9.3+Content+-+Vector+Theory

Add Vector Theory content to this page. See specific details on the home page. This unit has been split into parts A & B. Content between the two must be cohesive as it is one continuous, connected material. They are split simply to better organize the page.
 * Quick Instructions**

**Vector Theory (Unit 9B) Learning Targets**
> Write vectors as a linear combination of unit vectors.
 * Represent vectors as directed line segments and write the component forms of vectors.
 * Perform vector addition, subtraction and scalar multiplication and represent them graphically.
 * toc
 * Use vectors to model and solve real-life problems.
 * Evaluate the dot product of two vectors.
 * Prove and apply properties of the dot product.
 * Find the angle between two vectors.
 * Write a vector as a sum of component vectors using a projection

**‍Ways to Describe a Vector ‍**
A vector is a quantity that has both mass and direction. ‍There are five ways to describe a Vector: ‍
 * 1)** Initial point and a terminal point: from A (tail) to B(head)

math u= math Vectors in Component form describe the distance the vector travels horizontally (c) and vertically (d). Ex: math u=<5,6> math The vector travels 5 to the right and 6 up.
 * 2)** Component form:

math math where math math
 * 3)** Magnitude and direction
 * u|
 * u|^2=a^2+b^2

math \hat{u}=u/|u| math A unit vector is a vector with a magnitude of 1.
 * 4)** Unit Vector

math u= math math u=ai+bj math math u=<-4,8> math math u=-4i+8j math
 * 5)** ij form

**Equivalent vectors:**
Two vectors in the same direction with the same magnitude. Ex: math v=<3,1> math math a=<3,1> math math v=a math

This image represents equivalent vectors.

**Scalar Multiplication:**
This will produce a vector with a magnitude that is a multiple of the original vector and in the same direction. math u=<-3,1> math math 2u=<-6,2> math

**‍Vector Addition:‍**
Adding two vectors creates a new vector in a new direction. Adding vectors is similar to adding matrices. Just add across. math u=<-2,-4> math math v=<3,2> math math t=u+v=<-2,-4>+<3,2> math math t=<1,-2> math

‍This is an image of adding vectors:‍

**Vector Subtraction:**
Subtracting vectors is similar to adding them except the direction is going in the opposite direction. Treat the problem like you're adding the negative vector. math u=<-2,-4> math math v=<3,2> math math t=u-v=<-2,-4>+<-3,-2> math math t=<-5,-6> math

This is an image of subtracting vectors:

**Vector Dot Product**
The dot product of vectors is used in a variety of ways. One would need it for the formula to calculate the angle between two vectors and the formula to calculate the projection of a vector onto another vector. It is defined as the following…

Given, vector math u < a_1,b_1> math and vector math v < a_2, b_2> math math u\bullet v math = (dot product of vector // u // and vector // v // ) = "some value" = math (a_1)(a_2)+(b_1)(b_2) math

Example: Given //u// <1,2> and //v// <3,4> math u\bullet v = (1)(3) + (2)(4) = 3 + 8 = 11 math

Dot product should give you a value not a vector.

**Angle Between Vectors**
There are many ways to find the angle between two vectors. But, the most efficient way is to use an equation.

math \cos\theta=(u\bullet v)/(|u||v|) math

This video explains it pretty well as well as the use of dot product: []

**Vector Projection**
Vector projection is used to find the "projection" of a vector onto another vector (like if one were to shine a light onto one of the vectors, how much of it would cover the other vector).

To start, the component of //u// onto //v// is w//. //This is to say that the portion of //u// that is on //v// can be valued as w (some numerical value). To find w, one would do the following...

This, however, is just a value that describes the how much of //u// is on //v// and not the direction. Projection allows us to find the "component of //u// onto //v// in the __direction of //v//__." It is denoted as, in this case, math proj_{v}u math To accomplish the direction part of the projection, one would multiply the formula to get the component of //u// onto //v//(stated above) by the unit vector of v. It would turn out to be this:

**‍Orthogonal Vectors‍**
Orthogonal vectors are two non-zero vectors whose dot product equates to 0. This is another way to say two vectors are perpendicular or 90 degrees apart, but because vectors can work in three dimensions, we call it orthogonal. math u\bullet v=0 math

In other words, math (a_1)(a_2)=-(b_1)(b_2) math

This works because math \cos\theta=u\bullet v/(|u||v|) math

Because the angle is 90 degrees, math \cos\theta=0 math

So, math math math math math u\bullet v=0 math
 * u||v|\ast \cos\theta=u\bullet v
 * u||v|\ast 0=u\bullet v

Orthogonal vectors can have infinite solutions. Ex: math u=<3,-7> math math u\bullet v=0 math math v= math let math b=p math math u\bullet v=7a-3b=0 math math 3a=7p math math a=7p/3 math math v=<7p/3,p>, \forall p \epsilon \mathbb{R}, p\not= 0 math

**Application Problems**
In this unit, there are three basic types of problems that are down with vectors: the true direction and velocity of an airplane, weight (tension), and work.

__Airplane Application__
Given: A plane flies due North at 150 mph (airspeed) Crosswind is at 10 mph S 20° W

1. Write the velocity of vector // v // of the airplane relative to air. Velocity describes the direction of the airplane; therefore, the component form of the vector describing the direction of the airplane is needed.

If a plane flies“due North”, it is implied that it is going straight up (north). Therefore, there is no horizontal component to the vector describing the plane’s direction because the plane has no horizontal movement at all if it is going straight up. So, the vector describing the the plane’s direction is…

vector //v// = <0,150>

2. Write the wind vector. Knowing that the airspeed of the crosswind is 10 mph, in "vector words", the airspeed would be called the magnitude of the vector that describes it (vector with magnitude 10). "S 20° W" states that the wind is blowing 20° west of due South, which is also 250° from 0°. Now, the following picture describes what exactly is happening to the plane...

By looking at the diagram, the direction of the wind is 250° from 0°. This allows one to write the component form of the wind, and it goes like this...

vector //w// =  x = 10cos(250°) y = 10sin(250°)

So...the wind vector is //w//<10cos(250°),10sin(250°)>.

3.Find the velocity of vector //t// (vector of the true path of the plane (after being affected by the wind)).

This can be done just by adding the two vectors because by adding the two vectors, one can find the vector that results from it. In this case, the vector would describe the true path of the plane.

Know: //v// <0,150> and //w//<10cos(250°),10sin(250°)> //v// + //w// = <0+10cos(250°),150+10sin(250°) ≈ //t// <-3.42, 140.60>

4.Find the true speed and direction of the plane. First, the true speed of the plane can also be looked as the magnitude of the vector describing the plane's path. This can be accomplished by using the Pythagorean Theorem (a^2+b^2=c^2) with the horizontal and vertical components of the vector describing the plane's path.

math \sqrt{(10cos250^{\circ})^{2}+(150+10sin150^{\circ})^{2}} math ≈ 140.64 Therefore, the true speed of the plane is 140.64 mph.

Now, moving on to direction, the direction portion implies some sort of angle because the angle will tell one which way the plane is pointing towards. So, by using old trigonometric functions, and by using the horizontal and vertical components of //t// again, the direction of the plane can portrayed as the following... math \tan^{-1}\frac{10sin250^{\circ}+150}{10cos250^{\circ}} = \Theta = 271.39^{\circ} math

__‍Weight (Tension) Application‍__
Moving on, the second type of problem using vectors involves tension.

Given:

Find the tension on the two cables (blue).

Again, by using old trignometric functions, the following can be taken from this diagram. math x_{1} = \left | T_{1} \right |cos150 math math x_{2} = \left | T_{2} \right |cos20 math math y_{1} = \left | T_{1} \right |sin150 math math y_{2} = \left | T_{2}\right |sin20 math Note: The angle for Tension 1 is, in this case, 150° because in Pre-Calculus, angles are view as going from the initial ray (0°) to some terminal ray. Therefore, the angles lying in the 2nd Quadrant would be the supplement of the reference angle. Since there are more than two variables here, it would be helpful to set up a system and solve it. In this diagram, it can be seen that this structure (cables holding blocks) must be stable; otherwise, the block will not be held up or it would be moving around. Knowing this, all of the forces must add up and equal zero because they must cancel each other out. Therefore, the following two systems can be set up.

Sum of the forces in x-direction: math x_{1}+x_{2}+150cos270 = 0 math Sum of the forces in y-direction: math y_{1}+y_{2}+150sin270 = 0 math

Note: There is an additional force for both directions. These two additional forces are there because they describe the forces created by the block. After all of this, one could use a matrix to solve for the two tensions. It would be like this: math \begin{bmatrix} \left.\begin{matrix} cos150& cos20& \\ sin150& sin20& \end{matrix}\right|\begin{matrix}-150cos270 \\ -150sin270 \end{matrix} \end{bmatrix} = \begin{bmatrix}184.0 \\ 169.6 \end{bmatrix} math

__Work Application__
The last type of problems with vectors are work problems. The general formula for work is this: math W = F\cdot D math where "W" is the work required, "F" is the force used, and "D" is the direction.

Given: A person is pulling a wagon with a force (on handle) of 50 lbs. at a 20° angle, and he pulls the wagon a distance of 10 ft.

Find the work done.

Well, first one would need to write the vector describing the work... The vector describing "F" would be //F//<50cos20°,50sin20°> (just using trignometric functions and rearranging for x = and y =). Since the distance traveled was just 10 feet horizontally, the vector describing it would be //D//<10,0> (no vertical shift). Therefore, the work done will be the dot product of //F// and //D,// which would be... math W = F\cdot D = ((50cos20*10) + 50sin20) \approx 469.85 math foot lbs.

**Projectile Motion**
A projectile has a horizontal and vertical component to its velocity.

The horizontal motion is not affected by gravity, so it is constant. The x-t equation would then be: math x(t)=(|v_o|\cos\theta)t+x_o math

The vertical motion is affected by gravity, so we must account for this using the //g// constant where g=-9.8 m/s^2 or -32 ft/s^2. The v-t equation would then be: math y(t)=(|y_o|\sin\theta)t+y_o+1/2gt^2 math

Ex: A ball is thrown from a rooftop, 53 feet above the grown with an initial velocity of 26 ft/s and at an initial angle of 33 degrees above the horizontal.

The x-t equation would be: math x(t)=(26cos33^{\circ})t math

The y-t equation would be: math y(t)=(26sin33^{\circ})t+53+1/2(-32)t^2 math

When y(t)=0, the ball has hit the ground. The equation will produce negative y values, but these don't apply in reality. To find how far the ball travels, we must set y(t) to 0 and solve for time: math 0=-16t^2+(26sin33^{\circ})t+53 math By simplifying we can find t. math t=2.316 s math We now plug the new t back into the x-t equation. math x(2.316)=(26cos33^{\circ})(2.316) math The answer should be: math x(2.316)=50.5 ft math




 * Primary authors of this page (as of 06/02/12): Jonathan C. and Will Han**