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Math Help - Vector Help

  1. #1
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    Vector Help

    I need help with two things:

    1. suppose I had the vector R=(x,y,z), how do i take R^2, where x,y,z are integers
    2. Using vector R, how do i take DelR in this form from above
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  2. #2
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    Quote Originally Posted by Rollo87 View Post
    I need help with two things:

    1. suppose I had the vector R=(x,y,z), how do i take R^2, where x,y,z are integers

    Mr F says: Do you mean the dot (scalar) product {\color{red}R \cdot R } or the cross (vector) product {\color{red}R \times R}? Either way, your class notes or textbook will surely have the process and there thousands of google hits that show you how.

    2. Using vector R, how do i take DelR in this form from above Mr F says: Do you mean {\color{red} \nabla \cdot R} (divergence of R) or {\color{red} \nabla \times R} (curl of R)? Again, either way, your class notes or textbook will surely have the process and there thousands of google hits that show you how.
    ..
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    I'll give you more detail (should've just done this in the first place). This is the actual quesiotn:

    1. Consider the separation vector
    r - r ' from the point r' = (x1 ', x2 ', x3 ')= (2, 8, 7) to the point

    r
    = (x1, x2, x3)= (4, 6, 8)expressed in Cartesian coordinates.
    (a) Determine the magnitude (
    r - r' ), and construct the unit vector (r - r ')(hat)
    =
    r - r' /|r - r' |.
    (b) Show that
    “DEL(r - r ')^2= 2 (r - r').

    Obviouly constructing the unit vector is very simple. And yes i know how to take both the dot product and cross-product but my problem is more that I'm not sure what the question wants me to do exactly. Also it's the dot product of Del with (r-r') so it's the divergence I'm looking for and I know how to do this but all the stuff i've read gives it in terms of partial derivatives which is fine but these vectors are integers they are not functions of x,y or z.

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  4. #4
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    Quote Originally Posted by Rollo87 View Post
    I'll give you more detail (should've just done this in the first place). This is the actual quesiotn:


    1. Consider the separation vector
    r - r ' from the point r' = (x1 ', x2 ', x3 ')= (2, 8, 7) to the point
    r


    = (x1, x2, x3)= (4, 6, 8)expressed in Cartesian coordinates.
    (a) Determine the magnitude (r - r' ), and construct the unit vector (r - r ')(hat)
    = r - r' /|r - r' |.
    (b) Show that “DEL(r - r ')^2= 2 (r - r').

    Obviouly constructing the unit vector is very simple. And yes i know how to take both the dot product and cross-product but my problem is more that I'm not sure what the question wants me to do exactly. Also it's the dot product of Del with (r-r') so it's the divergence I'm looking for and I know how to do this but all the stuff i've read gives it in terms of partial derivatives which is fine but these vectors are integers they are not functions of x,y or z.



    I think what you're having the most trouble with is actually reading the question. What you need to see here is that r - r' is just one vector.

    r is a vector from O to (4, 6, 8) and r' is another vector from O to (2, 8, 7). -r' will be the vector going from (2, 8, 7) to O. If this gets translated so that the starting end is at the origin, it will go from O to (-2, -8, -7). So r - r' is the sum of the vectors r and -r' and represents the vector going from (4, 6, 8) to (-2, -8, -7), which can be found by adding the ordinates, giving (2, -2, 1).

    a) Now that we have the vector r - r' = (2, -2, 1), we can work out it's magnitude (length) by summing the squares of the i, j, k components and then taking the square root.

    i.e. |r - r'| = sqrt[2^2 + (-2)^2 + 1^2]
    = sqrt(4 + 4 + 1)
    = sqrt(9)
    = 3

    We can disregard the negative answer because when dealing with vectors, the modulus is the LENGTH of the vector, and we can't have a negative length.

    Now we can find (r - r')hat by dividing each component of r - r' by the length of the vector. What we are really doing here is finding a vector in the direction of r - r', but with a length of 1 unit.

    So (r - r')hat = (2/3, -2/3, 1/3).

    b) doesn't make any sense because 2(r - r') is a vector and the divergence of (r-r') is a scalar, and like you say, r-r' is not a function of x, y and z... Are you sure you copied down the question correctly?
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    thanks for that, i didn't have any problem with part a it was simple. However I agree with you on part b. I'm sure i copied it correctly and forgetting the fact that we can't take the divergence of the vector without simply getting zero for everything, it's too vague anyway as to what he expects us to do for the square of the vector. He then goes on to do the following:
    (c) For extra credit: Show that
    “ DEL(1/ |r - r' |)= - (r - r')(hat)/

    |r - r'|^2.

    Which clearly again results in a scalar being equal to a vector.

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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Rollo87 View Post

    (b) Show that
    “DEL(r - r ')^2= 2 (r - r').

    Obviouly constructing the unit vector is very simple. And yes i know how to take both the dot product and cross-product but my problem is more that I'm not sure what the question wants me to do exactly. Also it's the dot product of Del with (r-r') so it's the divergence I'm looking for and I know how to do this but all the stuff i've read gives it in terms of partial derivatives which is fine but these vectors are integers they are not functions of x,y or z.

    Be careful here.

    Do you mean \nabla (\bold r- \bold r')^2 or \nabla\cdot(\bold r-\bold r')^2?

    There is a huge difference between the two statements. From what I can tell, we are supposed to be looking for \nabla (\bold r-\bold r')^2, NOT \nabla\cdot(\bold r-\bold r')^2.

    --Chris
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    Quote Originally Posted by Chris L T521 View Post
    Be careful here.

    Do you mean \nabla (\bold r- \bold r')^2 or \nabla\cdot(\bold r-\bold r')^2?

    There is a huge difference between the two statements. From what I can tell, we are supposed to be looking for \nabla (\bold r-\bold r')^2, NOT \nabla\cdot(\bold r-\bold r')^2.

    --Chris
    Yep you're right we're looking for \nabla (\bold r-\bold r')^2. Sorry i think i did write dot product before. So how does this change things?
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