## across Group Title 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. 2 years ago 2 years ago

1. timo86m Group Title

try khan academy under linear algebra section. Maybe it there.

2. athe Group Title

Use $\left[ \vec A\times \left[ \vec B\times \vec C \right] \right]=\vec B \cdot \left( \vec A \cdot \vec C \right) - \vec C \cdot \left( \vec A \cdot \vec B \right)$

3. TuringTest Group Title

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

4. athe Group Title

Try use BAC-CAB and identity: $\left[ \vec A \times\left[ \vec B \times \vec C \right]\right] +\left[ \vec C \times\left[ \vec A \times \vec B \right]\right]+\left[ \vec B \times\left[ \vec C \times \vec A \right]\right] =0$

5. athe Group Title

$\vec \nabla \times (\vec F \times \vec G)= -\vec G \times (\vec \nabla \times \vec F) -\vec F \times (\vec G \times \vec \nabla)$

6. experimentX Group Title

|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

7. mukushla Group Title

lol

8. experimentX Group Title

lol .... what happened to my drawing!!

9. mukushla Group Title

also see this

10. experimentX Group Title

oh that's easier ...