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Write a vector as the combination of 2 other vectors

  1. Dec 10, 2014 #1
    1. The problem statement, all variables and given/known data
    Write the vector u=(2,3,-1) as the sum of two vectors, one parallel to v=(1,0,-3) and the other orthogonal to v=(1,0,-3)

    2. Relevant equations
    orthogonal vector would imply that v"dot"w = 0 ( dot product)

    Parallel vector would imply that v=kz
    OR the crossproduct of both is 0
    3. The attempt at a solution

    My attempt went as follow:
    I defined 2 vectors that w+z = u where u=(2,3,-1)

    Vector w is orthogonal to v and therefore --> w"dot"v = 0 --> w1*1+w2*0+w3*-3=0

    And that z= kv which would give z= (1*k,0*k,3*k)

    I thought that I could define any vector orthogonal that would give 0, for instance (3,1,1) would give 0.

    However the system always ends up being inconsistent, I don't know where I might be going wrong :/
     
  2. jcsd
  3. Dec 10, 2014 #2

    jedishrfu

    Staff: Mentor

    For u = av + bw then w must be in the same plane as u and v and in addition it must be perpendicular to v as stated in your problem. The a and b are scalars.

    Try drawing a picture and see if you can figure out w from the geometry and your vectors.
     
  4. Dec 10, 2014 #3
    two questions from this: is it obligatory to use a scalar in that equation for the w?

    2) I don't see the point in understanding that w must be in the same plane as u and v, I mean I don't see how it helps me solve the problem.

    Thing is, I chose an arbitrary vector for w that would satisfy the orthogonal requirement. However, using it, I get an a=-7 and b=3 which gives me a z parameter that is wrong :/
     
  5. Dec 10, 2014 #4

    jedishrfu

    Staff: Mentor

    There is a whole plane of vectors that are orthogonal to v but you must choose the one that is in the same plane as u and v right?

    Try to visualize it and you'll see all three vectors must be in the same plane.

    Another way to think of it is vector addition is a parallelogram with v and w as sides and with the diagonal between them being u the sum of the vectors.
     
  6. Dec 10, 2014 #5
    Is it kinda like spanning? I mean two vectors that add up to any vector in a plane?
     
  7. Dec 10, 2014 #6

    jedishrfu

    Staff: Mentor

    Yes that sounds right.
     
  8. Dec 10, 2014 #7
    but how would I go then to find my problem, for spanning I usually use a matrix to find my coefficient, but I don't know if that would apply in this case as it has to satisfy requirements, no?
     
  9. Dec 10, 2014 #8

    jedishrfu

    Staff: Mentor

    I'm thinking you should look at u - k.v = w and v.w=0
     
  10. Dec 10, 2014 #9

    Mark44

    Staff: Mentor

    For this problem you need to find the vector projection of u in the direction of v, which is denoted in some texts as ##\vec{Proj_v u}##. If you've been assigned this problem, it's a fairly safe bet that this concept has been presented in class. Once you have that vector, find a vector perpendicular to the vector projection, so that the two vectors you found add vectorially to u.

    Draw a sketch. The two vectors you're looking for are the sides of a right triangle. Forget about matrices and spanning sets and focus on the geometry here.
     
  11. Dec 10, 2014 #10
    Well I can see where the projection comes from, how you use it geometrically. Only thing is that , in class, my teacher used a set of equation from the orthogonal and parallel requirement, so I assumed I had to do it that way, and that way I don't understand.
     
  12. Dec 10, 2014 #11

    jedishrfu

    Staff: Mentor

    You have u and you have the project of u on v so you have the lengths of two sides of a triangle with w being the third side.
     
  13. Dec 11, 2014 #12
    I realize I can do it with the projection, but we didn't use projection for that in class, so I believe my teacher would prefer if we solved it with system of equation(s). This is where I get confused, I don't know hjow to get it that way.
     
  14. Dec 11, 2014 #13

    jedishrfu

    Staff: Mentor

    I'd do it the way you know how and ask the teacher about the other method later. You might discover that that method doesn't even apply in this case.
     
  15. Dec 11, 2014 #14

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You wrote the equation u=w+z. Multiplying it with v ( dot product) you get ##\vec v \cdot \vec u = \vec v \cdot \vec w + \vec v \cdot \vec z ##. The first product is zero. z=kv. Find k.
     
  16. Dec 11, 2014 #15

    Mark44

    Staff: Mentor

    The problem works out pretty nicely if you organize your information. You're given vector u and v. The vectors you want are kv and w.

    The three constraints yield a consistent set of equations so that you can find the scalar k and the coordinates of w.

    1. kv + w = u
    2. kv is perpendicular to w
    3. Pythagorean theorem on the lengths of kv, w, and u

    BTW, the method I mentioned earlier will also work, but as it hasn't been presented yet, probably shouldn't be used.
     
  17. Dec 11, 2014 #16

    haruspex

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    Science Advisor
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    2016 Award

    Hard to believe there's a simpler way than ehild's above.
     
  18. Dec 11, 2014 #17
    I just did this ( realized where my mistake was)

    (2,3,-1)=a(1,0,3)+b(3,c,1) where the second vector comes from w1*1+)-3W3=0 --> W1=3W3 so I can just say 3,c,1,

    Solve for 3 variables with 3 equation. I just didn't use the c but took any number at first, without understanding why ^^
     
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