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Basic Multivariable Proof Equating Determinant to dot/cross product

  1. Jun 3, 2013 #1
    Hello folks! Just a concise introduction of myself before I get to the task at hand: I'm new to these forums although I have been surfing them frequently for the past 5 years! I am not a math major and quite frankly, my skills in the subject are limited. Be that as it may, my fascination for math has inspired me to pursue a career in either pure math or computational science. My name is Mark and I'm an undergrad Uni student.

    K, now on to the task at hand: I am not familiar with symbolic/notation input on these forums so I apologize in advance for my crude way of displaying the question in a pictorial format. Simply put, my goal in the picture is to equate the determinant of vectors A, B and C to the dot product of vector A and (vector B cross-product vector C).

    Wow that sounded stupid! Lol, please excuse my naïveté. I have a STRONG feeling I am doing something incorrectly. Can anyone please steer me in the right direction?:

    [Broken]
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jun 3, 2013 #2

    lurflurf

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    Well clearly (using your notation)
    $$\mathop{det} ( \vec{A} , \vec{B} , \vec{C} )=\vec{A} \cdot ( \vec{B} \times \vec{C})$$

    as there is only one function that has the properties each side has, they must be equal.
    This could be shown different ways, but I like.
    $$\mathop{det} ( \tau \vec{A} , \tau \vec{B} , \tau \vec{C} )=\mathop{det} ( \vec{A} , \vec{B} , \vec{C} )$$
    and
    $$\mathop{det} ( \vec{i} , \vec{j} , \vec{k} )=1$$
    where $$\tau$$ is any rotation
    These properties determine det uniquely
    since $$\vec{A} \cdot ( \vec{B} \times \vec{C})$$ has these properties as well the two are equal.
     
  4. Jun 3, 2013 #3

    lurflurf

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    It looks like you are trying to prove it using the coordinate formula, that (while crude) will work also.
     
  5. Jun 3, 2013 #4
    Well, blimey, that was simple enough! Looks like what I did was overkill:blushing:

    Unfortunately, I did not know that [itex]\tau[/itex] stood for rotation.
     
  6. Jun 3, 2013 #5

    Fredrik

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    Welcome to PF, Mark. Five years is a long to time to lurk without posting.

    ##\tau## is not a standard notation for rotation. I think R is the most common notation, but I'm not sure if it can be called "standard". Here I guess ##\tau## looks prettier, since you're using uppercase Latin letters for vectors.

    Your own approach works, but the calculation ends much earlier. I didn't try to understand everything you did, but here's how I'd do it.
    $$
    \begin{vmatrix}a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\\ \end{vmatrix} =a_1\underbrace{\begin{vmatrix}b_2 & b_3\\ c_2 & c_3\end{vmatrix}}_{\displaystyle=(\vec B\times\vec C)_1} -a_2\underbrace{\begin{vmatrix}b_1 & b_3\\ c_1 & c_3\end{vmatrix}}_{\displaystyle=-(\vec B\times\vec C)_2} +a_3\underbrace{\begin{vmatrix}b_1 & b_2\\ c_1 & c_2\end{vmatrix}}_{\displaystyle=(\vec B\times\vec C)_3} =\vec A\cdot(\vec B\times\vec C)
    $$
    This is the link to the FAQ post about how to make the math look pretty:

    https://www.physicsforums.com/showpost.php?p=3977517&postcount=3

    If you hit the quote button next to my post, you can see how I did what I did. Same thing with lurflurf's post.
     
    Last edited: Jun 3, 2013
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