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Force vector mulitplication = work

  1. Nov 16, 2011 #1
    1. The problem statement, all variables and given/known data
    F = 3x-y+2z in newtons
    the line that it follows is -x+y+2z

    What is the work done

    2. Relevant equations

    3. The attempt at a solution
    I am guessing
    3x^2 -y^2+2z^2=w
  2. jcsd
  3. Nov 16, 2011 #2


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    In your equations above, do x y and z represent unit vectors in those respective directions? Otherwise this doesn't make much sense.

    If they do, then your answer is not quite right. What is the definition of the dot product in Cartesian coordinates?
  4. Nov 16, 2011 #3
    I would assume that the x, y, and z are vectors.

    As to the dot product, I have no clue. It would have been nice if it was ever covered....
  5. Nov 16, 2011 #4


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    One way to evaluate the dot product of two vectors A and B (boldface denotes vectors) is to evaluate it component-wise. In other words, if:

    A = Axi + Ayj + Azk

    B = Bxi + Byj + Bzk

    where i, j, and k are unit vectors (vectors of magnitude 1) in the x, y, and z directions respectively, and the subscripted quantities are x, y, and z components, then the dot product is:

    A · B = (Axi + Ayj + Azk) · (Bxi + Byj + Bzk)

    Now, just using the ordinary distributive property of multiplication to expand out this product of two trinomials, you'd get nine terms in total. However, most of these terms (in fact all of the so called "cross" terms) vanish. The reason is that (once again, due to the definition of the dot product) the dot product of two perpendicular vectors is 0. Hence i · j = 0, j · k = 0, i · k = 0, and all permutations thereof. This means only three terms in the product survive, namely the three terms in which like components are multiplied together (x with x, y with y, and z with z):

    A · B = AxBxi·i + AyByj·j + AzBzk·k

    Now, another consequence of the definition of a dot product is that the dot product of a vector with itself is just equal to the magnitude of that vector, squared. Hence, i·i = |i|2 etc. But since these are unit vectors, their magnitudes are 1, and we have i·i = j·j = k·k = 1. Therefore, the result of the dot product is:

    A · B = AxBx + AyBy + AzBz

    THIS equation above is the one you should remember about how to compute the dot product of two vectors given their components. Everything above it was just an explanation of where it comes from. Notice that the result of the dot product of two vectors is a scalar. That's why it's also known as the scalar product. It's the type of vector multiplication that results in a scalar, as opposed to that other type of vector multiplication called the "cross product" or "vector product", which results in a vector. Of course, all of this is math you should already have been taught before being assigned physics problems like this one.
  6. Nov 16, 2011 #5
    So Work = -3x-y+4z

    I had part of a class way back in Trig that vaguely talked about vector multiplication but since it was the day before the final it was considered a "bonus" lesson. Yeah, my physics professor, and the word teach don't exactly mix.

    Or do the alpha characters just go away, and thus I would have zero as the answer
  7. Nov 16, 2011 #6


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    Well, unless if x, y and z are unit vectors (i.e. they correspond to what I called i, j, and k), your vectors don't really make sense.

    Alternatively, rather than being constant, maybe the vectors are functions of position (x,y,z) in 3D space so that F(x,y,z) = 3x - y + 2z. But this is less likely, because if it is what is meant, then you're still missing one or more unit vectors to tell you what the direction of F is. This would also mean you wouldn't be able to solve the problem with a simple dot product. In any case, you have to figure out what you've been given in your problem. If it helps, unit vectors are often typeset with a "hat" symbol (^, which is properly called a "caret") on top of them. E.g. [itex] \mathbf{\hat{x}}, \mathbf{\hat{y}} [/itex] and [itex] \mathbf{\hat{z}} [/itex]. OR equivalently [itex] \mathbf{\hat{i}}, \mathbf{\hat{j}} [/itex] and [itex] \mathbf{\hat{k}} [/itex]. Whichever you prefer.
  8. Nov 16, 2011 #7
    If, and I do stress the IF I am understanding your explanation of the dot product then it would follow like this:

    Force = 3x-y+2z (technically the way it was stated is "component 3 is in newtons")
    Line = -x +y+2z (also stated as "- is in meters")

    Find the work, ergo W=F*d

    I am assuming that when my professor put line, he meant straight, and that was the distance.

    Therefore it should look something like

    Force * line = work
    which equals zero

    P.S. all of the x, y, and z had the "carat" on top of them.
  9. Nov 16, 2011 #8


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    That's exactly right. Take another look at the very last equation in my previous post, and you'll see that it says the following. To compute the dot product of the two vectors:

    1. multiply their x-components together
    2. multiply their y-components together
    3. multiply their z-components together
    4. Take the sum of these three products.

    So, you have done it correctly! Since the dot product is zero, the force and displacement vectors must be perpendicular to each other.

    Okay, then this all makes perfect sense. Those are UNIT VECTORS. In a Cartesian coordinate system, any vector can be resolved into three components. If you look carefully at the expression, you'll see that what it is saying is that the vector is the sum of three individual vectors, one of which points entirely in the x-direction, one of which points entirely in the y-direction, and one of which points entirely in the z-direction. Each "unit vector" has the corresponding component as its coefficient. In other words, the unit vectors (which have length 1) have each been "scaled" to the right length for that component of the vector. That's what the notation means.
  10. Nov 16, 2011 #9
    Yay!! Thanks for the explanation :D
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