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H/W help on Geometric vs Component Vector Addition

  1. Jan 15, 2012 #1
    1. The problem statement, all variables and given/known data

    Find y component of vector C from its length and the angle it makes with the x axis, that is, from geometry. Express the y component of vector C in terms of C and [itex]\phi[/itex].

    2. Relevant equations

    Vector addition using geometry:

    1) C = [itex]\sqrt{A^{2}+B^{2}-2ABcos(c)}[/itex]. Law of Cosines

    2) [itex]\phi[/itex] = sin[itex]^{-1}[/itex][itex]\frac{Bsin(c)}{C}[/itex]. Law of Sines

    Vector addition using components:

    3) C[itex]_{x}[/itex] = A + Bcos[itex]\vartheta[/itex].

    4) C[itex]_{y}[/itex] = Bsin[itex]\vartheta[/itex]



    3. The attempt at a solution

    Since I'm supposed to use geometry, I'm limited to using equations 1 or 2 (Law of cosine and sin)

    I could easily determine C[itex]_{y}[/itex] by making a right triangle and forming the equation: [sin[itex]\phi[/itex]=[itex]\frac{Cy}{C}[/itex]] = C[itex]_{y}[/itex] = Csin[itex]\phi[/itex]


    However, that would be using the component addition method. I don't see how I would use geometry (Law of sine or cosine) to produce the same answer of Csin[itex]\phi[/itex].

    The only thing I could come up with was taking SAS (side angle side) [itex]\sqrt{C}[/itex][itex]_{y}[/itex][itex]^{2}[/itex] = [itex]\sqrt{C^{2}+C_{x}^{2}-2C_{x}Ccos\phi}[/itex] , but then how does that equate to
    C[itex]_{y}[/itex] = Csin[itex]\phi[/itex] ?


    Am I supposed to find the side using the Law of Cosine equation? If so, then

    Have I described the correct way of proceeding with the problem? If so, then

    Did I come up with the correct Law of Cosine forumula? If so, then

    Am I supposed to put it in the same form that I would have gotten using component addition? ( C[itex]_{y}[/itex] = Csin[itex]\phi[/itex]) ? If so, then

    How does [itex]\sqrt{C^{2}+C_{x}^{2}-2C_{x}Ccos\phi}[/itex] turn into Csin[itex]\phi[/itex] if they are essentially equal to each other and represent C[itex]_{y}[/itex]?
    Last edited: Jan 15, 2012
  2. jcsd
  3. Jan 15, 2012 #2


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    That should be [itex]\sqrt{C^{2}+C_{x}^{\,2}-2C_{x}Ccos\phi}\,.[/itex] You have Cy under the radical. It should be Cx .
  4. Jan 15, 2012 #3
    Thanks, appreciate the response. Fixed the typos, but still confused for the original reason. I had the correct equation on paper, just typed it incorrectly in the message post. I had actually derived the equation with the given terms from scratch just to make sure it was the correct equation.

    So there's no confusion on the Law of Cosine equation itself, I'm just confused on how I'm supposed to use the Law of Cosines to get Csin[itex]\phi[/itex].
    Last edited: Jan 16, 2012
  5. Jan 16, 2012 #4
    Maybe restating the question will make it more clear what i'm trying to seek help on exactly.

    The only answer that works in my MasteringPhysics homework is Csin[itex]\phi[/itex], but I don't know how to get to this answer geometrically (using Law of Cosines). How would this be accomplished?
  6. Jan 16, 2012 #5


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    I'm puzzled by this thread.

    What is it that you are given?

    If you're given the magnitude of vector, C, and angle, ɸ, then it seems that the preferred way to find Cy is simply Cy = C sin(ɸ).

    If you're given the magnitudes of vectors, A and B, and angle, ɸ, then using the Law of Cosines and the Law of Sines makes some sense.

    If you're given the magnitudes of vectors, A and B, and angle, θ, then Cy is simply equal to By and is given by Cy = By = B sin(θ). Of course, if the answer should be in terms of C and ɸ, then ... well, come to think of it, that would be crazy !

    Are you sure you're not given the magnitudes of vectors, A and B, and angle, θ, and then supposed to find ɸ and the magnitude of C
  7. Jan 16, 2012 #6

    I like Serena

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    Hmm, since you want Cy geometrically, that means you need to draw a triangle that includes Cy as one of its sides.
    There are 2 obvious triangles that have Cy as one of its sides.

    The first triangle is by definition the one that has Cx, Cy, and C as its sides and that has an angle of 90 degrees between Cx and Cy.

    The other triangle is the one that has Bx, Cy, and B as its sides, and that also has an angle of 90 degrees between Bx and Cy.

    If we pick the first triangle, then using the law of sines, you would get:
    $${C \over \sin 90^0} = {C_y \over \sin \phi}$$

    Note that this basically boils down to the definition of the sine:
    $$\sin(\phi)={C_y \over C}$$
  8. Jan 16, 2012 #7

    Simon Bridge

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    At the risk of stating the bleeding obvious - isn't that [itex]C_y=C\sin(\phi)[/itex]??? from the definition of the sine. OK - sine rule works too ... [edit]oh - serenaphile beat me to it :/
    Last edited: Jan 16, 2012
  9. Jan 16, 2012 #8

    So what you are saying is that Law of Cosines is completely irrelevent for this problem, geometrically? And between Law of Cosine and Law of Sin, only Law of Sin can be used to find Cy in this particular triangle?


    Okay, I didn't realize before that I was looking at a SAA scenario, thus allowing for Law of Sins of course. The first thing I had noticed was SAS, being C,Φ,Cx .
    As per my microsoft paint job with the big green text identifying SAS for Law of Cosine...I didn't stop to think that perhaps I could just use SAA, law of sins.
    Last edited: Jan 16, 2012
  10. Jan 16, 2012 #9

    Simon Bridge

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  11. Jan 16, 2012 #10
    Oh, I looked back in my notes and saw that I actually did attempt to use Law of Sines at first, but it didn't seem to work because of a confusion, let me explain:

    Componentwise addition uses the x and y components in their "trig" form.

    Therefore, I was then trying to use Law of Sines on the Cx,Cy,C triangle, but instead of using C and Cx, I was using their "trig" form counter part and trying to evaluate this:

    [itex]\frac{sin\phi}{Csin\phi}[/itex] = [itex]\frac{sin90}{C}[/itex]

    Which turned into [itex]\frac{Csin\phi}{Csin\phi}[/itex] = 1

    and 1 = 1 is no substitute for Cy = Csin[itex]\phi[/itex]!!

    Would this not work out correctly because there is a mixing "trig form" components and actual vector "magnitude components"..similar to an apples and oranges type of deal?

    For instance, would it work out correctly if I would have the Law of Sines set up like this:

    [itex]\frac{Csin\phi}{sin\phi}[/itex]= [itex]\frac{\sqrt{C^{2}+C_{x}^{2}-2C_{x}Ccos\phi}}{sin90}[/itex] ?

    In otherwords, if I were good enough at evaluating trigonometric equations, I should find that this:

    [itex]\frac{Csin\phi}{sin\phi}[/itex]= [itex]\frac{\sqrt{C^{2}+C_{x}^{2}-2C_{x}Ccos\phi}}{sin90}[/itex]

    would equal: Csin[itex]\phi[/itex] somehow, right?
    Last edited: Jan 16, 2012
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