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Vectors and vector components

  1. Sep 8, 2007 #1
    ^^ ^^
    1.) Given the two vectors b=x+y and c=x+z find b+c, 5b+2c



    2.) r+s = (r1+s1, r2+s2, r3+s3), if r and s are vectors with components, r=(r1,r2,r3) and
    s=(s1,s2,s3)



    3.) Is this correct for b+c?
    b+c= (bx+cx,by,cz)



    To me that doesn't look right, but I think I'm applying the rules correctly. I don't know how to approach 5b+2c because I don't know now to apply the scalars 5 and 2 to the vector components. How is that subquestion approached?

    thanks alot

    Edit: the ^ signs are supposed to appear over the x+y and the x+z within the b and c vectors.
     
    Last edited: Sep 8, 2007
  2. jcsd
  3. Sep 8, 2007 #2
    no that's not right.
    you don't need to multiply the components x,y by b. Just add the components directly.

    When you do 5b+2c first figure out what 5b is, just multiply each component by 5. The same with 2c, and then add the components like you do for b+c
     
  4. Sep 8, 2007 #3

    HallsofIvy

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    Surely, this is next best thing to trivial! Ignore the fact that these are vectors.
    If these were just algebraic expressions and you are told that b= x+y, c= x+ z, what would you get substituting those into b+ c and 5b+ 2c?

    ?? I don't see a question here!
    Are you asking if, given that r= (r1,r2,r3) and s= (s1,s2,s3), is r+s= (r1+s1, r2+s2, r3+s3)? It certainly is! That's the definition of addition of vectors in component form.

    Well, it can be if cy= 0 and bz= 0. In general, if b= (bx,by,bz), c= (cx,cy,cz) then
    b+ c= (bx+cx,by+cy,bz+cz). If it happens that b= (bx,by,0), c= (cx,0,cz) then
    b+c= (bx+cx,by,cz)- but only in that case.



    Well what "rules" are you talking about? Can you quote them?
     
  5. Sep 8, 2007 #4
    How do you mean to add the components directly? Can't like terms only be added? i.e. x's with x's etc.
     
  6. Sep 8, 2007 #5
    Halls of ivy, i was given a template to follow and 2.) was 'any given formulas'; that is what I was given. You are correct in saying that it is a definition.

    So my first question is wrong your saying? cy wasn't even given in the question...
     
  7. Sep 8, 2007 #6

    learningphysics

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    As Halls said, just treat them like ordinary algebra variables. They want you to get b+c and 5b+2c... in terms of x and y. How would you do it if they weren't vectors but just normal variables..
     
  8. Sep 8, 2007 #7
    changing the form of the vectors a bit might help you see how to add them:
    b=1x+1y+0z, c=1x+0y+1z
    so for b+c=? just replace b by it's equation and c by it's equation and add.
     
  9. Sep 8, 2007 #8
    Ok, I guess I thought it was more complicated then it was.

    b+c= 2x+y+z
    5b+2c=7x+5x+2z

    the next part was to do bc as a dot product, the formula for this is AB=ABcos(theta)
    since no angle is given, the answer is simply:
    x^2+xz+yx+yz
    ya?
     
  10. Sep 8, 2007 #9
    no that's not right it's actually that: ||A||*||B||*cos(theta) = A dot B
    where ||A|| = distance of A to the origin: [itex]\sqrt{x^2+y^2+z^2}[/itex]

    you can use the definition that if you have
    [tex]\vec{u}=u_x+u_y+u_z[/tex]
    [tex]\vec{v}=v_x+v_y+v_z[/tex]
    the dot product of u and v =
    [tex]u_x*v_x+u_y*v_y+u_z*v_z[/tex]

    so you just multiply the x,y,z components and add them together.
     
    Last edited: Sep 8, 2007
  11. Sep 8, 2007 #10

    learningphysics

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    Your answer is correct, though be careful about the right notation and all that...

    So this would work:
    [tex]\vec{x}\cdot\vec{x} + \vec{x}\cdot\vec{z} + \vec{y}\cdot\vec{x} + \vec{y}\cdot\vec{z}[/tex]

    or this one:

    [tex]|\vec{x}|^2 + \vec{x}\cdot\vec{z} + \vec{y}\cdot\vec{x} + \vec{y}\cdot\vec{z}[/tex]

    Because the angle between [tex]\vec{x}[/tex] and itself is 0, so the cosine of 0 is 1, and the dot product of x with itself becomes [tex]|\vec{x}|^2[/tex]
     
  12. Sep 8, 2007 #11
    But in A dot B you need to know the angle between A and B, in this case the angle between the vector's x+y and x+z.

    Usually you should use A dot B = ||A||||B||cos(theta) when the vectors are given with numbers and an angle.
     
  13. Sep 8, 2007 #12

    learningphysics

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    Yes, if he knows the angles he should use them. I assumed the angles weren't given. In which case, I don't see what's wrong with just writing [tex]\vec{A}\cdot\vec{B}[/tex]

    I'm assuming x,y and z are arbitrary vectors... not necessarily in the x direction y direction or z direction. The question doesn't specify them being any specific type of vectors...
     
  14. Sep 8, 2007 #13
    hm..well using the other definition of dot product I get x^2 as the solution since (x*x+y*0y+0z*z)=x^2

    and I think that book uses x/y/z instead of i/j/k but if it's wrong it should still be x^2 I think.
     
  15. Sep 8, 2007 #14

    learningphysics

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    If they are arbitrary vectors then it won't be x^2.

    But I think you're right about them being i/j/k, since he mentions the "^" in which case the answer is 1. |i|^2 = 1.

    Sorry if I confused the OP. :redface:
     
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