Question-219185

Question Number 219185 by fantastic last updated on 20/Apr/25

Answered by MrGaster last updated on 20/Apr/25

▽∙(F^→ ×G^→ )=∂_i (ε_(ijk) F_j G_k ) =ε_(ijk) (∂_i F_j )G_k +ε_(ijk) F_j (∂_i G_k ) =G_k (ε_(kij) ∂_i F_j )−F_j (ε_(jik) ∂_i G_k ) =G^→ ∙(▽×F^→ )−F^→ ∙(▽+G^→ ) determinant (((▽∙(F^→ ×G^→ )=G^→ ∙▽×F^→ −F^→ ∙▽×G^→ (a)✓))) ▽∙(gF^→ )=∂_i (gF_i ) =g▽∙F^→ +▽g∙F^→ (d)✓ ▽∙(F^→ ×G^→ )=G^→ ∙▽×F^→ −F^→ ∙▽×G^→ (a)✓ ▽∙(gF^→ )=∂_i (gF_i ) ∴d✓

$$\bigtriangledown\centerdot\left(\overset{\rightarrow} {{F}}×\overset{\rightarrow} {{G}}\right)=\partial_{{i}} \left(\epsilon_{{ijk}} {F}_{{j}} {G}_{{k}} \right) \\ $$$$=\epsilon_{{ijk}} \left(\partial_{{i}} {F}_{{j}} \right){G}_{{k}} +\epsilon_{{ijk}} {F}_{{j}} \left(\partial_{{i}} {G}_{{k}} \right) \\ $$$$={G}_{{k}} \left(\epsilon_{{kij}} \partial_{{i}} {F}_{{j}} \right)−{F}_{{j}} \left(\epsilon_{{jik}} \partial_{{i}} {G}_{{k}} \right) \\ $$$$=\overset{\rightarrow} {{G}}\centerdot\left(\bigtriangledown×\overset{\rightarrow} {{F}}\right)−\overset{\rightarrow} {{F}}\centerdot\left(\bigtriangledown+\overset{\rightarrow} {{G}}\right) \\ $$$$\begin{array}{|c|}{\bigtriangledown\centerdot\left(\overset{\rightarrow} {{F}}×\overset{\rightarrow} {{G}}\right)=\overset{\rightarrow} {{G}}\centerdot\bigtriangledown×\overset{\rightarrow} {{F}}−\overset{\rightarrow} {{F}}\centerdot\bigtriangledown×\overset{\rightarrow} {{G}}\left({a}\right)\checkmark}\\\hline\end{array} \\ $$$$\bigtriangledown\centerdot\left({g}\overset{\rightarrow} {{F}}\right)=\partial_{{i}} \left({gF}_{{i}} \right) \\ $$$$={g}\bigtriangledown\centerdot\overset{\rightarrow} {{F}}+\bigtriangledown{g}\centerdot\overset{\rightarrow} {{F}} \\ $$$$\left(\mathrm{d}\right)\checkmark \\ $$$$\cancel{\bigtriangledown\centerdot\left(\overset{\rightarrow} {{F}}×\overset{\rightarrow} {{G}}\right)=\overset{\rightarrow} {{G}}\centerdot\bigtriangledown×\overset{\rightarrow} {{F}}−\overset{\rightarrow} {{F}}\centerdot\bigtriangledown×\overset{\rightarrow} {{G}}\left({a}\right)\checkmark} \\ $$$$\cancel{\bigtriangledown\centerdot\left({g}\overset{\rightarrow} {{F}}\right)=\partial_{{i}} \left({gF}_{{i}} \right)} \\ $$$$\therefore{d}\checkmark \\ $$

Answered by Nicholas666 last updated on 20/Apr/25