AravindG
  • AravindG
is there any easy way to findcross product AxAxA rather than finding product AxA and multi[lying by A??
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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2bornot2b
  • 2bornot2b
I would like the answer to this question
radar
  • radar
Since A has implied an exponent of 1. Simply ad the exponents. AxAxA=A^3 \[a ^{2}\times a ^{1}=a ^{3}\] etc.
2bornot2b
  • 2bornot2b
I think asker is talking about vectors, not scalars.

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AravindG
  • AravindG
lol not that its cross product
AravindG
  • AravindG
no
AravindG
  • AravindG
i am asking cross product in functions
radar
  • radar
Never mind lol.
JamesJ
  • JamesJ
AxA = 0 because the A is parallel to itself. Hence AxAxA = 0 also
JamesJ
  • JamesJ
Make sense?
AravindG
  • AravindG
no i didnt ask that
JamesJ
  • JamesJ
What's your question then.
AravindG
  • AravindG
i am asking abot ordered pairs
JamesJ
  • JamesJ
You mean this identity? \[ A \times B \times C = B(A\cdot C) - C(A \cdot B) \]
JamesJ
  • JamesJ
\[ A \times (B \times C) \]

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