text: Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
  
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  #1 2011-03-29 00:43:57
  
  bobbym
  
  Offline
  
  Solving vector problems with geogebra
  
  Take a look at the problem over here:
  
  http://www.mathisfunforum.com/viewtopic … 17#p168917
  
  We want to solve for P and Q. We want to solve this using Geogebra.
  
  Define the origin as A
  
  Start by putting in the initial vectors. On the command line enter:
  
  u = (0,-5)
  
  Now create a point at (-8 , 0 ) called B. Get the angle of B and A of 15 degrees. That point will be called B'. Draw a vector from A to B' it should be called v. It will have a magnitude of 8 N.
  
  Add the two vectors u and v by inputting u + v in the command line. It should be called w or w'. Negate that vector to reflect it into the first quadrant by inputting:
  
  w = -w'
  
  This is the reflection vector.
  
  Put a point at ( 0 , 8 ) called C. Create an angle of 20 degrees using C and A. That will reate the point C'. Draw a vector from A to C'. Relabel C' to Q.
  
  Create a vector from A to ( 8 , 0 ) and call it P. Get the resultant of P and Q by adding in the command the name of two vectors they will be small letters. On mine they are a,z
  
  a + z
  
  The resultant is called b. Recolor this b vector red. Now move P ( the head of the horizontal vector ) towards A until it covers the reflection vector from the 3rd quadrant.
  
  Now if you drag C (0,8) towards A the red vector c will shorten. Keep adjusting P and C until the red vector exactly superimposes over the reflection vector.
  
  When you eyeballed it as best you can measure the distance between A and Q and A and P. I get Q = 7.55 and P = 5.18 That is quite close.
  
  In mathematics, you don't understand things. You just get used to them.
  I have the result, but I do not yet know how to get it.
  All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
  
  #2 2011-04-19 22:56:58
  
  alexa.pete9
  Member
  
  Offline
  
  Re: Solving vector problems with geogebra
  
  The "number" is correct for your solution to part one of the first problem, but I don't like what you have written.
  
  The resultant vector is
  
  (120 + 200COS52, 200SIN52)= (243.132, 157.602)
  
  The magnitude of this vector is
  
  Sqrt(243.1322 +157.6022) = 289.744
  
  The angle measurement of BAC is 52 degrees and the vector AD is the resultant vactor. The angle you want is BAD and you know the lengths of the sides of the triangle BAD so you can find the angle you need using the Law of Coaines.
  
  #3 2011-04-20 05:21:16
  
  bobbym
  
  Offline
  
  Re: Solving vector problems with geogebra
  
  Hi alexa.pete9;
  
  http://www.mathisfunforum.com/viewtopic.php?id=15250
  
  It is not possible to say you do not like what I did. I did not do it, Geogebra did. That is the point.
explanation: The document is a discussion thread from a math forum, where users are discussing the solution to a vector problem using Geogebra. The first post (#1) provides a detailed step-by-step guide to solving the problem using Geogebra, and the second post (#2) critiques the solution, providing an alternative method and pointing out the correct calculations. The third post (#3) is a response to the critique, defending the use of Geogebra.
  
  The document contains a question and an answer, as the initial post presents a problem and a solution, and the subsequent posts provide further discussion and critique.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
  
  You are not logged in.
  
  #1 2011-03-29 00:43:57
  
  bobbym
  
  Offline
  
  Solving vector problems with geogebra
  
  Take a look at the problem over here:
  
  http://www.mathisfunforum.com/viewtopic … 17#p168917
  
  We want to solve for P and Q. We want to solve this using Geogebra.
  
  Define the origin as A
  
  Start by putting in the initial vectors. On the command line enter:
  
  u = (0,-5)
  
  Now create a point at (-8 , 0 ) called B. Get the angle of B and A of 15 degrees. That point will be called B'. Draw a vector from A to B' it should be called v. It will have a magnitude of 8 N.
  
  Add the two vectors u and v by inputting u + v in the command line. It should be called w or w'. Negate that vector to reflect it into the first quadrant by inputting:
  
  w = -w'
  
  This is the reflection vector.
  
  Put a point at ( 0 , 8 ) called C. Create an angle of 20 degrees using C and A. That will reate the point C'. Draw a vector from A to C'. Relabel C' to Q.
  
  Create a vector from A to ( 8 , 0 ) and call it P. Get the resultant of P and Q by adding in the command the name of two vectors they will be small letters. On mine they are a,z
  
  a + z
  
  The resultant is called b. Recolor this b vector red. Now move P ( the head of the horizontal vector ) towards A until it covers the reflection vector from the 3rd quadrant.
  
  Now if you drag C (0,8) towards A the red vector c will shorten. Keep adjusting P and C until the red vector exactly superimposes over the reflection vector.
  
  When you eyeballed it as best you can measure the distance between A and Q and A and P. I get Q = 7.55 and P = 5.18 That is quite close.
  
  In mathematics, you don't understand things. You just get used to them.
  I have the result, but I do not yet know how to get it.
  All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
  
  #2 2011-04-19 22:56:58
  
  alexa.pete9
  Member
  
  Offline
  
  Re: Solving vector problems with geogebra
  
  The "number" is correct for your solution to part one of the first problem, but I don't like what you have written.
  
  The resultant vector is
  
  (120 + 200COS52, 200SIN52)= (243.132, 157.602)
  
  The magnitude of this vector is
  
  Sqrt(243.1322 +157.6022) = 289.744
  
  The angle measurement of BAC is 52 degrees and the vector AD is the resultant vactor. The angle you want is BAD and you know the lengths of the sides of the triangle BAD so you can find the angle you need using the Law of Coaines.
  
  #3 2011-04-20 05:21:16
  
  bobbym
  
  Offline
  
  Re: Solving vector problems with geogebra
  
  Hi alexa.pete9;
  
  http://www.mathisfunforum.com/viewtopic.php?id=15250
  
  It is not possible to say you do not like what I did. I did not do it, Geogebra did. That is the point.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a discussion thread from a math forum, where users are discussing the solution to a vector problem using Geogebra. The first post (#1) provides a detailed step-by-step guide to solving the problem using Geogebra, and the second post (#2) critiques the solution, providing an alternative method and pointing out the correct calculations. The third post (#3) is a response to the critique, defending the use of Geogebra.
  
  The document contains a question and an answer, as the initial post presents a problem and a solution, and the subsequent posts provide further discussion and critique.
  
  Label: question-and-answer