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Simplifying expression with vectors

  1. Dec 29, 2015 #1
    1. The problem statement, all variables and given/known data
    To simplify:
    1. ##\mid \vec b -\frac{\vec a*\vec b}{\vec a*\vec c} * \vec c \mid##
    2. ##\mid \vec b -\frac{(\vec a*\vec c)*(\vec b*\vec b)-(\vec a*\vec b)*(\vec b*\vec c)}{(\vec a*\vec b)*(\vec c*\vec c)-(\vec a*\vec c)*(\vec b*\vec c)} * \vec c \mid##
    Vectors may be any dimensional!
    2. Relevant equations
    ##\vec a*\vec b=\sum(a_i*b_i)##
    ##\mid \vec b \mid=\sqrt{\sum(b_i^2)}##
    3. The attempt at a solution

    1. ##\sqrt{(\vec b*\vec b)*(\vec a*\vec c)^2-2*(\vec a*\vec b)*(\vec b*\vec c)*(\vec a*\vec c)+(\vec c*\vec c)*(\vec a*\vec b)^2}/(\vec a*\vec c)## is it correct? Can it be more simplified ? Can it be written as one product under square root ?To simplify it I wrote vectors as 3-dimensional vectors and opened braces: ##b2^2*a3^2*c3^2+b1^2*a3^2*c3^2+b2^2*a2^2*c3^2+2*b1*a1*b2*a2*c3^2+b1^2*a1^2*c3^2+2*b1^2*a2*a3*c2*c3+2*a1*b2^2*a3*c1*c3+b3^2*a3^2*c2^2+2*b1*a1*b3*a3*c2^2+a2^2*b3^2*c2^2+b1^2*a2^2*c2^2+b1^2*a1^2*c2^2+2*a1*a2*b3^2*c1*c2*+b3^2*a3^2*c1^2+2*b2*a2*b3*a3*c1^2+a1^2*b3^2*c1^2+b2^2*a2^2*c1^2+a1^2*b2^2*c1^2*-2*b1*a1^2*b2*c1*c2-2*b1*a1^2*b3*c1*c3-2*b1*a1*b2*a3*c2*c3-2*b1*a1*a2*b3*c2*c3-2*b1*b2*a2^2*c1*c2-2*b1*b2*a2*a3*c1*c3-2*b1*a2*b3*a3*c1*c2-2*b1*b3*a3^2*c1*c3-2*a1*b2*a2*b3*c1*c3-2*a1*b2*b3*a3*c1*c2-2*b2*a2^2*b3*c2*c3-2*b2*b3*a3^2*c2*c3## but I can not factorisate it!
    2. ##\sqrt{(\vec b*\vec b)-2*(\frac{(\vec a*\vec c)*(\vec b*\vec b)-(\vec a*\vec b)*(\vec b*\vec c)}{(\vec a*\vec b)*(\vec c*\vec c)-(\vec a*\vec c)*(\vec b*\vec c)})*(\vec b*\vec c)+(\vec c*\vec c)*(\frac{(\vec a*\vec c)*(\vec b*\vec b)-(\vec a*\vec b)*(\vec b*\vec c)}{(\vec a*\vec b)*(\vec c*\vec c)-(\vec a*\vec c)*(\vec b*\vec c)})^2}## is it correct? Can it be more simplified ? Can it be written as one product under square root ?
     
    Last edited: Dec 29, 2015
  2. jcsd
  3. Dec 29, 2015 #2
    You should really differentiate your dot products from ordinary multiplications; I'm seeing stars after all the asterisks o_O.
    Anyways, it helps greatly if you recognise that [itex]|\vec{b}| = \sqrt{(\vec{b}\cdot \vec{b})}[/itex], and not rush to expand everything.
     
  4. Dec 29, 2015 #3

    haruspex

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    Not sure it's a simplification, but are you familiar with the triple vector product?
     
  5. Dec 29, 2015 #4
    I know it, but how it helps?

    What should I do then?
     
  6. Dec 29, 2015 #5
    Keep the vector forms.
    As an example, let's say I have a vector [itex]\vec{c} = \alpha \vec{a} + \beta\vec{b} [/itex], and I want to find its norm. Then its straightforward for me to apply
    [tex]|\vec{c}|^{2} = \left(\alpha \vec{a} + \beta\vec{b}\right) \cdot \left(\alpha \vec{a} + \beta\vec{b}\right) = \alpha^{2} |\vec{a}|^{2} + 2\alpha\beta \left(\vec{a} \cdot \vec{b}\right) + \beta^{2} |\vec{b}|^{2} [/tex]

    Edit: Okay, I see that's essentially what you have there, although you have a typo in the first answer that threw me off.
     
  7. Dec 29, 2015 #6

    haruspex

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    I'm not seeing anything different here from what olgerm tried in the OP.
    @olgerm, there is an interesting relationship between the two expressions (Q1 and Q2). Can you see how to get the second from the first?
     
  8. Dec 29, 2015 #7
    Yeah, my bad, I was somewhat thrown off by a typo he made in his first answer, so I assumed he had messed up somewhere.
    If you really want to "simplify" it further, you can ostensibly try to factor out the denominator. That lets you use the triple vector product as Haruspex suggested.
     
  9. Dec 29, 2015 #8
    ##\mid \vec b -B * \vec c \mid##
    1. ##B=\frac{\vec a*\vec b}{\vec a*\vec c}##
    2. ##B=\frac{(\vec a*\vec c)*(\vec b*\vec b)-(\vec a*\vec b)*(\vec b*\vec c)}{(\vec a*\vec b)*(\vec c*\vec c)-(\vec a*\vec c)*(\vec b*\vec c)}##
    ##\sqrt{(\vec b*\vec b)-2*B*(\vec b*\vec c)+B^2*(\vec c*\vec c)}##
     
  10. Dec 29, 2015 #9
    Does anybody know, can these be more simplified ,so that simplified equation includes only dot products, or not?
     
  11. Dec 29, 2015 #10

    Fredrik

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    The dot product symbol is written as \cdot, and the product of two real numbers x and y is written as xy. So you should write ##(\vec a\cdot\vec c)(\vec b\cdot\vec b)## instead of ##(\vec a*\vec c)*(\vec b*\vec b)## for example.

    Is there a hint about what the simplified expressions should look like? I mean, it's straightforward to rewrite them as expressions that don't involve sums of vectors, but should those expressions be considered simpler? How do you know if what you get is simple enough?
     
  12. Dec 29, 2015 #11
    1. The problem statement, all variables and given/known data
    To simplify:
    1. ##\mid \vec b -\frac{\vec a \cdot \vec b}{\vec a \cdot \vec c} \vec c \mid=\sqrt{(\vec b \cdot \vec b)(\vec a \cdot \vec c)^2-2(\vec a \cdot \vec b)(\vec b \cdot \vec c)(\vec a \cdot \vec c)+(\vec c \cdot \vec c)(\vec a \cdot \vec b)^2}/(\vec a \cdot \vec c)##
    2. ##\mid \vec b -\frac{(\vec a \cdot \vec c)(\vec b \cdot \vec b)-(\vec a \cdot \vec b)(\vec b \cdot \vec c)}{(\vec a \cdot \vec b)(\vec c \cdot \vec c)-(\vec a \cdot \vec c)*(\vec b \cdot \vec c)} \vec c \mid =\sqrt{(\vec b \cdot \vec b)-2(\frac{(\vec a \cdot \vec c)(\vec b \cdot \vec b)-(\vec a \cdot \vec b)(\vec b \cdot \vec c)}{(\vec a \cdot \vec b)(\vec c \cdot \vec c)-(\vec a \cdot \vec c)(\vec b \cdot \vec c)})(\vec b \cdot \vec c)+(\vec c \cdot \vec c)(\frac{(\vec a \cdot \vec c)(\vec b \cdot \vec b)-(\vec a \cdot \vec b)(\vec b \cdot \vec c)}{(\vec a \cdot \vec b)(\vec c \cdot \vec c)-(\vec a \cdot \vec c)(\vec b \cdot \vec c)})^2}##
    I want in as simplified form as possible, but i do not know how simple is that!
    For example may it be square root of product of 3 dot products?


    To simplify first one I opened braces and expanded it:
    ##b2^2*a3^2*c3^2+b1^2*a3^2*c3^2+b2^2*a2^2*c3^2+2*b1*a1*b2*a2*c3^2+b1^2*a1^2*c3^2+2*b1^2*a2*a3*c2*c3+2*a1*b2^2*a3*c1*c3+b3^2*a3^2*c2^2+2*b1*a1*b3*a3*c2^2+a2^2*b3^2*c2^2+b1^2*a2^2*c2^2+b1^2*a1^2*c2^2+2*a1*a2*b3^2*c1*c2*+b3^2*a3^2*c1^2+2*b2*a2*b3*a3*c1^2+a1^2*b3^2*c1^2+b2^2*a2^2*c1^2+a1^2*b2^2*c1^2*-2*b1*a1^2*b2*c1*c2-2*b1*a1^2*b3*c1*c3-2*b1*a1*b2*a3*c2*c3-2*b1*a1*a2*b3*c2*c3-2*b1*b2*a2^2*c1*c2-2*b1*b2*a2*a3*c1*c3-2*b1*a2*b3*a3*c1*c2-2*b1*b3*a3^2*c1*c3-2*a1*b2*a2*b3*c1*c3-2*a1*b2*b3*a3*c1*c2-2*b2*a2^2*b3*c2*c3-2*b2*b3*a3^2*c2*c3## but I can not factorise it! Can somebody (who for example has Wolfram Mathematica) factorise it for me?
     
  13. Dec 29, 2015 #12

    Fredrik

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    I think it's a mistake to introduce the components of the vectors into the calculation. Note by the way that you're not allowed to make assumptions about how many of them there are.

    What you did in post #8 looks like a good start. The last term under the square root can be simplified a lot. (I have only looked at problem 1). Edit: Uhh, I think I spoke to soon. It can't be simplified the way I thought. I will think about it some more.
     
  14. Dec 29, 2015 #13
    I know ,but for example if I get ## (a_1*b_1+a_2*b_2+a_3*b_3)*(a_1*c_1+a_2*c_2+a_3*c_3)*(c_1*b_1+c_2*b_2+c_3*b_3) ## then I can just check is ## (\vec a \cdot \vec b) (\vec a \cdot \vec c)(\vec b \cdot \vec c)## equal to
    or not.
     
  15. Dec 29, 2015 #14

    SammyS

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    Rather than that, substitute ##\ \vec b - \vec c \ ## in for ##\ \vec b \ ## in ##\displaystyle \ \left |\, \vec b -\frac{\vec a\cdot \vec b}{\vec a\cdot \vec c} \, \vec c \, \right| \ ##.

    It seems to me that the vector triple product gives a form which is at least visually simplified.
     
  16. Dec 29, 2015 #15

    Fredrik

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    I don't see anything better than to use the vector triple product as discussed above. Look up that term at Wikipedia or something, find a formula that contains an expression that looks a bit like what you have, and then try to use that formula. I'm not sure why that result should be considered "simpler" though.
     
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