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Vector question confusion

  1. Apr 24, 2017 #1
    1. The problem statement, all variables and given/known data
    Hi guys, I have am having a problem with the questions below.
    upload_2017-4-24_19-0-22.png

    what do they mean by show ## B_{perp}=B sin\theta ## I mean I know it a silly question, and my first thought is just use socahtoa, or am I missing something. The question from there on is straight forward enough,. I just have a feeling I am missing something or am I over thinking it completely?

    2. Relevant equations


    3. The attempt at a solution
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Apr 24, 2017 #2

    Doc Al

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    I think that's all there is to it.
     
  4. Apr 24, 2017 #3

    FactChecker

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    Do you have another way to calculate B? If so, you should show that it gives the same answer.
     
  5. Apr 24, 2017 #4
    I was thinking of showing it with an orthogonal projection, would this be correct?
     
  6. Apr 24, 2017 #5

    Ray Vickson

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    Try it out, to see whether it works.
     
  7. Apr 24, 2017 #6

    FactChecker

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    The exercise does say that the first step is to decompose B into the projection on A and the rejection (orthogonal projection) on A.
     
  8. Apr 28, 2017 #7
    first thanks for the responses I have used the orthogonal projection, ]which I feel is the best way to show this relationship.
     
  9. May 7, 2017 #8
    Hi guys, my perivous comment is bogus ^^^ I have been looking ove this question again and I now fully confusing myself. The way I thought I could show it which is I think i worng now is that I could say that:

    ##|| Proj_ba+ \perp Proj_ba=b##

    By rearanging this in to: ##|| b-Proj_ba= \perp Proj_ba## I would be able to show what they wanted, when I did it before for some reason I thought that ##|b|(1-cos\theta)=|b| sin \theta## I have no idea why throught this at the time but I know this to be wrong. Can someone please give some advice on that matter thank you in advance.
     
  10. May 7, 2017 #9

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    I'm not sure about your notation, but it is true that a||b + a⊥b = a, where a||b is the projection of vector a onto b and a⊥b is the vector rejection of a on b. So if you can calculate a||b, then you can use that to calculate a⊥b and prove what the exercise asked for.
     
  11. May 7, 2017 #10
    That what I oringinally throught but how could you, I cant seem to figure how to. I mean I know how the projection vector, ##a_||b## using dot product but how I convert to sine is what confusing me at the moment. Erm am I on the right lines with this:


    ##|a_{||b}+a_{\perp b}|=|a|## ##\rightarrow## ##|a_{||b}|+|a_{\perp b}|=|a|##,

    If this correct the ##a_{||b}##=##|a|cos\theta##, but taking the mod of this is where it get confusing, what is the mod of this?
     
  12. May 7, 2017 #11

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    If you can calculate cos(θ) then you can get θ and then sin(θ).
    No. Use the Pythagorean theorem instead. a||b and a⊥b are at right angles.
    It is not true. The mod is already there. |a||b| = |a|cos(θ).
     
  13. May 7, 2017 #12
    Im sorry, but I am now more confuse to how ##arccos(\frac{|a_{||b}|}{|a|})## to get the desired result. I really not seeing somthing here.

    I know this not the way you are saying but can you use the face that ##tan x= sinx/cosx ## & ##cosx=a_{||b}/|a|##?


    So now I am thinking this, when you said pythagours

    ##(1) |B_{||}|^2+|B_{\perp}|^2=|B|^2##& ##(2) B_{||}=Bcos\theta##

    sub 1 into to 2 and rearrange for ##|B_{\perp}|^2##

    I then get: ##B{\perp}=\sqrt(|B|^2)\sqrt(1-cos\theta^2)##
     
    Last edited: May 7, 2017
  14. May 7, 2017 #13

    FactChecker

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    θ = arccos( |a||b| / |a| )
    a||b is a vector. If you used |a||b| that would be correct, but sort of a more complicated way to do things. The goal is to find θ, which would be the same as x. So it is simpler to just use θ = arccos( |a||b| / |a| )
    yes. They form a right triangle.
    no. Look at figure 3 in post #1. B|| and B are vectors that point in different directions. B|| points in the direction of A. That's why I think the notation B||A is better than B||. Likewise BA is better notation than B. That notation would help you keep track of what vector they are parallel or perpendicular to (which is A).
    They are the vectors shown in figure 3 of post #1.
     
  15. May 7, 2017 #14
    Ok im still a bit hazy. So the method i described is incorrect? Ok i cn see that B has been projected on to A, so the sclar projection is ##B_{|| A}=|B|cos\theta## so that why I throught I could use that in pythagours as the sclar projection would be the length so then i could you that with the length of the vect ##B## to show the result.
     
  16. May 7, 2017 #15

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    I'm still not sure that we are talking about the same thing. ##B_{|| A}## is a vector and ##|B|cos\theta## is not. So they are not equal. The scalar projection of B onto A is | ##B_{|| A}## | =##|B|cos\theta##. The modulus symbols are necessary. Please don't omit them.
     
  17. May 8, 2017 #16
    First I apologise for my poor latex and also thank you being patient with me. I have finally figure out the problem. Thank you for your help it was much apprecited.
     
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