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Determine components of a vector without knowing the angle

  1. Oct 14, 2015 #1
    1. The problem statement, all variables and given/known data

    (I roughly translate the problem statement from German)
    Given the vectors a = (1,-2,3) and b = (1,1,1), divide the vector a in two components a1 (parallel to b) and a2 (perpendicular to b).

    2. Relevant equations

    In a previous question of the problem, I found that:
    a.b = 2
    a x b = (-5, 2, 3)
    |a| = √(14) ≈ 3.74
    |b| = √(3) ≈ 1.73

    So far, our class has been mentioning vector addition, multiplication by a real number, components of a vector, scalar product, vector product, as well as diverse properties of the operations I just listed (commutativity, distributivity, associativity and homogeneity).

    3. The attempt at a solution

    My problem here is that the angle between the vectors is not given in the problem. I found several ways to calculate it with the use of arccos, but since that was not yet mentioned in my class, I am reluctant to use it.

    What I know so far is that:
    (I'll call ∂ the angle between a and b)
    a1 = |a| cos ∂
    a2 = |a| sin ∂
    a1 x b = 0 (because a1 and b are parallel, their cross product is equal to 0)
    a2.b = 0 (because a2 and b are perpendicular, their scalar product is equal to 0)
    since a.b = |a|.|b|.cos ∂, I also find that cos ∂ = 2/√(14).√(3) ≈ 0.31
    I also know that a = √(a1^2 + a2^2) since the angle between a1 and a2 is π/2 rad.

    What am I missing to solve this problem? I just started a program of physic after many years without maths, so I need to refresh a bit :)


    Thank you very much for your answers, I appreciate it.


    Julien.
     
    Last edited: Oct 14, 2015
  2. jcsd
  3. Oct 14, 2015 #2

    HallsofIvy

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    Staff Emeritus
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    First you can find the angle because [itex]\vec{a}\cdot\vec{b}= |\vec{a}||\vec{b}| cos(\theta)[/itex].

    But you don't need to find the angle, directly, because component of [itex]\vec{a}[/itex] parallel to [itex]\vec{b}[/itex] has length [itex]|\vec{a}| cos(\theta)[/itex] and then replace that cosine with [itex]\frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}[/itex] to get [itex]\frac{\vec{a}\cdot\vec{b}}{|\vec{b}}[/itex] as the length. Of course the vector component parallel to [itex]\vec{b}[/itex] is that times a unit vector in the [itex]\vec{b}[/itex] direction.
     
  4. Oct 14, 2015 #3

    DEvens

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    Remember the definition of dot product. Remember it both in terms of components of the vector, and in terms of the cosine of the angle between the two vectors.
     
  5. Oct 14, 2015 #4

    Mark44

    Staff: Mentor

    Slight correction: definitions.
    DEvens lists both definitions: the component definition, and the coordinate-free definition; i.e. ##\vec{a} \cdot \vec{b} = |\vec{a} | |\vec{b} | \cos(\theta)##.
     
  6. Oct 14, 2015 #5

    Ray Vickson

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    Look up "orthogonal projection"; see, eg., https://en.wikipedia.org/wiki/Vector_projection or
    https://en.wikibooks.org/wiki/Linear_Algebra/Orthogonal_Projection_Onto_a_Line
     
  7. Oct 14, 2015 #6
    To begin, thank you very much for all of your answers. That is really helpful!

    I understand now the concept better, but I must still be making a mistake somewhere:

    Since a.b = |a|.|b|.cos ∂ & a1 = |a|.cos ∂,
    a.b = |b|.a1
    So a1 = (a.b)/|b| = 2/√3 (if my first answers for the scalar product and for the magnitude were correct)

    But there is still something wrong, because it is not a unit vector, and I cannot seem to find the values of the vector that respect the property of two perpendicular vectors a1.b = 0. Did I get the wrong result, or am I misunderstanding what I am supposed to find?


    Sorry to ask again, I am sure it is pretty easy and I must quickly integrate it in my brain :)
     
  8. Oct 14, 2015 #7
    Maybe I am misunderstanding what a unit vector is supposed to be... Here I have a definition that says "A unit vector is any vector with a magnitude of 1." But what I found was that a1 = 2/√3. And when I try to figure out what would be a1(x,y,z), I go like that:

    Since a1 and b are parallel, there exists a real number k so that a1 = kb;
    a1 = k(1 + 1 + 1) = 3k
    k = a1/3 = (2/√3)/3 = (2√3)/9
    thus defining the vector a1 = ((2√3)/9,(2√3)/9,(2√3)/9)

    Ehem... Not only does this result look very strange, but it is also incompatible with the property of parallel vectors a1.b = 0:
    a1.b = (2√3)/9 + (2√3)/9 + (2√3)/9 = (6√3)/9 = (2√3)/3 ≠ 0!

    Is it in the calculus that I went wrong? Or is it my reasoning that is flawed? From another way of thinking that does not really rely on any theorem, I instinctively come to a result a1 = (1/3,1/3,1/3). Not only does that verify the property of parallel vectors a1.b = 0, but such vector would also have a magnitude of 1!


    Thank you very much if you can clear up my confusion, I appreciate your help.


    Julien.
     
    Last edited: Oct 14, 2015
  9. Oct 14, 2015 #8

    Mark44

    Staff: Mentor

    No. a1 should be a vector, not a scalar.
    No again. a1 should be a scalar multiple of b, so a1 = k<1, 1, 1> = <k, k, k>.
    I think you might be confusing the ideas of a vector vs. the magnitude of a vector. IOW, confusion between v and |v|.
    This vector has the right direction, but not the right length (magnitude).
    Right, because a1 is supposed to be parallel to b, not perpendicular to it. If these two vectors were perpendicular, their dot product would be zero.
     
  10. Oct 15, 2015 #9
    Thank you Mark for your answer. There must indeed be some confusion in my formulas. I still don´t get how to obtain a vector out of a scalar product in a fraction, in that case I am stuck at a1 = (a.b)/IbI.
     
  11. Oct 15, 2015 #10
    Is the magnitude I calculated |a1| = 2√3 wrong? Because whatever I try, I always come back to this very same result :(

    |a1| = |a| cos ∂
    a.b = |a|.|b|.cos ∂
    cos ∂ = |a1|/|a|
    Therefore, a.b = |a|.|b|.(|a1|/|a|) and |a1| = 2√3

    That is a slight variation from the method suggested by Hallsofivy previously, and it gives me the same magnitude. I really dont get where is the mistake there..
     
  12. Oct 15, 2015 #11

    Mark44

    Staff: Mentor

    This formula doesn't make sense as you have written it. The left side is a vector, but the right side is a scalar.
    The corrected formula is ##|\vec{a_1}| = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}##.
    This gives you the length of the projection of a in the direction of b, i.e., the length of a1. To get the actual vector a1, multiply |a1| by a unit vector with the same direction. IOW, multiply by ##\frac{\vec{a_1}}{|\vec{a_1}|}##.
    Yes, it's wrong. See above.
    It's hard to follow your thinking here.
    First equation: Not useful, since you don't know |a1| and you don't know ##\theta##.
    Second equation: Can be used to find ##\theta##, since you know a and b.
    Third equation: Not useful, since you don't know ##\theta##.
    Fourth equation: The equation is a true statement, but I don't know what you did to arrive at |a1| = 2√3. You can use what I said at the top of my post here to find |a1|, and then multiply by a suitable unit vector to get a1 itself.
     
  13. Oct 15, 2015 #12
    I mistyped it actually, the answer I was getting is 2/√3, because the homework asked earlier to calculate the dot product a.b and the length of |b| (amongst other things).

    a.b = a1b1 + a2b2 + a3b3 = 1 -2 + 3 = 2
    |b| = √(b1^2+b2^2+b3^2) = √(1 + 1 + 1) = √3

    So in the equation |a1| = a.b/|b|, I just replace a.b by 2 and |b| by √3.

    I will take a deep look again at what you wrote and try to figure it out. I'm trying to understand why the length |a1| should be multiplied by a unit vector.


    Thank you very much for your patience, I appreciate your efforts of explaining me the process. I'm working hard to soon not get stuck on such problems anymore. :)


    Julien.
     
  14. Oct 15, 2015 #13

    Mark44

    Staff: Mentor

    Yes, that's the right value for |a1|.
    To make a vector with the right length (2/√3) and pointing in the right direction. In other words, to get what you're calling a1.
     
  15. Oct 15, 2015 #14
    Wow that's great, I was scratching my head over and over again because I thought that was wrong.

    Okay, I think I understand:
    the "suitable unit vector" u to obtain a1 is a vector that would have the same direction as b and a magnitude of 1. For the magnitude of u to be 1, I must divide each of the component of b by |b| (= √3).

    The unit vector u in direction of b would then be u = (1/√3,1/√3,1/√3).

    Am I correct so far? It kinda look strange.. Then next step is to obtain a1 is:

    a1 = |a1|.u = (2/3, 2/3, 2/3)

    I hope I'm on the right track :)
     
  16. Oct 15, 2015 #15

    Mark44

    Staff: Mentor

    Yes, everything looks good.
     
  17. Oct 15, 2015 #16
    Thanks a lot for your help, I appreciate it.


    J.
     
  18. Oct 15, 2015 #17
    Sorry to come back, I have a short additional question: to determine a2, can I just make a2 = a - a1? I'm wondering because I get very complicated results.
     
  19. Oct 17, 2015 #18

    Mark44

    Staff: Mentor

    Yes. Since a = a1 + a2, you can subtract a1 from both sides to get your equation.
     
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