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AravindG

  • 4 years ago

is there any easy way to findcross product AxAxA rather than finding product AxA and multi[lying by A??

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  1. 2bornot2b
    • 4 years ago
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    I would like the answer to this question

  2. radar
    • 4 years ago
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    Since A has implied an exponent of 1. Simply ad the exponents. AxAxA=A^3 \[a ^{2}\times a ^{1}=a ^{3}\] etc.

  3. 2bornot2b
    • 4 years ago
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    I think asker is talking about vectors, not scalars.

  4. AravindG
    • 4 years ago
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    lol not that its cross product

  5. AravindG
    • 4 years ago
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    no

  6. AravindG
    • 4 years ago
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    i am asking cross product in functions

  7. radar
    • 4 years ago
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    Never mind lol.

  8. JamesJ
    • 4 years ago
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    AxA = 0 because the A is parallel to itself. Hence AxAxA = 0 also

  9. JamesJ
    • 4 years ago
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    Make sense?

  10. AravindG
    • 4 years ago
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    no i didnt ask that

  11. JamesJ
    • 4 years ago
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    What's your question then.

  12. AravindG
    • 4 years ago
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    i am asking abot ordered pairs

  13. JamesJ
    • 4 years ago
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    You mean this identity? \[ A \times B \times C = B(A\cdot C) - C(A \cdot B) \]

  14. JamesJ
    • 4 years ago
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    \[ A \times (B \times C) \]

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spraguer (Moderator)
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