Week Four – Vector Space

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There are a number of examples in the text that today’s lecture expects you to type in and practice. If in future, you do not know how to program, it will be because you were lazy in simply typing in code and running them. “The fault, dear Brutus, is not in our stars,
But in ourselves, that we are underlings.” A word enough for the wise!
There is a test next week on Chapter One. Those who refuse to practise are going to have a cold shower on a harmattan morning!

19 comments on “Week Four – Vector Space

  1. Tobi Ogunrinde says:

    Good day sir, please how can we access the textbook?

  2. eniola says:

    Good day sir,
    page 70 solved problem 1.3
    according to the tell tale signs
    A dummy index must occur twice. But it DOES NOT have to occur in another term.

    but in the solved problem we have
    × = k
    a*b = eijk aj bk ek

    i know from Vector Product in Component Form that

    × = k
    ei * ej = eijk ek

    aj is correct also but the bk i think it should be bi and not bk.

    i dont know if am the one getting it wrong but according to the tell tale sign if am to solve that problem the results would be

    a * b = eijk aj bi ek
    the tell tales sign 3 made a *b = eijk aj bk ek wrong to me dont know if it is correct sir.

  3. ben says:

    from solved problem 39 the result of ⋅ ( ⊗ ) is a vector.
    from solved problem 40 the result of × ( ⊗ ) is a tensor.
    I don’t understand how this was simplified to result in a vector and tensor respectively.

    • ben says:

      from solved problem 39 the result of u ⋅ (v ⊗ w) is a vector.
      from solved problem 40 the result of u × (v ⊗ w) is a tensor.
      I don’t understand how this was simplified to result in a vector and tensor respectively.

      • oafak says:

        The first expression, when fully expanded in component form is the same as (w⊗v)u which is obviously a vector. The tensor nature of the second expression will become more obvious to you when we are deep into the next chapter.

  4. George says:

    Good Morning sir,
    Tried the mathematica codes on page 12 and kept getting error
    First no arrow automatically appeared after typing Plot range and Ticks but was able to bypass that with a cut and paste trick collecting arrow from Plot Style But it still didn’t run.
    It said
    Visualisation’Core’ParametricPlot3D: –Message text not found — ({1, y, z, Null})
    General : further Output of Visualisation’Core’ParametricPlot3D::invfuncs will be suppressed during this calculation
    And so many more

    • oafak says:

      To give me a copy of your code so I can see your problems is doable. We are NOT at that point yet. The way to type in the arrow is to follow the minus sign with > (the “greater than sign”, above the period sign). You are at the first step of simply copying my working code and getting the pain of executing it while hunting down your typos. I have typed the code in just now without changing anything. It works just fine. I will urge you to go back again and compare what you typed to what was given. The code works fine!

  5. Damilare says:

    Solved problem 1.16,page 78… Anybody can help with how they got to the next step of the solution(the part I wrote the note)

    1.16 Show that ( × ) ⋅ ( × ) × ( × ) = ( × ⋅ )^2

    ( × ) ⋅ ( × ) × ( × ) = ( × ) ⋅ [( × ) ⋅ ] − ( × ) ⋅ [( × ) ⋅ ]

    I don’t really get Wat happened here….seems to me like a part of the equation was just taken

    = ( × ) ⋅ [( × ) ⋅ ]
    = ( × ⋅ )(( × ) ⋅ )
    = ( × ⋅ )( ⋅ × )
    = ( × ⋅ )^2

  6. Fidipe says:

    Nice content
    170404529

  7. Oyepeju Oluwafemi - 160407026 says:

    Good afternoon sir. I have a question regarding equation(65) in page 39 of Chapter 1. I do not understand why
     -a_{1}\sin\alpha + a_{2}\cos\alpha became  a_{1}(\bold{\xi}_{2} \cdot \bold{e}_{1}) + a_{2}(\bold{\xi}_{2} \cdot \bold{e}_{2})
    Especially how  -a_{1}\sin\alpha became  a_{1}(\bold{\xi}_{2} \cdot \bold{e}_{1})

    • oafak says:

      Hello Oyepeju,
      Before I take on your question, let me commend you on the the Latex coding you did. I leave the edited code at the end to show you the few changes I made so that we can continue communicating this way with ease. Please note that the beginning of Latex is not the double dollar sign here but [late x], the end is [/late x]. In what I typed, I deliberately put a space into these delimiters to prevent the code from creating the equations in order to show you the raw code.
      Now to your question:
      Notice that  \bold{\xi}_{2}, \bold{e}_1 and  \bold{e}_{2} are unit vectors. Computing their scalar products is now a matter of knowing the angles between the pairs of them. Back to the graphics, you find that the angle between  \bold{\xi}_{2} and  \bold{e}_{1} is  \alpha+\pi/2 while the angle between  \bold{\xi}_{2} and  \bold{e}_{2} is  \alpha. Now remember that

      (1)   \begin{align*} \cos(\alpha+\pi/2)=\cos\alpha\cos\left(\pi/2\right)-\sin\alpha\sin\left(\pi/2\right). \end{align*}

      Do I need to remind you that \sin\left(\pi/2\right)=1 and that \cos\left(\pi/2\right)=0?
      Your latex input corrected: [late x] -a_{1}\sin\alpha + a_{2}\cos\alpha [/late x] became [late x] a_{1}(\bold{\xi}_{2} \cdot \bold{e}_{1}) + a_{2}(\bold{\xi}_{2} \cdot \bold{e}_{2}) [/late x]
      Especially how [late x] -a_{1}\sin\alpha[/late x] became [late x] a_{1}(\bold{\xi}_{2} \cdot \bold{e}_{1}) [/late x]

  8. Arasi Emmanuella 170407510 says:

    Good evening sir.I was going through chapter 1 again and I saw that I didn’t understand how we arrived at equation 61 under coordinate transformation .Please I will really appreciate if you can explain sir.

    • oafak says:

      Once you realize that \bold{\xi}_i, ~i=1,2,3 as well as \bold{e}_\alpha, ~\alpha = 1,2,3 are sets orthonormal vectors, then the scalar products will always lead to Kronecker deltas: \bold{\xi}_i\cdot\bold{\xi}_j=\delta_{ij}, and \bold{e}_\alpha\cdot\bold{e}_\beta=\delta_{\alpha\beta}. Use the substitution property of the Kronecker deltas and the fact that scalars can be carried along as they move around the equations; there is nothing really difficult in equation 61. If you have any particular step you can still not follow, let me know.

  9. Nmesoma Udojike says:

    Good day sir, (a+b).(a-b)x(b+a)
    The first two terms yields the difference of two squares ||a||^2 – ||b||^2 how can the last term (b+a) act on it

    160407013

    • oafak says:

      In a triple product, you are NOT allowed to perform the dot products first! That principle is elementary and it is the reason why we do not bracket off the cross products because the order of the products obviously starts from the vector product! Overcome that deficiency in your knowledge and use the result in Q1.4b. Notice that you can swap the cross and the dot without changing the result:

      (1)    \begin{align*} \bold{(a+b)\times(a-b)\cdot(a+b)}&=-2 \bold{a\times b\cdot (a+b)}\\ &=-2\bold{ (a\times b)\cdot a}-2\bold{(a\times b) \cdot b}\\ &=0 \end{align*}

      The last equality coming from the fact that triple products of linearly dependent vectors vanish. If we write the original question as,

      (2)    \begin{align*} \left[ \bold{(a+b),(a-b),(a+b)}\right] \end{align*}

      we can conclude right away that the answer is zero because of linear dependence! No need to expand at all!

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