across
  • across
A student asked me to prove\[\nabla\times\left(\vec F\times\vec G\right)=(\vec G\cdot\nabla)\vec F-(\vec F\cdot\nabla)\vec G+(\nabla\cdot\vec G)\vec F-(\nabla\cdot\vec F)\vec G,\]and I ended up having to stop midway through because of how long and tedious the process is. Would someone mind helping me find a proof of it? It's a well-known vector operator identity.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
try khan academy under linear algebra section. Maybe it there.
athe
  • athe
Use \[\left[ \vec A\times \left[ \vec B\times \vec C \right] \right]=\vec B \cdot \left( \vec A \cdot \vec C \right) - \vec C \cdot \left( \vec A \cdot \vec B \right)\]
TuringTest
  • TuringTest
somebody seems to have made sense of this, though I can't complete this proof I don't think: http://www.physicsforums.com/showthread.php?t=459660

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athe
  • athe
Try use BAC-CAB and identity: \[\left[ \vec A \times\left[ \vec B \times \vec C \right]\right] +\left[ \vec C \times\left[ \vec A \times \vec B \right]\right]+\left[ \vec B \times\left[ \vec C \times \vec A \right]\right] =0\]
athe
  • athe
\[\vec \nabla \times (\vec F \times \vec G)= -\vec G \times (\vec \nabla \times \vec F) -\vec F \times (\vec G \times \vec \nabla)\]
experimentX
  • experimentX
|dw:134380766|dw:1343807791809:dw|9966:dw| \[ \hat i \left( {\partial \over \partial y} (F_xG_y - F_yG_x) - {\partial \over \partial z} (F_zG_x - F_xG_z) \right) + \\ \hat j \left( {\partial \over \partial z} (F_yG_z - F_zG_y) - {\partial \over \partial x} (F_xG_y - F_yG_x) \right) + \\ \hat k \left( {\partial \over \partial x} (F_zG_x- F_xG_z) - {\partial \over \partial y} (F_yG_z - F_zG_y) \right) \] \[ = \hat i F_x \left ( {\partial G_y \over \partial y} + {\partial G_z \over \partial z} \right ) + \hat j F_y \left ( {\partial G_z \over \partial z} + {\partial G_x \over \partial x} \right ) + \hat k F_z \left ( {\partial G_x \over \partial x} + {\partial G_y \over \partial y} \right ) \\ - \left (\hat i G_x \left ( {\partial F_y \over \partial y} + {\partial F_z \over \partial z} \right ) + \hat j G_y \left ( {\partial F_z \over \partial z} + {\partial F_x \over \partial x} \right ) + \hat k G_z \left ( {\partial F_x \over \partial x} + {\partial F_y \over \partial y} \right )\right ) \] \[ = \hat i F_x \left ( {\partial G_x \over \partial x} +{\partial G_y \over \partial y} + {\partial G_z \over \partial z} \right ) + \hat j F_y \left ( {\partial G_y \over \partial y}+ {\partial G_z \over \partial z} + {\partial G_x \over \partial x} \right )\\ + \hat k F_z \left ( {\partial G_x \over \partial x} + {\partial G_y \over \partial y}+{\partial G_z \over \partial z} \right ) - \left( \hat i F_x \left( \partial G_x \over \partial x \right ) +\hat j F_y \left( \partial G_y \over \partial y \right ) + \hat k F_x \left( \partial G_z \over \partial z \right )\right ) - \\ \left (\hat i G_x \left ( {\partial F_y \over \partial y} + {\partial F_z \over \partial z} + {\partial F_x \over \partial x} \right ) + \hat j G_y \left ( {\partial F_z \over \partial z} + {\partial F_x \over \partial x} +{\partial F_y \over \partial y} \right ) + \hat k G_z \left ( {\partial F_x \over \partial x} + {\partial F_y \over \partial y} +{\partial F_z \over \partial z}\right ) \right ) \\ + \left( \hat i G_x \left ( \partial F_x \over \partial x \right ) + \hat j G_y \left ( \partial F_y \over \partial y \right ) + \hat k G_z \left ( \partial F_z \over \partial z \right )\right ) \] \[ \left ( \hat i F_x + \hat j F_y + \hat k F_z \right ) \left ( \nabla \cdot \vec G \right ) - \left ( \vec F \cdot \nabla \right )\vec G \\ -\left( \hat i G_x + \hat j G_y + \hat k F_z \right ) \left ( \nabla \cdot \vec F \right ) + \left( \vec G \cdot \nabla \right ) \vec F\] hence the required result follows
anonymous
  • anonymous
lol
experimentX
  • experimentX
lol .... what happened to my drawing!!
anonymous
  • anonymous
also see this
1 Attachment
experimentX
  • experimentX
oh that's easier ...

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