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Basic vectors

  1. Sep 16, 2012 #1
    i just started physics and i missed the first day. i'm trying to do these vector problems but i can't quite understand it and my book isn't helping.

    given that A+2B = x1j +y1j and 2A-B = x2i + y2j what is A?

    given that A + B =x1i + y1j and A-B = x2i + y2j what is A?

    given that A +B =x1i + y1j and A-B = x2i + y2j what is B?
     
  2. jcsd
  3. Sep 16, 2012 #2

    Simon Bridge

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    Welcome to PF;
    There's a bunch of people with these problems coming in today ... so having attended the class does not seem to have helped ;)

    If you look at those equations, they are in pairs with A-B on one side and A+B on the other ... to get the A out, just add them. To get the B out, subtract them.

    I'll show you:

    Lets say A+B=xi+yj and A-B=wi+zj

    then I can put v1= A+B and v2=A-B
    then v1+v2 = (A+B)+(A-B) = (xi+yj)+(wi+zj) do you see now?


    v1+v2 = A+B+A-B = 2A = (x+w)i+(y+z)j

    therefore A = (1/2)[(x+w)i+(y+z)j]
     
  4. Sep 16, 2012 #3

    Simon Bridge

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    What we are doing is lining the relations up and adding the columns just like for a sum of two regular numbers.

    [tex]
    \begin{array}{ccccccccc}
    & A & + & B & = & x\hat{\imath} & + & y\hat{\jmath}&\\
    +(& A & - & B & = & w \hat{\imath}& + & z\hat{\jmath}&)\\ \hline
    & 2A & & & = & (x+w) \hat{\imath}& + & (y+z)\hat{\jmath}&
    \end{array}
    [/tex]
     
  5. Sep 16, 2012 #4
    That makes so much more sense!
    Is this the right answer for "given that A + B =x1i + y1j and A-B = x2i + y2j what is A?"
    I got A= (x1+x2)/2i +(y1+y2)/2j
     
  6. Sep 16, 2012 #5
    Also for "A+2B = x1j +y1j and 2A-B = x2i + y2j what is A?" I got to this point and then got confused:

    3A+B=(x1+x2)i +(y1+y2)j and now i don't know how to get the A all alone
     
  7. Sep 17, 2012 #6

    Simon Bridge

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    When there is some constant multiplying the vector we want to get rid of, we have to add a multiple of the vectors together like this:

    [tex]
    \begin{array}{rcccccccl}
    & A & + & 2B & = & x\hat{\imath} & + & y\hat{\jmath}&\\
    +2\times(& 2A & - & B & = & w \hat{\imath}& + & z\hat{\jmath}&)\\ \hline
    & 5A & & & = & (x+2w) \hat{\imath}& + & (y+2z)\hat{\jmath}&
    \end{array}
    [/tex]

    notice how the +2x(... applies to the entire second row. I'm multiplying the second vector by 2 so that the second column will have a -2B in it.

    You won't always be asked to find A, you may be asked to extract the B instead. In that case you have to subtract instead of add:

    [tex]
    \begin{array}{rcccccccl}
    2\times( & A & + & 2B & = & x\hat{\imath} & + & y\hat{\jmath}&)\\
    -(& 2A & - & B & = & w \hat{\imath}& + & z\hat{\jmath}&)\\ \hline
    & & & 3B & = & (2x-w) \hat{\imath}& + & (2y-z)\hat{\jmath}&
    \end{array}
    [/tex]

    ... in this case I multiplied the top row by 2 so the first column would be 2A-2A=0.
     
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