Thursday, March 29, 2007

10 - 8 : Vector Equations of Lines in Space

10-8 : Vector Equations of Lines in Space



First of all let me explain the purpose of this lesson. It is very simple. If you have a line, and random line through space, and one point on that line, and a unit vector, you can find the equation of the position vectors to any point on that line. You must choose a distance, or "d" to work with. "d" is just a distance or magnitude, which represents the distance to the next point on the line you want to find.



Ok so i find that it is easier to explain this verbally rather than graphically, however I know some people would like to know precisely where everything im talking about is in a diagram so, I decided to make a diagram which u can refer to whenever your heart so desires!



















ok, i am not a very good "graph maker," so I color coded it so that I can try to make up for that with my explanation.





  • the red A is referred to as P0 in the book, and is the given point on the line.

  • the pink vector is referred to as vector P0, and is the vector to the given point.

  • the blue line is the line, and the unit vector that is used, is the unit vecotr of this line.

  • the yellow portion of the line is "d" or the distance from point A to point B.

  • the red B is the point which the position vector we are finding the equation for.

  • the orange vector is the actual position vector we are finding the equation of.



In essence, we are just adding vectors. Or in the case of the diagram, we are adding the pink vector to the yellow vector, in order to get the orange vector. Mathematically we would say, we are adding vector Po and vector PoP.

*note : Vector du = Vector PoP*


The method we use to figure out the equation of the position vector is :






  • Vector v = vector Po + d(unit vector)


This is when we are given the unit vector.




  • Vector v = (xo + c1*d) + (yo + c2*d) + (zo + c3*d)


This is when we must work with the cosines instead of being given the equation.

Sample problem when given the unit vector.

Problem : The point is (5,11,13) and the unit vector is 3/7 i + 6/7 j + 2/7 k. D = 5. Find vector PoP (v).

Work :
Vector v = (5 i +11 j +13 k) + 5(3/7 i + 6/7 j + 2/7 k)
Vector v = (5 i +11 j +13 k) + (15/7 i + 30/7 j + 10/7 k)
Vector v = 7.143 i + 15.286 j + 14.429 k

Answer : Vector v = 10.714 i + 47.143 j + 18.57 k


Now I will show you a problem where we are given the cosines

Problem : The point is (5,3,-1) and the cosines are : c1 = 6/11, c2 = -2/11, and c3 = 9/11. D=5.

Work :
vector v = (5 + (6/11)(5)) i + (3 + (-2/11)(5)) j + (-1 + (9/11)(5)) k
vector v = (5 + (30/11)) i + (3 + (-10/11)) j + (-1 + 45/11) k
vector v = 7.727 i + 2.091 j + 3.091 k

Answer : vector v = 7.727 i + 2.091 j + 3.091 k

Both of these methods are purely plug and chug. They are a sinch and aren't ridiculously long either. All in all they are pretty darn awesome.

Challenge Problem!!!

Problem : Timmy is looking at the great wall of china. He knows that the exact point he is staring at is located at (5,10,20). He knows that the unit vector for the wall is
(9/17) i + (12/17) j + (8/17) k. He wants to know the position vector from him to the point 10 feet to the right of the one he is looking at currently. Help Timmy find the position vector.

Work :
first set up the problem
vector v = (5 i +10 j +20 k) + 10 ((9/17) i + (12/17) j + (8/17) k)
distribute the distance
vector v = (5 i +10 j +20 k) + (90/17) i + (120/17) j + (80/17) k
group according to like terms
vector v = (5 i + (90/17) i) + (10 j + (120/17) j) + (20 + (80/17) k)
add like terms
vector v = 10.294 i + 17.059 j + 24.706

Answer : vector v = 10.294 i + 17.059 j + 24.706

This concludes MY lesson. But if you really want to indulge yourself in a very vector-like bliss, i suggest you check out http://mathforum.org/~klotz/Vectors/vectors.html.


This site literally has everything on vectors. It has formulas, but also explains them. It also has pretty good pictures and diagrams to go with them. So everyone, even Alvin and the chipmunks, should go check this out!



Here is a little bit of me : This is the cover the the Silversun Pickups debut cd, Carnavas. It is absolutely phenominal in every way. So I insist you go buy it now, or have me burn you a copy, because they are just that kool. They kill 2 stones with one bird, just like chuck norris.





REMINDER TO PAUL!!!! YOU ARE NEXT!!!!!!!!!!

Wednesday, March 28, 2007

Snakes, Direction Angles, Planes, and Direction Cosines

10.6: Vector Products of Two Vectors

We now have that long-awaited SECOND way to multiply vectors, so brace yourself. This multiplication process is called the "cross product," and it results in a vector answer, rather than a scalar one.





So basically, Vector C = Vector A X Vector B (the X represents "cross", so you'd read this as "vector C" is equal to Vector A cross Vector B). This process is NOT COMMUNITATIVE (authoritative use of capital letters, no?), so don't you dare assume that Vector B X Vector A will equal the same thing as Vector A X Vector B. If you do, by God, I will fight you. BUT, Vector A Cross Vector B WILL equal - Vector B Cross Vector A. Good to know.





Resultant Vector C will lie perpendicular to the plane that Vectors A and B reside on. What we must decipher is the direction in which Vector C is going; is it going toward us from the plane or away from us? We determine this by using the Right Hand Rule, and the steps for said rule are as follows


1) Cut a hole in the box


2) Put your ju.....oh wait, wrong steps, my bad.





1) For "Vector A Cross Vector B," Place your right-hand fingers along Vector A, and curl them towards Vector B


2) The direction in which your thumb is pointing represents the direction in which Vector C will travel.


3) For "Vector B Cross Vector A," simply do the opposite; place your fingers along Vector B, and curl them towards Vector A.


4) A good thing to remember is that if Vectors A and B are parallel, the product will be zero and the right hand rule consequently will fail miserably.





Other stuff to know:





Vector A Cross (Vector B + Vector C) = Vector A Cross Vector B + Vector A X Vector C





Here are the cross products of the unit coordinate vectors:


i x i=0 K x K=0 j x i=-k
k x i=j I x k=-j i x j=k

j x j=0 j x k=i k x j=-i




Example of Cross Products:





Find Vector C for Vector A Cross Vector B





Vector A= 8i+3j+4k


Vector B=9i+7j+2k





Vector A Cross Vector B= 72 i x i + 56 i x j + 16 i x k+ 27 j x i+ 21j x j + 6 j x k+ 36k x i + 28k x j + 8 k x k





Vector A Cross Vector B Reduced: 56k - 16j - 27k +6i +36j -28 i





Vector C = -22i +20j + 29k





Find Vector C for Vector A Cross Vector B Using Determinants:





Here is the basic rubric for determinants....so know it. It's important. I swear.






(i j k)

(Ax Ay Az)

(Bx By Bz)


Vector A Cross Vector B = (AyBz -AzBy)i - (AxBz - AzBx)j + (AxBy - AyBx)k



So then, let's apply it that rubric to determine Vector C (Using the same Vectors A and B)





(i j k)


(8 3 4)


(9 7 2)





Vector C = (6-28)i - (16 - 36)j + (56-27)k





Vector C = -22i + 20j +29k!








Alright, we're ALMOST done. Other stuff to know:





The Area of the parallelogram formed by Vectors A and B equals the absolute value of Vector A times the absolute value of Vector B times the Sine of the angle the two vectos form.





The Area of the triangle formed by Vectors A and B equals 1/2 times the absolute value of (Vector A times Vector B.)





And...THAT'S IT! Thank god.

Website: http://omega.albany.edu:8008/calc3/cross-product-dir/cornell-lecture.html





My personalization.......IS LAAAAARRRYYYY!






Leo, you're next. Peace out.

Monday, March 26, 2007

10-5 Planes in Space

We already know what a vector in a three dimensional setting looks like:




A plane in space is the set of all points that satisfy the equation: Ax + By + Cz = D
It looks like:

To find a vector ,normal or perpendicular, to the plane Ax + By +Cz = D, just plug the coefficients A, B, and C into the vector formula:




EXAMPLE

  • given 3x - 8y + 5z = 24

  • plug 3, -8, and 5 into the vector formula

  • your answer should be

You can also determine a plane by using a normal vector and a point, as shown below:





courtesy of Mr. French

To find a plane using a normal vector and a point P:


  • Place the vector's tail at the point P =

  • Pick another point and draw another vector to that point

  • Find the displacement vector (head - tail) =

  • *the vectors are perpendicular so their sum is 0

  • Use algebra so that

  • Then, and that's your answer!


EXAMPLE

given and P = (2,5,-1)


You will find that essentially, you only need the equation




These are pictures from a JACK'S MANNEQUIN concert I went to in February. They're my favorite band and everyone should go listen to their music. Their rooted from SOMETHING CORPORATE, which is my other favorite band, because the lead singer was originally from SoCo. Andrew McMahon, the lead singer, fought and beat cancer, which i think is amazing and he's amazing so everyone should go buy his music NOW!

JOEY YOU'RE UP NEXT!!!!

Thursday, March 22, 2007

Friday's Quiz Topics

Here’s a list of topics for Friday’s quiz:

Precalculus Quiz 10.1-4 Topics
Draw a three-dimensional vector and its “box”
Unit vectors: i, j, k
Adding vectors visually
Subtracting vectors visually
Dot/Scalar/Inner Product calculations – both formulas
Projection of a vector onto another vector – visual
Calculate vector projection of one vector onto another vector
Magnitude of a vector – calculation
Dot product – calculation
Calculate the angle between two vectors
Scalar projection calculation – two methods
Calculation of a unit vector
Vector projection calculation
Three-dimensional resultant forces
Angle between two three-dimensional vectors
Three-dimensional position vectors
Position vector to a point along a displacement vector


That’s it! The format’s the same as always – ½ Non-calculator, ½ Calculator. There are 10 questions on the Non-calculator portion and 12 questions on the Calculator section. I’ll be around briefly after school on Thursday, and in early on Friday morning.

I don’t know whether my life has been a success or a failure. But not having any anxiety about becoming one instead of the other, and just taking things as they came a long, I’ve had a lot of extra time to enjoy life.
—COMEDIAN HARPO MARX

Wednesday, March 21, 2007

9-8: Math Expectation

Well, we never really got a lecture on this section, so I'll be presenting just my own example of the lesson...

The mathematical expectation is the weighted average based on the expected probabilities. It is the weighted average for a random experiment each time it is run. It is a value that one calculates based on the outcomes for each event in a random experiment.

The mathematical expectation, or E, of a random experiment is the sum:

E = P(A1) a1 + P(A2) a2 + P(A3) a3 + . . . + P (An) an

for the n mutual exclusive events A1, A2, A3, ..., An in the experiment.

Outcomes = A1, A2, ..., An
Values of corresponding outcomes = a1, a2, ..., an

My Example:

Jane Doe is playing "Let's-Roll-the-Dice-and-See-How-Many-Points-I get" with her brother John Doe. Each roll of a single dice is considered an event. The points are as follows:
- Roll a 6: Win 72 points
- Roll a 4: Win 35 points
- Roll a 2: Win 27 points
- Roll an odd number: Lose 13 points

It is Jane's turn. Drawing a chart makes finding the mathematical expectation easier.
Odd # = 1, 3, 5 = 3/6 probability


>Note: Multiply Points for each with Probability to result in the last column.


>Then add up all the values you get from the Points * P(A) to finally get the mathematical expectation.
15.833 = average (after many many trials) points PER GAME

YAY! That sums it up. (Mr. French, that was my one original example that I made up, since no lecture was given and therefore, no example was shown to us in class.)

Here's a site for moredifferent examples/help on the topic: http://www.tpub.com/math2/86.htm


Soooo as a co-head of publicity for Leo Club, I think it's only appropriate for me to tell everyone that THE 2ND ANNUAL MR. FLINTRIDGE PREP is coming soon! It's on Friday, March 30th in the Norris Auditorium at 5:30. COME EVERYONE! It's going to be FUN and HYSTERICAL. Last year's was a big success, and this year's should be better!

These are the senior guys running for the crown this year:
Matt Adesuyan
Matt Brown
Sean & Mack Carroll (C-Unit)
Alex Jacobs
Ryan Lamont
Daniel Madore
James Mears
Casey Neumeier
Bryan Russell
Drew Titus
Jeff Wagner
Nick Weinstein
James Woolley
... It should and WILL be incredibly amazing SO I EXPECT EVERYONE TO COME!

Michelle you're technically supposed to be up next, but Amira took your spot. So Amira's up next, but she already posted her blog... so umm yeah... Kaori?

Tuesday, March 20, 2007

9-7 Functions of a Random Variable

This section is aimed at finding all of the probabilities of the possible events in a random experiment. EXCITING!

The random experiments in this section consist of repeating the experiment several times. The experiments also have only two possible outcomes.





In this equation,
N is the total times that an experiment is being done.
X is the number of times the event you want occurs in N repetitions.
A is the probability that the event we want does not occur in the N repetitions.
B is the probability that the event we want does occur in the N repetitions.

Because solving these types of equations shows the probabilities of all the possible outcomes, it is called a probability distribution.

The expressions that we are finding are terms in a binomial series. We get this series from expanding certain terms. For example, if the two probabilities we are using are .6 and .4, and we are repeating the experiment 5 times, we would be expanding the equation: (.6+.4)^5.

Since the probabilities apply to binomial series, we are specifically finding the binomial distribution. And, because there are only two possible outcomes, each trial is called a binomial experiment.

Example:
A butane candle lighter does not always light when you pull the trigger. Suppose that a lighter has a 60% probability of lighting on any one pull. You pull the triggier six times. Let P(x) be the probability it lights exactly x of those times. Show how to calculate P(4).

To solve this equation, all we have to do is plug the numbers into the equation!

N would be 6 because that is the total number of times we are pulling the trigger.
X would be 4 because we want to find out the probability of the lighter lighting 4 of the 6 times.
A would be .4 because that is the probability the lighter will not light.
B would be .6 because that is the probability the lighter will light.

So...
P(x)=6C4 * (.4^2)*(.6^4)
When we plug the equation into our calculator, we get
P(x)=.31104

There is a 31.1% chance that the lighter will light four of the six times.

Now, there is an easy way to calculate the probabilities for the rest of the outcomes. On your calculator, go STAT then EDIT. When the top of the L2 column is highlighted, press 2nd VARS. Go to binompdf. Inside the parenthesis, type in 6 (the total number of times we are performing the experiment) followed by a comma and then .6 (the probability the lighter will light).
You should get the values for all of the outcomes of the experiment.

Thanks to our handy dandy calculators, we can easily find all of the values in the binomial distribution, hooray!

This website is a little confusing because they use complicated notation, but if you study it long enough it makes sense.

http://mathworld.wolfram.com/BinomialDistribution.html

I have been tap dancing for most of my life, so here is a link to one of my favorite tap dancing videos...it's the awesome stair dance with Shirley Temple and Bill Bojangles!

http://youtube.com/watch?v=ImRGu4kjuZY

Gina, you're up next!

Section 10-3: Vectors in Space








This section covers...3D VECTORS!!! Dont worry, it sounds scarier than it looks. This section is actually really simple because it is almost exactly the same as section 10-2. The only difference is that you add an extra number to the equation. Instead of just i and j, we now have k to deal with as well. The graph of a 3D vector is also slightly different than a 2D vector. Instead of having just an x-axis and a y-axis, it has an x, a y, and a z-axis. A 3D vector graph also has octants in replace of quadrants. The first octant is the region in which x, y, and z are all positive.

Graph of 3D vector















*To find the magnitude of a 3D vector, you use the same equation that you did for a 2D vector, except you add k to it.

Example 1






Example 2





What would 20% of vector AB be?





CRAZYCOOL mATH WEbsites
Pope Leo X, you're up next!!!













Monday, March 19, 2007

10.2 Two-Dimensional Vector Practice

I know, I know- vectors. Not the most popular of all lessons, but getting these concepts down comes in pretty good handy in the real world (& in the computation of your math & physics grades). It’s all about mental paradigm, guys! We can do this! Here’s the trick: pretend that you have a friend named Hector. Now change that “H” to a “V” and you’re set. Vectors are our friends!

10.2 is basically about grasping basic concepts of 2D vectors, which will eventually be extended to 3D vectors!!!

Vector Anatomy & Concepts:


the 2 vector characteristics-





Vector Vocab:
-Vector Quantity: A quantity with magnitude & direction (like velocity, tension of a string, various forces)
-Scalar: A quantity with only magnitude (like height, speed, time)
-Magnitude = absolute value = length
Symbol:
According to the Pythagorean Theorem, if

then

-Unit Vector: A vector in terms of a single unit. It is the direction of a vector divided by its length.
-Position Vector: A vector that starts at the origin and ends at the point (x,y)


-Displacement Vector: the difference between an object’s initial & final positions.

Vector sum & vector difference:


Let’s try some * applications *…
Moustaches on a Plane
There are two moustaches on an x-y coordinate plane. A force from the origin of the plane is acting upon a curly English moustache at point A(4,11) and another on a Toothbrush moustache one at point B(12,5).






















This one's the best, just multiply the unit-vector answer by 10!

Also, as a side note, both of the vectors are position vectors, since they start from the origin.

Thaaaattt...should be it! Here's an extra help site: http://id.mind.net/~zona/mstm/physics/mechanics/vectors/vectors.html
or you can just ask me for help!

Just in case you guys didn't already know... I've got a concert tomorrow with the Glendale Youth Orchestra! If you're at all interested or willing to go out on a Tuesday night (7:30 precisely at the Alex Theater in Glendale), let me know- Beethoven 3 AND free tickets!!!



Oh, and Joey, you're up next. Hope ID went well for ya!