course PHY 201
June 14 around 10:00am
ph1 query 0Most queries in this course will ask you questions about class notes, readings, text problems and experiments. Since the first two assignments have been lab-related, the first two queries are related to the those exercises. While the remaining queries in this course are in question-answer format, the first two will be in the form of open-ended questions. Interpret these questions and answer them as best you can.
Different first-semester courses address the issues of experimental precision, experimental error, reporting of results and analysis in different ways and at different levels. One purpose of these initial lab exercises is to familiarize your instructor with your work and you with the instructor 's expectations.
Comment on your experience with the three lab exercises you encountered in this assignment or in recent assignments.
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Question: This question, related to the use of the TIMER program in an experimental situation, is posed in terms of a familiar first-semester system.
Suppose you use a computer timer to time a steel ball 1 inch in diameter rolling down a straight wooden incline about 50 cm long. If the computer timer indicates that on five trials the times of an object down an incline are 2.42sec, 2.56 sec, 2.38 sec, 2.47 sec and 2.31 sec, then to what extent do you think the discrepancies could be explained by each of the following:
• The lack of precision of the TIMER program.
To what extent to you think the discrepancies are explained by this factor?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
The Timer program has flaws itself, which gives it an uncertainty. The timer itself can only count some of the parts of a second and not all of the parts down to the millisecond, but the difference is not that huge but it is still an uncertainty.
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• The uncertain precision of human triggering (uncertainty associated with an actual human finger on a computer mouse)
To what extent to you think the discrepancies are explained by this factor?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
Every time you do this experiment you are not going to push the start and stop button on the timer at the same time every time, which will give room for different times the ball rolled down the wooden incline. So, the timer program has little to do with the differences of the seconds that the ball rolled down the incline.
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• Actual differences in the time required for the object to travel the same distance.
To what extent to you think the discrepancies are explained by this factor?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
Every time you do the experiment you are not going to put the ball at the exact location every time, which could add more time. Also, the exact place you click the time off is not going to be the exact location either, which could also leave room for uncertainty.
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• Differences in positioning the object prior to release.
To what extent to you think the discrepancies are explained by this factor?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
Where you put the ball on the incline is very important. The lower you put it down the incline, the less time it will take to go off of the incline, and the higher you put the ball, the more time it would take.
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• Human uncertainty in observing exactly when the object reached the end of the incline.
To what extent to you think the discrepancies are explained by this factor?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
Every time the ball rolls off the incline, you are not going to pin point exactly where and when the ball fell off, which will leave room for uncertainty.
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Question: How much uncertainty do you think each of the following would actually contribute to the uncertainty in timing a number of trials for the ball-down-an-incline lab?
• The lack of precision of the TIMER program.
To what extent to you think this factor would contribute to the uncertainty?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
I think that is the one that affects the uncertainty the least off all of the uncertainties.
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• The uncertain precision of human triggering (uncertainty associated with an actual human finger on a computer mouse)
To what extent to you think this factor would contribute to the uncertainty?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
I think that human triggering is the second most contributor to the uncertainty of the differences, because use humans are not perfect by no means.
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• Actual differences in the time required for the object to travel the same distance.
To what extent to you think this factor would contribute to the uncertainty?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
The actual differences, I think, are next to the last of the contributors, because the actual differences do not have much uncertainty with it compared to the rest of the contributors.
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• Differences in positioning the object prior to release.
To what extent to you think this factor would contribute to the uncertainty?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
I think that the different positioning of the object is the third most contributor to the uncertainty, because no matter where you put the ball, you are still going to have greater doubt with uncertainty of observing and human triggering.
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• Human uncertainty in observing exactly when the object reached the end of the incline.
To what extent to you think this factor would contribute to the uncertainty?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
I think that human uncertainty of observing exactly when the object reached the end of the incline is the greatest factor in contributing to uncertainty. No matter where you put the ball, the actual differences of the times, when you clicked the off button on the timer, and the precision of the timer, the precise time the ball ran off of the incline will be greater than any other factor contributing to uncertainty.
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Question: What, if anything, could you do about the uncertainty due to each of the following? Address each specifically.
• The lack of precision of the TIMER program.
What do you think you could do about the uncertainty due to this factor?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
I do not think you could do anything to the precision of the TIMER program, because you cannot change a program, unless you try another program.
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• The uncertainty precision of human triggering (uncertainty associated with an actual human finger on a computer mouse)
What do you think you could do about the uncertainty due to this factor?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
I do not think you could change this factor either, because the finger on the mouse could not change.
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• Actual differences in the time required for the object to travel the same distance.
What do you think you could do about the uncertainty due to this factor?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
The actual differences could change with different numbers used, but only with the discretion of the timer, which is not that accurate.
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• Differences in positioning the object prior to release.
What do you think you could do about the uncertainty due to this factor?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
The different positioning of the object prior to release could be improved, but only with the discretion of the timer and human triggering which are not that accurate.
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• Human uncertainty in observing exactly when the object reached the end of the incline.
What do you think you could do about the uncertainty due to this factor?
your answer: vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
I do not think that you could improve this uncertainty, unless you are extremely vigilant, and even if you are extremely vigilant you are not going to press the button at the same time for every released object at the end of the incline.
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Question: If, as in the object-down-an-incline experiment, you know the distance an object rolls down an incline and the time required, explain how you will use this information to find the object 's average speed on the incline.
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Your solution:
When doing a object-down-an-incline experiment, you can find out how fast the objects’ average speed is down the incline. First, you measure how long the object took to reach the end of the incline and that is your distance. Second, the time it took to reach the bottom of the incline is the time required for that object. Then, you divide distance traveled into time elapsed to get your average speed of the object.
confidence #$&* 3
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Question: If an object travels 40 centimeters down an incline in 5 seconds then what is its average velocity on the incline? Explain how your answer is connected to your experience.
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Your solution:
If the object travels a distance of 40 cm’s down the incline in 5 seconds, then the average velocity would be: 40 cm / 5 seconds = 8 cm’s per second the object traveled. I used the same amount of centimeters, but mine did not time and velocity was not as high.
confidence #$&* 3
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Question: If the same object requires 3 second to reach the halfway point, what is its average velocity on the first half of the incline and what is its average velocity on the second half?
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Your solution:
If the halfway point is 20 cm’s and it traveled 3 seconds to reach the halfway point then the average velocity on the first half is: 20 cm’s / 3 seconds = 6.67 cm’s per second. If the first took 3 seconds then the second took 2 seconds, so the average velocity of the second half is: 20 cm’s / 2 seconds = 10 cm’s per second.
confidence #$&* 3
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Question: `q According to the results of your introductory pendulum experiment, do you think doubling the length of the pendulum will result in half the frequency (frequency can be thought of as the number of cycles per minute), more than half or less than half?
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Your solution:
When the length of the pendulum from my fingers got longer and longer the cycles got longer and longer by the minute. I think that when you double the length the result will be less than half of the frequency, because the longer the string goes that less and less frequencies you are going to have.
confidence #$&* 2
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Question: `q Note that for a graph of y vs. x, a point on the x axis has y coordinate zero and a point on the y axis has x coordinate zero. In your own words explain why this is so.
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Your solution:
The two axis’s cross in the (0,0) point. The x axis is going from right to left, so no matter what the x coordinate is as long as the y coordinate is 0, it is always going to stay on the x axis. The y axis is going from top to bottom, so not matter what the y coordinate as long as the x coordinate is 0, it is always going to stay on the y axis.
confidence #$&* 3
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Question: `q On a graph of frequency vs. pendulum length (where frequency is on the vertical axis and length on the horizontal), what would it mean for the graph to intersect the vertical axis (i.e., what would it mean, in terms of the pendulum and its behavior, if the line or curve representing frequency vs. length goes through the vertical axis)? What would this tell you about the length and frequency of the pendulum?
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Your solution:
This would tell me that the pendulum has no length, because it has to have length to conduct the experiment. You cannot go through the vertical axis if the length is increasing from left to right, because you have to have some sort of the string length to have frequencies and to conduct the experiment. The length cannot be a negative number, because you cannot cut a negative length of string for the pendulum.
confidence #$&* 2
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Question: `q On a graph of frequency vs. pendulum length, what would it mean for the graph to intersect the horizontal axis (i.e., what would it mean, in terms of the pendulum and its behavior, if the line or curve representing frequency vs. length goes through the horizontal axis)? What would this tell you about the length and frequency of the pendulum?
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Your solution:
I think that no matter what the length is the frequency is never going to reach zero. Every time you swing the pendulum back and forth that is one frequency. You cannot have a negative frequency in a cycle, so this is not possible.
confidence #$&* 2
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Question: `qIf a ball rolls down between two points with an average velocity of 6 cm / sec, and if it takes 5 sec between the points, then how far apart are the points?
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Your solution:
If the ball rolled between the two points at the average velocity of 6 cm / sec for 5 seconds then the distance between the points is 30 cm’s: 6 cm / sec * 5 sec = 30 cm’s in length.
confidence #$&* 3
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Given Solution:
`aOn the average the ball moves 6 centimeters every second, so in 5 seconds it will move 30 cm.
The formal calculation goes like this:
• We know that vAve = `ds / `dt, where vAve is ave velocity, `ds is displacement and `dt is the time interval.
• It follows by algebraic rearrangement that `ds = vAve * `dt.
• We are told that vAve = 6 cm / sec and `dt = 5 sec. It therefore follows that
• `ds = 6 cm / sec * 5 sec = 30 (cm / sec) * sec = 30 cm.
The details of the algebraic rearrangement are as follows:
• vAve = `ds / `dt. We multiply both sides of the equation by `dt:
• vAve * `dt = `ds / `dt * `dt. We simplify to obtain
• vAve * `dt = `ds, which we then write as{}`ds = vAve *`dt
Be sure to address anything you do not fully understand in your self-critique.
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Your solution:
OK
confidence #$&* 3
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Question: `qYou were asked to read the text and some of the problems at the end of the section. Tell your instructor about something in the text you understood up to a point but didn't understand fully. Explain what you did understand, and ask the best question you can about what you didn't understand.
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Your solution:
I am still unclear about significant figures and what is not a significant figure and what is. Can you explain a little better than the book about what significant figures are?
The information below was not included in your document:
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SOME COMMON QUESTIONS:
QUESTION: I didn’t understand how to calculate uncertainty for a number such as 1.34. When given examples we had problems such as 1.34 ±0.5 and with that we had a formula (0.5/1.34)*100. So I do not understand how to compute uncertainty when no estimated uncertainty is given.
INSTRUCTOR RESPONSE:
The +- number is the uncertainty in the measurement.
The percent uncertainty is the uncertainty, expressed as a percent of the number being observed.
So the question in this case is simply, 'what percent of 1.34 is 0.5?'.
0.5 / 1.34 = .037, approximately. So 0.5 is .037 of 1.34.
.037 is the same as 3.7%.
I recommend understanding the principles of ratio, proportion and percent as opposed to using a formula. These principles are part of the standard school curriculum, though it does not appear that these concepts have been well mastered by the majority of students who have completed the curriculum. However most students who have the prerequisites for this course do fine with these ideas, after a little review. It will in the long run save you time to do so.
There are numerous Web resources available for understanding these concepts. You should check out these resources and let me know if you have questions.
QUESTION: I understood the main points of changing the different units, but I’m not sure when in the problem I should change the number to 10 raised to a certain power. In example 1-8 I did not understand why they changed 70 beats/min to 2 x 10^9 s.
2 * 10^9 is about the number of seconds in 70 years.
70 beats / min were not changed to 2 * 10^9 seconds; in changing the beats / minute to beats in a lifetime, there was a step where it was necessary to multiply by 2 * 10^9 seconds.
The example actually used 80 beats / min as a basis for the solution. This was converted to beats / second by the calculation
80 beats / min * 1 minute / (60 seconds), which would yield about 1.33 beats / second.
This was then multiplied by 2 * 10^9 seconds to get the number of beats in a lifetime:
2 * 10^9 seconds * 1.33 beats / second = 3 * 10^9 beats.
In the given solution 80 beats / min * 1 minute / (60 seconds) was not actually calculated; instead 80 beats / min * 1 minute / (60 seconds) was multiplied by 2 * 10^9 seconds in one step
80 beats / min * 1 minute / (60 seconds) * 2 * 10^9 seconds = 3 * 10^9 beats.
In your instructor's opinion the unit 'beats' should have been left in the result; the text expressed the result simply as 3 * 10^9, apparently ignoring the fact that the unit 'beats' was included in the quantities on the left-hand side.
Also the text identified this number as 3 trillion. In the British terminology this would be correct; in American terminology this number would be 3 billion, not 3 trillion.
COMMENT:
I thought that these problems were pretty basic and felt that I understood them well. However, when I got to questions 14 (determine your own mass in kg) and 15 (determining how many meters away the Sun is from the Earth), I did not understand how to complete these. I know my weight in pounds, but how can that be converted to mass in kilograms? I can look up how to convert miles to meters, but is this something I should already know?
INSTRUCTOR RESPONSE:
Both of these questions could be answered knowing that an object with a mass of 1 kg has a weight of 2.2 lb, and that an inch is 2.54 centimeters. This assumes that you know how many feet in a mile, and that the Sun is 93 million miles away. All these things should be common knowledge, but it doesn't appear to be so.
For my own weight I would reason as follows:
I weigh 170 lb and every kg of my mass weighs 2.2 lb. I'll have fewer kg of mass than I will pounds of weight, so it's reasonable to conclude that my mass is 170 / 2.2 kg, or about 78 kg.
More formally 170 lb * (1 kg / (2.2 lb) ) = 170 / 2.2 kg = 78 kg, approx.. (technical point: this isn't really right because pounds and kilograms don't measure the same thing--pounds measure force and kg measure mass--but we'll worry about that later in the course).
Converting 93 million miles to kilometers:
93 million miles * (5280 feet / mile) * (12 inches / foot) * (2.54 cm / inch) * (1 meter / (100 cm) ) = 160 billion meters (approx.) or 160 million kilometers.
QUESTION
What proved to be most tricky in the problems portion was the scientific notation. I am somewhat familiar with this from
past math classes, but had trouble when dealing with using the powers of 10. I had trouble dealing with which way to move my decimal according to the problems that were written as 10^-3 versus 10^3. Which way do you move the decimal when dealing with negative or positive powers of 10?
INSTRUCTOR RESPONSE
Using your numbers, 10^3 means 10 * 10 * 10 = 1000.
When you multiply a number by 1000 you move the decimal accordingly. For example 3.5 * 1000 = 3500.
10^-3 means 1 / 10^3 = 1 / (10 * 10 * 10) = 1 / 1000.
When you multiply by 10^-3 you are therefore multiplying by 1 / 1000, which is the same as dividing by 1000, or multiplying by .001.
For example 3.5 * 10^-3 = 3.5 * .001 = .0035.
As another example 5 700 000 * 10^-3 would be 5 700 000 * (1 / 1000) = 5 700.
From these examples you should be able to infer how the decimal point moves.
You can also search the Web under 'laws of exponents', 'arithmetic in scientific notation', and other keywords.
There isn't a single site I can recommend, and if I did find a good one its URL might change by the time you try to locate it. In any case it's best to let you judge the available materials yourself.
When searching under 'arithmetic in scientific notation' using Google, the following appear as additional suggested search phrases:
scientific notation
exponents
scientific notation metric prefixes
significant digits
multiply with scientific notation
scientific notation decimal
scientific notation lessons
addition and subtraction with scientific notation
scientific notation metric system
'scientific notation lessons' might be a good place to look.
QUESTIONS AND RESPONSES
1)In the text question five asks for the percent uncertainty of a measurement given 1.57 m^2
I think that we figure this by an uncertainty of .01/1.57m^2 = .6369 or approximately one. ??????Am I correct in how I
calculate this??????? Can I asuume that if the number given was 1.579 then we would calculate it by .001/1.57 = .1 % approximately or am I incorrect?????
You're on the right track.
There are two ways to look at this.
1.57 m^2 represents a quantity which rounds off to 1.57, so presumably lies between 1.565 and 1.575.
This means that the quantity is within .005 of 1.57.
.005 / 1.57 = .003, approx., so the uncertainty is .003 of 1.57, which is the same as 0.3%, of 1.57.
Another way to look at it:
1.57 could be interpreted to mean a number between 1.56 and 1.58. The uncertainty would then be .01, which is .01 / 1.57 = .006, or 6%, of 1.57.
2)In the text question number 11 the book asks what is the percent uncertainty in the volume of a sphere whose radius is
r=2.86 plus or minus .09.
I know that the Volume of a sphere is 4/3 pi r^3, so I calculated the volume to be 4/3 pi (2.86)^3 = 97.99 and to get the
percent uncertainty I tried to divide 0.09/97.99 * 100 =.091846, but the book answer is 9% ??????I am not sure what i am doing wrong here?????????????????
Again there are two ways to approach this.
I believe the book tells you that the uncertainty in the square of a number is double the uncertainty in the number, and the uncertainty in the cube of the number is trip the uncertainty in the number.
An uncertainty of .09 in a measurement of 2.86 is .09 / 2.86 = .03, approx., or about 3%. As you state, you cube the radius to find the volume. When 2.86 is cubed, the resulting number has three times the uncertainty, or about 9%.
Another approach:
Calculate the volume for r = 2.86.
Then calculate the volume for r = 2.86 - .09 = 2.77.
You will find that the resulting volumes differ by about 9%.
You could just as well have calculated the volume for r = 2.86 + .09 = 2.95. Again you would find that the volume differs from the r = 2.86 volume by about 9%.
STUDENT QUESTION
When reading the section about the scientific notation some of the answers were written in powers of 10 and some were just
written regularly. How do I know when to turn my answer into a power of 10 or to leave my answer as is?
INSTRUCTOR RESPONSE
Good question.
Convenience and readability are the main factors. It's a lot less typing or writing to use 438 000 000 000 000 000 000 than 4.38 * 10^20, and it's easier for the reader to understand what 10^20 means than to count up all the zeros.
For readability any number greater than 100 000 or less than .001 should probably be written in scientific notation.
When scientific notation is first used in a calculation or result, it should be used with all numbers in that step, and in every subsequent step of the solution.
QUESTION
In my problems (I am working from the University Physics text- exercise 1.14) they are asking for the ratio of length to
width of a rectangle based on the fact that both of the measurements have uncertainty. ?????Is there anything special you
have to do when adding or multiplying numbers with uncertainty?????? I know that there are rules with significant figures,
but I don’t understand if the same is true for uncertain measurements.
INSTRUCTOR RESPONSE:
For example:
If there is a 5% uncertainty in length and no significant uncertainty in width, then area will be uncertain by 5%.
If there is a 5% uncertainty in length and a 3% uncertainty in width, then it is possible for the area result to be as much as 1.05 * 1.03 = 1.08 times the actual area, or as little as .95 * .97 = .92 times the actual area. Thus the area is uncertain by about 8%.
This generalizes. The percent uncertainty in the product or quotient of two quantities is equal to the sum of the percent uncertainties in the individual quantities (assuming the uncertainties are small compared to the quantities themselves).
(optional addition for University Physics students): The argument is a little abstract for this level, but the proof that it must be so, and the degree to which it actually is so, can be understood in terms of the product rule (fg) ' = f ' g + g ' f. However we won't go into those details at this point.
QUESTIONs RELATED TO UNIVERSITY PHYSICS (relevant only to University Physics students)
I understand everything but the part on measuring the individual i j k vectors by using cosine.
INSTRUCTOR RESPONSE
It's not completely clear what you are asking, but I suspect it has to do with direction cosines.
The vector A = a_1 i + a_2 j + a_3 k makes angles with the directions of the x axis, the y axis and the z axis.
Let's consider first the x axis.
The direction of the x axis is the same as the direction of the unit vector i.
The projection of A on the x direction is just a_1. This is obvious, but it can also be found by projecting the A vector on the i vector.
This projection is just | A | cos(alpha), where alpha is the angle between A and the x direction.
Now A dot i = A = (a_1 i + a_2 j + a_3 k) dot i = A = a_1 i dot i + a_2 j dot i + a_3 k dot i = a_1 * 1 + a_2 * 0 + a_3 * 0 = a_1.
It's also the case that A dot i = | A | | i | cos(alpha). Since | i | = 1, it follows that A dot i = | A | cos(alpha), so that
cos(alpha) = A dot i / | A | = a_1 / sqrt( a_1 ^ 2 + a_2 ^ 2 + a_3 ^ 2 ).
Making the convention that alpha is the angle made by the vector with the x direction, we say that cos(alpha) is the direction cosine of the vector with the x axis.
If beta and gamma are, respectively, the angles with the y and z axes, reasoning similar to the above tells us that
cos(beta) = a_2 / sqrt( a_1 ^ 2 + a_2 ^ 2 + a_3 ^ 2 ) and
cos(gamma) = a_3 / sqrt( a_1 ^ 2 + a_2 ^ 2 + a_3 ^ 2 ).
cos(alpha), cos(beta) and cos(gamma) are called the 'direction cosines of the vector A' with respect to the three coordinate axes.
Recall that alpha, beta and gamma are the angles made the the vector with the three respective coordinate axes.
If we know the direction cosines and the magnitude of the vector, we can among other things find its projection on any of the coordinate axes.
STUDENT QUESTION (University Physics)
Chapter 1 wasn’t bad of course I had to read in detail the vector section there is little confusion on what is meant by
antiparallel. Does that mean that you wouldn’t displace anything if the magnitude was equal only the direction was different?
Also when handwritten vectors are written above the say A the arrow is only in one direction (to the right) not the direction
traveled?
INSTRUCTOR RESPONSE
I don't have that reference handy, but my understanding of the word 'antiparallel' is two vectors, one of which is in the direction exactly opposite the other.
If two vectors are antiparallel, then their dot product would equal negative of the product of their magnitudes:
The angle theta between antiparallel vectors v and w would be 180 degrees, so v dot w = | v | * | w | * cos(180 deg) = - | v | * | w | .
Please feel free to include additional comments or questions:
You left off the last part of the document. I've inserted it above.
Your question is related to questions answered in this part of the Query. After reviewing the information relevant to the question, you are welcome to ask additional questions.
&#Good work. See my notes and let me know if you have questions. &#
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