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Homework Help: Find the vector and find the magnitude of the vector

  1. Mar 12, 2013 #1
    1. The problem statement, all variables and given/known data

    Vectors A and B are shown in the figure. Vector C is given by C=B-A. The magnitude of vector A is 15.0 units, and the magnitude of Vector B is 5.00 units.


    Part a) Find the vector C?

    Part b) What is the magnitude of vector C?

    2. Relevant equations


    3. The attempt at a solution

    I drew these lines on the diagram and then solve for the sides using the angles with the SOHCAHTOA Method. Then after I found the sides I use the Pythagorean theorem to solve for side C. I am assuming vector C is the line connected with vector A and B.




    Part b) the magnitude of vector C is 14.523.

    Q: How to do part a? I don't understand what it means by finding vector C. Also is the above correct?
  2. jcsd
  3. Mar 12, 2013 #2

    Simon Bridge

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    finding vector C is what you did when you worked out the components. You cannot do part (b) without it. You can answer the question by drawing the vector on the diagram and/or by writing out the components with the i and j unit vectors.

    Did you see that the two vectors use special triangles so you don't need a calculator?
  4. Mar 12, 2013 #3
    I'm a bit confused.

    The value I got, 14.523 units, is that the answer for part b, which is to find the magnitude?

    So if I understand what you're saying, part a is to either draw the vector on the diagram or write out the components. Example answer is Vector C = 6.27i + 13.1j units.
  5. Mar 13, 2013 #4

    Simon Bridge

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    That's a vector all right - well done! The magnitude of that vector would be 14.523 units... the answer to (b).
  6. Apr 5, 2013 #5
    Hey quick question, how do I solve this using the physics method and not the pre-calculus method like I did above using SOHCAHTOA?
  7. Apr 5, 2013 #6

    Simon Bridge

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    sohcahtoa is just a mnemonic to remember the basic trigonometric relations.

    trigonometry (the art of measuring triangles) is a basic tool of physics ... there is more to it than sohcahtoa.

    In this case I think I suggested the special triangles.
    The two in the diagram were 45-45-90, and 30-60-90 ...
    The first one is the triangle you get when O=A, and the second is when H=2O
    The lengths of the sides have ratios O:A:H = 1:1:√2 and 1:√3:2

    i.e. when O=A then H=O√2

    - for vector A: Ay=-Ax=|A|/√2
    - for vector B: By=-|B|/2, Bx=-|B|(√3/2)

    One advantage of this approach is that you can write:
    |A|/√2 = 15/√2

    ... and stop there - it's exact: no need to go into decimals.
    Leave the heavy calculator work to the end avoids accumulating rounding-off errors.
    It also makes your working easier to troubleshoot if you make a mistake.

    Another approach is to draw out the vectors head-to-tail so C=B-A = B+(-A)
    the magnitude of C is the length of the third side of the resulting triangle - which you can get from the cosine rule. The sine rule gives you the other angles - which gives you the direction of C.
    However - it is easiest to do this one in terms of components.
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