Vectors Questions Homework - Fundamentals of Physics

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SUMMARY

The discussion focuses on solving vector problems from the "Fundamentals of Physics" textbook, specifically questions 53 and 54 regarding the angle between two vectors A and B of equal magnitudes. For question 53, the angle must be approximately 1.1458 degrees when the magnitude of A+B is 100 times greater than A-B, derived using the Law of Cosines and geometric principles. In question 54, the angle can be calculated using the formula arctan(1/n), where n represents the ratio of the magnitudes of A+B to A-B. The participants confirm the correctness of the calculations and the geometric interpretation of vector addition and subtraction.

PREREQUISITES
  • Understanding of vector addition and subtraction
  • Familiarity with the Law of Cosines
  • Basic knowledge of trigonometry, specifically arctangent functions
  • Concept of geometric representation of vectors, including rhombus properties
NEXT STEPS
  • Study the Law of Cosines in detail for vector applications
  • Explore vector addition and subtraction using geometric methods
  • Learn about the properties of rhombuses and their relevance in vector problems
  • Practice solving similar vector problems involving angles and magnitudes
USEFUL FOR

Students studying physics, particularly those focusing on vector analysis, as well as educators looking for examples of vector problem-solving techniques in a classroom setting.

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Homework Statement


Ok I did all the chapter 3 questions in Fundamentals of Physics, but I could not get 2 of them.

53. Two vectors A and B have precisely equal magnitudes. For the magnitude of A+B to be 100 times greater that the magnitude of A-B, what must be the angle between them?

54. Two vectors A and B have precisely equal magnitudes. For the magnitude of A+B to be n times greater that the magnitude of A-B, what must be the angle between them?

Homework Equations


Law of Cosines.

The Attempt at a Solution


I am pretty sure the angle has to be small for 53. I know that the magnitude of A is equal to the magnitude of B. Using the Law of Cosines I get cos:smile:=1-mag(A-B)/2A^2. For A+B I get cos:-p=1-50mag(A-B)/A^2. 180-cos:-p=cos:smile:. I don't know where to go after this. I messed around with the equations, but I can't ever get rid of A or B. The second one I have not tried yet, because if I can't the first, then how am I supposed to get the second. THANKS for the help!
 
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Do you know the parallelogram geometric view of addition and subtraction of vectors? Note that in the case that all four sides are equal, you are dealing with a rhombus, for which the diagonals bisect each other in a right angle.
 
Oh got it so, that makes a right triangle with legs 50mag(A-B) and another leg mag(A-B)/2. aTan(1/100)=.5729. Doubling that i get 1.1458 degrees. Right?

For the second one I get aTan(1/n). Right?

Thanks slider142!
 
Yep. Good job!
 

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