Question

(A.dell)R=A plzzzzz prove it...
where A is any costant vector and R=xi+yj+zk

Answer 1

Did you mean
\[ (A * \nabla)r\]

Answer 2

replace r with R

Answer 3

(A dot dell)R

Answer 4

Let \[ A = a_x i + a_y j + a_z k \\
A* \nabla = (a_x i + a_y j + a_z k)* \left ( i\frac \partial {\partial x} +
j\frac \partial {\partial y} + k\frac \partial {\partial z}\right ) \\
= a_x\frac \partial {\partial x} + a_y\frac \partial {\partial y} + a_z\frac \partial {\partial z}\]
Now we have
\[ \left ( a_x\frac \partial {\partial x} + a_y\frac \partial {\partial y} + a_z\frac \partial {\partial z} \right ) ( i x + jy + kz) = i(a_x + 0 + 0) + j(0 + a_y + 0) + k(0 + 0 + a_z) \\
= ia_x + ja_y + ka_z = A\]

Answer 5

if A is constant delta* A=0

Answer 6

I guess it was
\[ \vec A * \nabla \text{ not } \nabla*\vec A\]
sure it would have been true then

Answer 7

ur right @experimentX

Answer 8

it's the same thing

Answer 9

i think A cant be a constant

Answer 10

Well, i meant the nabla is before or after A matters
though i'm not so sure ...

Answer 11

dot product is commutative

Answer 12

isn't it supposed to be operator??
i'm not sure though ...

Answer 13

if oyu have A=(3,3,3),you will get (3,3,3)=(0,0,0)

Answer 14

lol ... now i'm confused!!

Answer 15

ax,must depend x,ay y, az z do you agree?

Answer 16

yes

Answer 17

but isn't
\[ \nabla *(ix+jy+kz) = 3\]
while
\[ (ix+jy+kz)*\nabla = \text { some operator }\]
I think you might encounter them a lot in quantum mechanics