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I need to find a unit vector perpendicular to vector b

  1. Apr 7, 2006 #1

    danago

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    Hey. Here is the question:
    [tex]
    \underline{b}= -8\underline{i} - 6\underline{j}
    [/tex]

    I need to find a unit vector perpendicular to vector b.

    Ive come up with the following:
    [tex]
    10 \times |\underline{x}| \cos 90 = \underline{b} \bullet \underline{x}
    [/tex]

    I dont know if what ive done is even close to what i need to do, but from there, im completely stuck.

    Any help greatly appreciated.
    Dan.
     
  2. jcsd
  3. Apr 7, 2006 #2

    nrqed

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    Well, the equation you wrote does not give you any information since cos 90 =0.

    YWrite your unknown vector as [itex] a {\vec i} + b {\vec j} [/itex] and then impose that the dot product of this with your vector above gives zero (write out the scalar product explicitly in terms of *components*, not in terms of magnitude and angle). You will get one equation for two unknowns so there will be an infinite number of solutions. Just pick a value for a (ANY value, except zero) and solve for b. Then you can normalize your vector by dividing it by its magnitude.
     
  4. Apr 7, 2006 #3
    Clearly the solution is in the x-y plane. Let's say the solution is [tex]\textbf{z}=x\textbf{i}+y\textbf{j}[/tex]. You want to solve [tex] \textbf{z} \cdot \textbf{b} = -8x-6y = 0[/tex], subject to the constraint [tex]x^2+y^2 =1[/tex]. How would you normally go about solving these?
     
  5. Apr 7, 2006 #4

    danago

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    thanks for both of those posts, but im not really understanding them. nrqed, you referred to scalar product, and i wouldnt have a clue what that means.

    And euclid, im really lost with what youre trying to say sorry.

    Thanks for attempting to help me anyway.
     
  6. Apr 7, 2006 #5
    Two vectors a and b are orthogonal (by definition) if their dot product is zero.

    If [tex]\textbf{a} = a_1 \textbf{i}+a_2\textbf{j}[/tex] and [tex]\textbf{b} = b_1\textbf{i}+b_2\textbf{j}[/tex], then their dot product is
    [tex]\textbf{a}\cdot \textbf{b} =a_1b_1+a_2b_2[/tex]

    The length of the vector [tex]\textbf{a}[/tex] is
    [tex]|a|=\sqrt{a_1^2+a_2^2}[/tex].

    Hence your probem is to find a vector [tex]\textbf{z}[/tex] such that [tex]\textbf{z}\cdot \textbf{b} = 0[/tex] and [tex]|\textbf{z}|=1[/tex]. Simply follow the definitions to get the system of equations above.

    BTW, nrqed's method is more efficient than solving the two equations directly. It works because if [tex]\textbf{a}[/tex] and [tex]\textbf{b}[/tex] are orthogonal, so are [tex]c\textbf{a}[/tex] and [tex]\textbf{b}[/tex] for any scalar c.
     
    Last edited: Apr 7, 2006
  7. Apr 7, 2006 #6

    nrqed

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    Sorry about the confusion... "scalar product" and "dot product" are two terms representing the same thing. You seem to already know about this type of product between two vectors because this is essential what you wrote as [itex] {\undeline b} \cdot {\underline x} [/itex] in your first post.
    However, have you learned that there are *two* ways to calculate this product? One is using the form you wrote, the other way involves multiplying components and adding them, as Euclid wrote. Have you seen this? It's this other way of calculating a dot product that you need to solve this problem. The equation you wrote is correct but not useful for this type of problem.

    Hope this helps.

    Patrick
     
  8. Apr 7, 2006 #7

    danago

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    ahhh i think i understand now.

    So from that, i can write two equations:
    [tex]-8a-6b=0[/tex]

    and

    [tex]a^2+b^2=1[/tex]

    The first equation as another way of writing the dot product, and the second equation because the solution is a unit vector, then i solve them as simultaneous equations?

    I came up with the final vector:
    [tex]\textbf{z}=-0.6\textbf{i}+0.8\textbf{j}[/tex]

    That sound about right?
     
  9. Apr 7, 2006 #8
    Yup, that's right!
     
  10. Apr 7, 2006 #9

    danago

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    Yay. Thanks so much to both of you for the help. Makes sense to me now :)
     
  11. Apr 8, 2006 #10

    nrqed

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    That's perfect!

    Notice that there is one other solution (makes sense, right? I mean if you have a vector, it is possible to get *two* different unit vectors which will be perpendicular to it...one at 90 degrees on one side and one on the other side). That other solution comes from when you use a^2+b^2 =1 and you isolate a, let's say, you can take two different roots.

    Good for you!

    I was glad to help but you did most of the work.

    Regards

    Patrick
     
  12. Apr 8, 2006 #11

    danago

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    thats because when i write [itex]a^2+b^2=1[/itex] in terms of b i get
    [itex]b=\pm\sqrt{1-a^2}[/itex] right? which means that the second solution would be [itex]\textbf{z}=0.6\textbf{i}-0.8\textbf{j}[/itex], the negative of [tex]\textbf{z}[/tex]?

    Makes perfect sense to me. Thanks alot :)
     
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