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Homework Help: Vector dot product homework help?

  1. Jun 3, 2010 #1
    if A = 2i^+6k^
    find A.B
    if A.B=0 then what do u conclude from it?

    one more question
    if A = 6i^+9k^
    and B=-4i^+4j
    then find A+B,A-B,A.B
  2. jcsd
  3. Jun 3, 2010 #2
    Do you know how to calculate the dot product of two vectors?
  4. Jun 3, 2010 #3
    dot product= ABcos(theeta)

    now try doing
  5. Jun 3, 2010 #4


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    Homework Helper

    Actually, I wouldn't use that way here because the angles are not provided in the problem. The other way of calculating dot products will be much easier.
  6. Jun 4, 2010 #5
    so can i get that easier method to solve this problem
  7. Jun 4, 2010 #6
    i m just confused because of the diferent unit vectors! in both equations
  8. Jun 4, 2010 #7
    Look up dot product from a textbook or online.
  9. Jun 4, 2010 #8


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    Science Advisor

    These are two different questions!
    A.B= 2(6)+ 6(5)

    If A.B= 0 then, using the formula jyothsna pb gave: [itex]A. B= |A||B| cos(\theta)[/itex] so at least one of these is true: |A|= 0, |B|= 0, [itex]\theta= \pi/2[/itex].

    one more question
    if A = 6i^+9k^
    and B=-4i^+4j
    then find A+B,A-B,A.B[/QUOTE]
    A- B= (6-(-4))i+ (9- 4)j, etc.

    If you are expected to do these as homework, certainly the definitions of A+ B, A- B, and A.B must be in your textbook!
  10. Jun 4, 2010 #9
    A- B= (6-(-4))i+ (9- 4)j, etc.

    If you are expected to do these as homework, certainly the definitions of A+ B, A- B, and A.B must be in your textbook![/QUOTE]

    i no it is the dot product question
    but the confuzing thing is that in A=the units r i^and k^ but in B=the units r i^and j^
    how can we solve it when the units r different?
  11. Jun 4, 2010 #10
    i^ , j^ n k^ are perpendicular vectors so as per the formula i gave their dot product ( whatever b the product) will b 0.

    that is i.j , i.k , j.k are all 0
    now try doing
    remember its scalar product so it does not have production
  12. Jun 4, 2010 #11


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    Homework Helper

    They're not really "units," but "unit vectors." But the name doesn't matter as much as knowing what they are.

    So here's your problem: you have the following two vectors:
    [tex]\vec A=2\hat i+6\hat k[/tex]
    [tex]\vec B=6\hat i+5\hat j[/tex]
    and you need to take the dot product. To do this, you need to use the distributive property. Think about this: if you were multiplying
    [tex](m + n)(p + q)[/tex]
    you would know how to expand that out, right? You'd get
    [tex]mp + mq + np + nq[/tex]
    Stop me if you're not 100% comfortable with that.

    To take the dot product of two vectors, you do the same thing, except that you replace regular multiplication with the dot product. So in the example above, if m,n,p,q were vectors instead of regular variables, what you get would be
    [tex](\vec m + \vec n)\cdot(\vec p + \vec q) = \vec m\cdot\vec p + \vec m\cdot\vec q + \vec n\cdot\vec p + \vec n\cdot\vec q[/tex]
    Does that make sense? Please ask if it doesn't.

    If that does make sense, try doing the same thing with your two vectors [tex]\vec A[/tex] and [tex]\vec B[/tex]. You should get four terms, and each one should contain a product of two unit vectors, like [tex]\hat i\cdot\hat i[/itex] or [tex]\hat j\cdot\hat k[/tex], etc.

    Now you need to evaluate those products of unit vectors, i.e. replace them with their numeric values. The relevant identities have been given already in this thread, and you should also be able to look them up in your textbook or other reference. But you should also think about why they're true. They come from the definition
    [tex]\vec{A}\cdot\vec{B} = \vert A\vert \vert B\vert \cos\theta[/tex]
    Think about which direction the various unit vectors point in space and what the angles between them are.
  13. Jun 7, 2010 #12
    thanks a lot
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