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anonymous

  • one year ago

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  1. anonymous
    • one year ago
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    How can I solve the following determinant??Do I use dot product or cross product ?? \[\left[\begin{matrix}\vec a & \vec b & \vec c \\ \vec a . \vec a & \vec a . \vec b & \vec a . \vec c \\ \vec b . \vec a & \vec b . \vec b & \vec b . \vec c\end{matrix}\right]\] I have made a matrix but I mean to find the determinant, there's no tool for determinant in the equation box I did something like taking a and b out from the 2nd and 3rd row and it becomes 0

  2. Michele_Laino
    • one year ago
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    hint: I think that you have to develop the determinant along the first row of your matrix, what you should get is a vector

  3. anonymous
    • one year ago
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    |dw:1438410330020:dw|

  4. imqwerty
    • one year ago
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    take 'a' common from 2nd row and 'b' common from 3rd row nd then try to solve :)

  5. anonymous
    • one year ago
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    I was thinking of expanding but if I expand would I use dot product or cross poduct??I have no idea....it is also given that\[\vec a\] \[\vec b\] \[\vec c\] are coplanar

  6. anonymous
    • one year ago
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    damn im really bad at drawing

  7. imqwerty
    • one year ago
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    lol

  8. anonymous
    • one year ago
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    |dw:1438410602789:dw|

  9. Michele_Laino
    • one year ago
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    hint: if I develop the determinant along the first row, I get this: \[\Large \vec a\left\{ {\left( {\vec a \cdot \vec b} \right)\left( {\vec b \cdot \vec c} \right) - \left( {\vec a \cdot \vec c} \right)\left( {\vec b \cdot \vec b} \right)} \right\} - \vec b\left\{ {...} \right\} + \vec c\left\{ {...} \right\}\]

  10. imqwerty
    • one year ago
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    is the answer 0

  11. Astrophysics
    • one year ago
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    I don't really see what the problem is, is it just knowing how to do a determinant or something else? When you do take a cross product you are essentially taking the determinant.

  12. anonymous
    • one year ago
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    (AEI+BFG+CDH)-(CEG+AFH+BDI)|dw:1438410761644:dw|

  13. anonymous
    • one year ago
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    can I arrange \[(\vec a.\vec c)(\vec b . \vec b)=(\vec a . \vec b)(\vec b . \vec c)\] ??

  14. anonymous
    • one year ago
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    How can I solve \[(\vec a . \vec b)(\vec b . \vec c)-(\vec a . \vec c)(\vec b . \vec b)\] If I take the components it will take too long

  15. anonymous
    • one year ago
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    I think I solved it, since a b and c are coplanar, dotting 1st column with b cross c, second column with c cross a and third column with a cross b we get \[\left[\begin{matrix}\vec a .(\vec b \times \vec c) & \vec b .(\vec c \times \vec a) & \vec c.(\vec a \times \vec b) \\ \vec a .(\vec a .(\vec b \times \vec c)) & \vec a .(\vec b .(\vec c \times \vec a)) & \vec a .(\vec c . (\vec a \times \vec b)) \\ \vec b.(\vec a .(\vec b \times \vec c)) & \vec b .(\vec b .(\vec c \times \vec a)) & \vec b.(\vec c.(\vec a \times \vec b))\end{matrix}\right]\]

  16. anonymous
    • one year ago
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    becomes 0 ??

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