Gradient (dot) cross product of 2 differentiable vector functions

[rasanders22]

1. For two differentiable vector functions E and H, prove that (Delta (dot) (e X h) = H (dot) (delta X e) - e (dot) (Delta X h)

2. Cross product and dot product.

The Attempt at a Solution

First I took did the left side of the equation, I took the cross product of vectors e and h. I put in varables to make the math easier e= <x1,y1,z1> h=<x2,y2,z2>. Then I dotted them with the gradient function. I came out with ...

@=partial

@/@x(y1z2-z1y2)+@/@y(z1x2-x1z2)+@/@z(x1y2-y1x2)

Then I did the right side of the equations. I came up with the same answer except it was larger by a factor of 2.

So my answer was 2[@/@x(y1z2-z1y2)+@/@y(z1x2-x1z2)+@/@z(x1y2-y1x2)].

I tried plugging in some random values for my variable to see what would happen. I first tried e=<cos(x),sin(y),xy> and h=<sin(x),cos(y),yz>. I then plugged in random point (1,1,1) and compared the answers. The answer from the right side of the equation is twice as large as the left which is too be expected. I think my mistake was paramaterizing my z value. I did some other examples and made z a function of z only and everything zero's out. Would paramaterized functions not be considered differentialble vector functions?
Thank You

 

[ehild]

1. For two differentiable vector functions E and H, prove that (Delta (dot) (e X h) = H (dot) (delta X e) - e (dot) (Delta X h)

2. Cross product and dot product.

The Attempt at a Solution

First I took did the left side of the equation, I took the cross product of vectors e and h. I put in varables to make the math easier e= <x1,y1,z1> h=<x2,y2,z2>. Then I dotted them with the gradient function. I came out with ...

@=partial

@/@x(y1z2-z1y2)+@/@y(z1x2-x1z2)+@/@z(x1y2-y1x2)

Then I did the right side of the equations. I came up with the same answer except it was larger by a factor of 2.

That is wrong. Perform the derivations on both sides.

ehild