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Can someone please help with this notation? vectors

  1. Nov 19, 2011 #1
    1. The problem statement, all variables and given/known data
    I'm sure this is quite simple, but I don't understand the notation. This is from a supplemental book that we have to do outside of class--there isn't a matching textbook and I really don't understand what they want here.

    2. Relevant equations

    I noticed that if i square, sum, squareroot the left sides of the equations i get the right, so i am thinking that is a hypotenuse. So pythagorean theorem probably applies somewhere. But maybe not.


    3. The attempt at a solution
    I'm sorry, but I don't know what to put here. I know it looks like I'm being super lazy but I have already wasted way too much time trying to figure this out by reading and re-reading in my textbook. It's just not there. I really am not asking for someone to do my work, but if you could please give a clue as to what one of the equations may sound like read out loud it would help immensely in this search. I just found out our professor would like for us to turn in this 320 page workbook by the end of the semester. It wasn't on the syllabus so I didn't know. Since we had to buy six books I really didn't realize we were supposed to be doing all of this one, too. I only have about 100 pages of it done so far and we only have a couple of weeks left of class.

    Thank you.
     

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  3. Nov 19, 2011 #2
    The 'i's, 'j's, and 'k's with the little hats on top of them (my teacher calls them "i-hat" and "j-hat" and "k-hat") represent components. i=x component, j=y component, k=z component.
    So vector 'A' has an x component of positive 6 and a y component of positive 2.
    Now, all vectors have a MAGNITUDE and a DIRECTION.
    To find the magnitude of a vector, one uses the Pythagorean theorem with components. So to find 2*sqrt(10) as shown, they did the computation sqrt(2^2+6^2). That's how far the vector goes. (Hint: magnitudes are often expressed in absolute value signs)
    To find direction, you do a little trig. Direction is traditionally expressed as the angle from the positive x axis. So if you draw out this triangle of vector A and its components (which you definitely should do to better understand these problems), you find that you have the opposite and the adjacent length. arctan(6/2) will give you about 18 degrees (you use logic to deduce that it's in the first quadrant).

    Hopefully, this gives you enough info to go from there, assuming you are familiar with how vector operations work.
     
    Last edited: Nov 19, 2011
  4. Nov 19, 2011 #3
    thank you very much :smile:.

    so b and d are both negative in both the ihat and jhat value. can i put those in the third quadrant?


    question a is prompting for what vector operations could result in -38 units^2. so am i supposed to use the i hat as an imaginary number? i can't think of another way to get a negative square.
     
  5. Nov 20, 2011 #4

    Redbelly98

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    Homework Helper

    Yes.

    i-hat is different than the [itex]i = \sqrt{-1}[/itex] you are thinking of, in fact it has nothing to do with imaginary numbers. Instead, we have:
    i-hat is the vector (1,0), a vector of magnitude=1 pointing in the +x direction.
    j-hat is the vector (0,1), a vector of magnitude=1 pointing in the +y direction.​

    So when they write
    A = 6 i-hat + 2 j-hat​
    That is a kind of shorthand that means
    6·(1,0) + 2·(0,1)
    = (6,0) + (0,2)
    = (6,2)​

    The "units2" thing is a little confusing. They are implying that each vector, by itself, has units which they are simply calling "units" but it could be meters or m/s or any other units that are actually used in practice. In which case, "units2" could mean m2 or (m/s)2, etc.

    The key to this question is first to think of what different mathematical operations do you know about that could involve vectors? Just for example, we know that we can add vectors. But if we do that, the units would still be "units", whatever "units" is. For example, if we add two vectors that have units of m/s, then the result is still in m/s, not (m/s)2. So they can't be adding two of the vectors to get this -38 units2 result. Think about what other operations, other than adding, can you do with vectors?
     
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