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Published byDavis Duddleston Modified about 1 year ago

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Derivation of the Vector Dot Product and the Vector Cross Product

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Derivation of the Vector Dot Product u·v =∑ i u i v i = ∑ i u i e i ∑ i v j e j

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(u 1 e 1 + u 2 e 2 + u 3 e 3 ) (v 1 e 1 + v 2 e 2 + v 3 e 3 ) Kronecker Delta e i ·e j = δ ij = 1 when i = j 0 when i ≠ j

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= u 1 e 1 v 1 e 1 + u 1 e 1 v 2 e 2 + u 1 e 1 v 3 e 3 + u 2 e 2 v 1 e 1 + u 2 e 2 v 2 e 2 + u 2 e 2 v 3 e 3 + u 3 e 3 v 1 e 1 + u 3 e 3 v 2 e 2 + u 3 e 3 v 3 e 3

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= u 1 v 1 e 1 e 1 + u 2 v 2 e 2 e 2 +u 3 v 3 e 3 e 3 = u 1 v 1 + u 2 v 2 +u 3 v 3

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Vector Cross Product Einstein Notation u × υ = ε ijk e i u j υ k = Σ ijk ε ijk e i u j υ k = Σ i Σ j Σ k ε ijk e i u j υ k

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Levi-Civati Symbol ε = 0 unless i, j, k are distinct +1 if i, j, k is an even permutation of (1, 2, 3) -1 if i. j, k is an odd permutation of (1, 2, 3)

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Derivation of the Cross Product = (ε 121 u 2 v 1 + ε 122 u 2 v 2 + ε 123 u 2 v 3 + ε 131 u 3 v 1 + ε 132 u 3 v 2 + ε 133 u 3 v 3 ) e 1 + (ε 211 u 1 v 1 + ε 212 u 1 v 2 + ε 213 u 1 v 3 + ε 231 u 3 v 1 + ε 232 u 3 v 2 + ε 233 u 3 v 3 )e 2 + (ε 311 u 1 v 1 + ε 312 u 1 v 2 + ε 313 u 1 v 3 + ε 321 u 2 v 1 + ε 322 u 2 v 2 + ε 323 u 2 v 3 ) e 3

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Levi-Civati Symbol even 123, 231, 312 odd 321, 213, 132

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Derivation of the Cross Product = (ε 123 u 2 v 3 + ε 132 u 3 v 2 ) e 1 + (ε 213 u 1 v 3 + ε 231 u 3 v 1 ) e 2 + (ε 312 u 1 v 2 + ε 321 u 2 v 1 ) e 3 = (u 2 v 3 – u 3 v 2 )e 1 + (u 1 v 3 – u 3 v 1 )e 2 + (u 1 v 2 – u 2 v 1 )e 3

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