have that u equals 2√3, its x-component, comma, and its y-component, 2.2436

And that is what our values for u are.2448

Now, I want to point something out before we keep moving.2452

Notice right here: we were starting at sin(30) = uy/4, but we always get to this 4 times sin(30).2454

If you know the length of your vector, and you know what angle it is at, you can actually just hop to length of the thing,2463

times cosine (or sine, if it is side adjacent or side opposite, respectively) of the angle.2471

And that will just give you what that side adjacent or side opposite is, respectively.2478

So, we will end up doing that on the next one.2482

If it was a little bit confusing, notice the parallel to how we just did it with the u vector while we are working on the v vector.2484

But it is a really great way of being able to do this really, really quickly--2492

well, not really, really quickly, but it does help speed things up.2497

And it is a good trick, because you end up seeing this quite a lot.2500

So, we have 120 degrees here; we are at 120 degrees.2503

Now, I think it is going to be a little bit easier to figure out in terms of this angle here,2507

because it is easy to work under 90 degrees, so that is 60 degrees, because 120 + 60 = 180.2513

So, now we want to figure out the vx component, the horizontal component, and the vy, the vertical component.2521

So, vx is going to be equal to sine...it is 60...we have 60 in the angle,2529

and it is going to be side adjacent, so it is going to be the length, 6, times cosine, side adjacent of the angle involved.2536

However, there is one thing that we need to notice.2545

What we are figuring out here is the length of that side of a triangle, because the angle is inside of a triangle.2547

So, it is up to us to pay attention: is it going to be positive? is it going to be negative?2553

Normally, we have the x direction going negative as we go to the left.2556

That is negative x going this way, so that means we have to have a negative sign on our x-component for v.2561

Otherwise, it won't make sense, because we can see from the picture that it is going negative on the horizontal.2568

So, we have to pay attention to this; we have to notice this stuff happen, because otherwise it will be a mistake.2573

OK, -6 times cos(60)...cos(60) is 1/2, so we get -3; vx = -3; vy = 6.2577

This one is positive, because it is going up...sine of 60 is √3/2, so that gets us 3√3.2589

So, we have that the x-component is -3; the y-component is 3√3.2599

So, our v vector equals (-3,3√3).2606

Now, if we want to add these two together, u + v, it is simply a matter of adding them together.2613

Our u was (2√3,2); our v was (-3,3√3); we add them together; we have (2√3 - 3, 2 + 3√3).2620

And if we want to, we could get what that is approximately, in terms of a decimal number, although that is exact.2636

That is a perfect thing; this will just be approximate, because it is a decimal of square roots: (0.464,7.196).2641

All right, so at this point, we want to figure out what its length is.2651

The length of u + v is going to be equal to the square root of each of its components, squared and then added together.2654

So, its two components were 0.464, squared, plus 7.196, squared.2665

What does that end up coming out to be?2673

We work that out, and that ends up coming out to be...sorry, I couldn't find it in my notes...7.211.2676

It comes out to be approximately 7.211.2688

If we want to figure out what the angle is that it is at, the first thing we probably want to do is draw a quick diagram, so we can see it.2690

So remember: it was at this one right here, in terms of its components.2696

0.464 is just a little bit over the x-component, and then 7.196 up...so it is like this.2701

So, we can see that that should be what the angle is like.2707

So remember: tan(θ) is equal to the side opposite, divided by the side adjacent.2710

The side opposite in this case will be the vertical height, divided by the side adjacent of 0.464, the horizontal amount.2717

Take the arctan of both sides, the inverse tangent of 7.196 over 0464.2726

We plug that into our calculator or look it up in a table; and that comes out to be approximately 86.31 degrees.2736

Great; all right, the next one: what is the magnitude of the vector u = (4,3,12)?2744

And then, we want to figure it out by both the formula we were given and the Pythagorean theorem.2752

First, the formula we have--this is the easy part, a nice, handy formula.2757

It is going to be the square root of each of its components squared (4, 3, and 12), and all added together.2762

We work this out; we get √(16 + 9 + 144) =...add those together; 16 + 144 gets us 160; 160 + 9 gets us 169,2772

which simplifies to √169, is 13; so there is the length of our vector.2786

Now, we also want to figure out the Pythagorean theorem.2793

This is where we are going to understand why this formula--why this mystical formula actually works and makes sense all the time.2795

First, let's see where this vector would get plotted out to.2802

So, we are going to have to look at this three-dimensionally.2806

And while we haven't talked about three-dimensional coordinates before in this course,2807

you have probably seen them at some point previously in some course.2811

Here are our x, our y, and our z-coordinates.2813

We go out: (4,3,12); so 4 out on the x...a little way out on the x, a little less out on the y...2817

We are out here: 4, 3, 4, 3...and then we go up by 12; so our vector is like that.2826

Now, notice: we had to get out here to this place by the x and the y part first.2837

We could figure out what the length is here, and then we have a square angle here, as well.2843

So, let's figure out...we can break this down into two parts: what is happening in the (x,y) plane...2849

in the (x,y) plane we have (4,3) as the cross-section here.2855

I will color it: this part here is the same as this part here.2863

We figure that out; that is 3 there, as well; we are going to end up getting (by the Pythagorean theorem) √(32 + 42).2871

That equals √(9 + 16), equals √25, equals 5; great.2880

We have figured out what the lower part is on the bottom part.2889

Now, we can do this cross-section with the z-axis included.2892

So now, we look at a cross-section; using this cross-section, we can cut this, and we can see:2898

here is the thing we are trying to figure out, the length of this.2908

And we know that the z amount was 12; so a cross-section with the z-axis...this here maps to this part here.2912

And then, here, our purple part shows up here.2922

That was length 5, as we just figured out.2927

We use the Pythagorean theorem here; so once again, the value of our hypotenuse is going to be the square root of 122 + 52.2929

So, that is the square root of 144 + 25, or the square root of...add those together; we get 169, which equals 13.2938

That is the exact same thing that we saw over here when we used that formula.2948

Cool--it works out both ways.2952

Now, let's understand why it works out both ways.2953

Well, notice: the thing here for the purple line was the square root of 32 + 42.2956

It was the square root of the x2 plus the y2, the first component and second component put together, squared.2962

Now, notice: we end up just plugging in the purple part, because it makes up one of the parts of our triangle here.2969

We could have alternatively looked at the square root of 122, plus the purple part,2974

because that is what is going to go in there: the square root of 32 + 42,2983

because that is what the purple part ends up being;2988

and then, because we are going back to using the Pythagorean theorem, that whole thing is going to be squared, as well.2990

Well, that means √(122 plus...if we have the square root then being squared,2995

squared on top of square root, that cancels out, and we have + 32 + 42.3000

And look: that is the exact same thing that we have up here.3009

And so, that is where we are getting the ability to just put them all together--stack them all together.3012

It is because we have to figure out one cross-section after another after another.3016

But if we then plug in the way these cross-sections end up working out, each of the square roots3020

that would go in from a cross-section would get canceled by the next cross-section it goes into.3023

And so, ultimately, we end up getting this form of first component squared, plus second component squared,3027

plus third component squared, until we get to our last component squared.3032

And then we are adding them all and taking the square root, and that is why we have that formula for the magnitude.3035

All right, next: A box weighing 300 Newtons is hung up by two cables, A and B.3040

Using the diagram, figure out how much tension is in each cable.3045

The first thing to do is to understand what this means.3049

A lot of math problems get thrown at us like this, where they are actually pulling from physics.3052

And they are sort of assuming that we know things about physics that we might have no idea about.3056

This is a math course, not a physics course!3059

Let's first get an understanding of what this means.3061

If we have a box on a string--just imagine for a minute; now imagine that it weighs 100 Newtons.3065

And if you didn't know, a Newton is the unit of force and weight in the metric system.3072

They use kilograms for mass, but force, how hard something is being pushed or pulled--that is Newtons.3080

In the English system, the British imperial system, it is pounds; so pounds is used for weight and force,3085

although we actually have another unit for mass; but you almost never hear it.3092

It is called slugs, if you are using the British imperial system.3095

But Newtons are a way of measuring weight, which is just a question of how much gravity is pulling.3098

OK, so imagine that this thing is being pulled down by 100 Newtons of gravity.3104

We have 100 Newtons of force pulling down on this thing.3110

Well, if we have some rope or some string that is holding this thing up,3112

well, if it is being pulled down, it must be that the string is pulling back up.3116

Otherwise, the thing would fall to the ground.3120

So, how much is the string pulling up by?3123

Well, the string must have a tension pulling up of 100 Newtons in the opposite direction; so there are 100 Newtons going up and 100 Newtons going down.3124

Now, notice: as a vector, this would be a positive 100 Newtons, because it is in the up direction.3132

As a vector, this would be a negative 100 Newtons, in the down direction.3138

We take positive 100 Newtons, and we add that to -100 Newtons; we get 0.3142

This makes sense, because no force means no acceleration.3148

The thing is currently stopped; so as long as it doesn't have any acceleration to make it move somewhere, it is going to not pick up any motion.3155

What we have to have: we have to have no force out of it for it to not move.3161

Now, since it is hung up by two cables, it is perfectly reasonable to say, "Yes, if it is hung up, it is not currently moving anywhere."3165

It is not falling to the ground; it is not swinging left and right; it is just hanging there in space--it is sitting there.3171

So, that means that it must be a total of 0 for what is going down and what is going up.3175

We know that it is 300 Newtons going down; so now it is a question of how much is going up.3180

So now, we need to talk about these cables.3184

We can think about A as being a vector pulling out and away, because it is pulling up on that box.3187

Otherwise, it would be helping the box fall.3194

So, this is some vector A, and it has some force tension; so we will say A is equal to the tension in A.3196

And we will do the same thing over here with B; so now we have some vector B,3206

and B will be the amount of the tension in B, how much it is being pulled up by.3211

Now, we don't know what A and B are; that is what we are trying to figure out.3217

But we want to figure out how to get to them; so we start looking at this, and we say, "Well, I don't know a lot."3221

But they did tell us this information about the angles.3227

So, maybe we can break these vectors up into component forms, based on these angles.3230

So, if we have 40 degrees here, then we must have 90 - 40 here.3236

And since this is another right angle, that means we have 40 here again.3244

Similarly, with the same basic idea, since it was 70 up here, then we have 70 down here.3248

With that in mind, we can figure out what the pieces are here.3255

We know that A is the length of this hypotenuse; we don't know what the number is yet, but we are calling it A.3261

So, the vector A is going to be broken into the components: the horizontal amount is the side adjacent to 40,3267

so A, the length of the whole thing, times side adjacent of the angle;3273

and then, A times opposite, if we want to talk about the vertical part right here.3279

That will be Asin(40) over here.3285

Next, we know that we can talk about vector B; we can break it down into very much the same way.3289

That will be B times cos(70), because its side adjacent is 70; and B times its side opposite...3295

I'm sorry, not its side adjacent; its side adjacent is not 70, but the angle connected to side adjacent is 70 degrees.3304

And B times sin(70)...we can get this all just from basic trigonometry stuff.3308

Here is B; we can multiply B, our hypotenuse, and figure out what the sides opposite are, based on this.3314

We have two vectors here, A and B.3320

One other thing that we know is the force vector: what is the force on this box?3323

Well, it must be that the force on this box...is it moving up and down? Is it currently still?3329

Well, it is currently still; it is not moving anywhere; and it is not moving anywhere horizontally; and it is not moving anywhere vertically.3333

So, total force, in the end, once everything gets put together: it is not moving up and down; it is not moving left to right; so it must be (0,0).3342

And they did tell us one other piece of information, 300 Newtons; so the weight is equal to 03349

(because it is perfectly down, so it doesn't have any horizontal) and -300, because it is moving down; great.3357

At this point, we actually have enough information to solve it; let's work this thing out now.3362

Remember: we have weight = (0,-300); and we know that when we combine the weight with each of the cables--3367

once we put all of these forces together, all of the forces put together must come out to be a force of nothing,3378

because the box isn't moving anywhere--it is not being shoved around.3384

We can plug all of these things together; we have A + B + the weight vector is going to equal our total force of (0,0).3387

So, let's work out what that is: we have -A...3408

Oh, that was one thing I didn't say on the previous slide; I put it as Acos(40), because Acos(40) is the length of this side.3411

But it has to be negative, because remember: when we go to the left, it is negative in the x direction, so it is negative.3418

Both of the verticals are positive, because they are both pointed up; but only the B is pointed to the right in this--pointed positively horizontally, as well.3425

All right, we have (-Acos(40), Asin(40)), plus (Bcos(70),Bsin(70)).3433

I'm sorry; I am going to have to continue onto the next line.3449

Remember: this is just all one line put together...so + (0,-300).3452

In the end, we will end up equaling (0,0).3457

Let's combine this all together; we will switch to the color of green for everything together.3462

We have -Acos(40) + Bcos(70) + the 0 from the weight; that comes out to be the total for our first components.3467

And then, Asin(40) + Bsin(70) - 300 = (0,0).3480

So, at this point, we know that the first component on the left side of an equation3492

has to be the same as the first component on the right side of the equation.3496

The same thing: the second component on the left side must be the same thing as the second component on the right side.3499

So, we can break this down into two separate equations.3503

We have -Acos(40) + Bcos(70) = 0; so we can get B on its own or A on its own, and then plug that into the other one.3505

Let's solve for that first: Bcos(70) = Acos(40); at this point, we see that B is equal to A times cos(40), over cos(70).3519

We could plug that into a calculator right now and get some number out of it.3532

But then we would have to write all of these decimals; so we can just leave it like that for now.3535

Next, we will swap to a new color for solving this part.3538

We have Asin(40) + Bsin(70) - 300 = what was on the right-hand side, 0.3542

So, we can move the 300 over; we see that we have Asin(40) + Bsin(70) = 300.3554

Now, we see that B is the same thing as A times cos(40), divided by cos(70); so we have Asin(40) +...3564

we swap out our B: (Acos(40))/cos(70), times sin(70), still equals 300.3572

So, at this point, we pull out all of our A's; we have A times sin(40), plus cos(40)/cos(70), times sin(70), equals 300.3585

Notice that there is nothing we can cancel out there, because we don't have any exactly matching things.3599

But we can divide by it: 300 divided by sin(40) plus cos(40) over cos(70) times sin(70).3604

We can plug that all into a calculator, and it will come up and give us an answer.3622

And it will tell us that A is approximately equal to 36.40...oops, sorry, not 36.40, but times 3...3626

give me just a second...the magic of video...it comes out to be approximately 109.2.3640

And that is our value for A; to figure out our value for B, we have this handy thing right here.3648

B equals A, which was 109.2, times cos(40), divided by cos(70).3654

We work that out with our calculator, and we get approximately 244.6.3663

So, the tension in B is 244.6 Newtons, and the tension in A is 109.2 Newtons; great--those are our solutions.3670

All right, the last example: A plane has a compass heading of 75 degrees east of due north, and an airspeed of 140 miles per hour.3689

If the wind is blowing at 20 miles per hour, and towards 10 degrees west of due north, what is the plane's direction and speed, relative to the ground?3697

First, what does this first part mean?3704

We have a compass heading of 75 degrees east, going at 140 miles per hour airspeed.3705

Airspeed means your speed in the air--how fast you are moving, relative to the air right around you.3711

So, if that is the case, then our plane is moving 75 degrees east of due north.3716

Due north is this way, our vertical axis; so if that is the case, we need to curve 75 degrees down towards the east.3726

So, east is this way, so that is 75 degrees here.3735

If we want to figure out what is the thing in here, because we will probably want that for figuring out other things,3739

that is going to be 15 degrees; so it is 15 degrees for our normal θ that we are used to.3743

OK, so that is the direction that the plane is headed; it is going like that.3750

It is going off in this way; but then, that is its airspeed.3754

The air also is able to move; the air is going this way--the air is blowing the plane.3758

So, the plane is going like this, but at the same time it is being blown off-course slightly by the wind.3763

Or perhaps (hopefully) they have taken this into account, and it is not going off-course.3768

We have the wind blowing at 20 miles per hour, and towards 10 degrees west of due north.3772

So, once again, it is off of due north; and now it is just a little off, 10 degrees off, here.3778

And it is total of 20 miles per hour.3785

So, we have 140 miles per hour this way and 20 miles per hour, 10 degrees to the west of due north.3787

So, notice that the total angle for that is going to end up being 100 degrees, if we wanted to figure that out.3795

Or we could also look at it in terms of 80 degrees here, as well.3802

It depends on which one you think is easier; I am going to go with the one inside of the triangle,3809

and we will just have to remember to deal with the fact that our horizontal is going to be negative when we are working on the wind.3813

So, the plane--what is the velocity of the plane in the air?3818

To be able to figure out what the plane's direction and speed is relative to the ground,3823

we have to combine its motion in the medium with the medium's motion relative to the ground.3828

Its motion in the medium is the 75 degrees east of due north and an airspeed of 140 miles per hour, the blue part.3834

And the red part, the wind blowing, is the air relative to the ground; so we have to combine those two things.3840

So, the velocity of the plane in the air...its horizontal component is going to be the length of the vector, 140, times...3846

the angle we have is 15, so if we are talking about the horizontal, that is going to be side adjacent.3855

So, cos(15)...140 times sin(15) for the vertical, because that is side opposite...we work that out;3858

and that ends up coming out to be approximately 135.2 and 36.23.3867

And the units on both of those are miles per hour, because it is moving 135.2 miles per hour north and 36.23 miles east simultaneously.3875

And then, we have the wind: what is the velocity of the air itself?3884

The air is moving at a speed of 20; and if we want to figure out its horizontal component, it is going to be this part here,3889

at which point we say, "Oh, right; it is going negative, so let's put in that negative sign, because it is moving to the left."3896

We have to remember to catch those negatives.3903

Pay attention to if it is going to be a positive or a negative direction for everything.3906

So, that is going to be 20, the size of it, times cosine of 80 degrees, in this case;3909

and then positive 20, because this one is going positively, times sine of 80.3916

Work that one out with a calculator; and we get 131.7, 55.93.3923

Oops, I'm sorry--I wrote the entirely wrong thing.3932

I meant to write -3.47 (I read the wrong thing off of my notes), comma, 19.70; great.3934

All right, so if we want to figure out the combination of the two--if we want to figure out what the total motion of the plane,3944

relative to the ground, is, that is going to be the plane's motion relative to the air, plus the air's motion relative to the ground.3950

We just figured out what each one of those is: (135.2,36.23) was our plane's motion,3960

plus (-3.47,19.70); so in total, that gets us (131.5,55.93); great.3968

So, that is what the velocity vector is going to be--what the component form is.3981

However, it asked for speed and direction of the whole thing put together.3988

If that is the case, we want to figure out speed.3994

Well, speed is just the size of our velocity total vector.3997

That is going to be the square root of the first component of our total vector, 131.7, squared,4004

plus the second component, squared (55.93 squared).4009

We take the square root of that and figure it all out, using our calculator; and we end up getting 143.1 miles per hour.4013

The plane is actually going a little bit faster than its speed had been previously without the air connected to it.4021

However, its direction will also change.4028

To help us figure out direction, let's draw just a quick picture, so that we know what is going on.4030

We have this right here as our motion: 131.7 is fairly horizontal, and a little bit...between 1/3 and 1/2 of our amount horizontally up.4035

That is what our motion is like total.4048

My picture is totally not relative; 140 up here should be much longer than the 20 here;4052

and this 143.1 miles-per-hour long vector should be even longer.4057

But that is OK; we are just trying to get a sense of what is going on; they are sketches, not perfect drawings.4060

So, we are looking for this angle here: tan(θ) is going to be the side opposite, the vertical component, 55.93,4065

divided by the side adjacent, the horizontal component, 131.7.4075

We take the arctan of that; that gets us θ =...about 23 degrees.4080

Now, notice: that is what θ is equal to: θ equals 23 degrees.4085

All of this stuff was given in east of due north, so we have to put it in that same thing.4090

If it is 23 degrees as our θ here, that means it is 23 degrees going up from our positive x-axis, which was east.4095

Up would be towards the north; so we could phrase this as 23 degrees north of east.4102

Alternatively, if we wanted to use the exact same thing that they had done, when they all talked about east of due north,4112

90 - 23...if we want to figure out what this is here, 90 - 23 = 67.4119

So, we could also talk about it as 67 degrees east of north.4127

Either one would be fine; but we have to put it in the same format, because they didn't give us a θ previously.4138

We don't know where θ is based off of, so we have to make sure4143

that we are following this same pattern of east of due north, west of south, something like that.4145

We have to go in that same pattern.4150

All right, vectors are really, really useful; we will talk about them more when we talk about how matrices are connected to them.4152

But this is really great stuff here.4156

We have talked about a whole lot of things here, so if you had any difficulty understanding this lesson,4157

just try watching piece-by-piece, and just work examples, one after another.4162

There are a lot of things to digest here, but they all work together; and vectors are so useful.4165

All right, we will see you at Educator.com later--goodbye!4169