Menu Close

f-x-y-z-x-y-z-g-x-y-z-x-y-z-h-x-y-z-g-x-y-z-f-x-y-z-h-x-y-z-h-x-y-z-




Question Number 923 by 123456 last updated on 25/Apr/15
f(x,y,z)=x+y+z  g(x,y,z)=(x,y,z)  h(x,y,z)=g(x,y,z)−▽f(x,y,z)=???  ▽∙h(x,y,z)=?  ▽×h(x,y,z)=???
$${f}\left({x},{y},{z}\right)={x}+{y}+{z} \\ $$$$\boldsymbol{{g}}\left({x},{y},{z}\right)=\left({x},{y},{z}\right) \\ $$$$\boldsymbol{{h}}\left({x},{y},{z}\right)=\boldsymbol{{g}}\left({x},{y},{z}\right)−\bigtriangledown{f}\left({x},{y},{z}\right)=??? \\ $$$$\bigtriangledown\centerdot\boldsymbol{{h}}\left({x},{y},{z}\right)=? \\ $$$$\bigtriangledown×\boldsymbol{{h}}\left({x},{y},{z}\right)=??? \\ $$
Answered by 2closedStringsMeet last updated on 15/Aug/15
    h=(x,y,z)−(1,1,1)  ∴  ▽f=(1+0+0, 0+1+0, 0+0+1)  ▽∙h=3                          ∴  h=(x−1, y−1, z−1)  ▽×h= (0,0,0)
$$ \\ $$$$ \\ $$$$\boldsymbol{{h}}=\left({x},{y},{z}\right)−\left(\mathrm{1},\mathrm{1},\mathrm{1}\right)\:\:\therefore\:\:\bigtriangledown{f}=\left(\mathrm{1}+\mathrm{0}+\mathrm{0},\:\mathrm{0}+\mathrm{1}+\mathrm{0},\:\mathrm{0}+\mathrm{0}+\mathrm{1}\right) \\ $$$$\bigtriangledown\centerdot\boldsymbol{{h}}=\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\therefore\:\:\boldsymbol{{h}}=\left({x}−\mathrm{1},\:{y}−\mathrm{1},\:{z}−\mathrm{1}\right) \\ $$$$\bigtriangledown×\boldsymbol{{h}}=\:\left(\mathrm{0},\mathrm{0},\mathrm{0}\right) \\ $$$$ \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *