As we know cross product of any two vectors yields a vector perpendicular to plane containing both the vectors so is it same for the vector operator del crossed with a vector ∇ × F (curl of vector field F). if not why?
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$\begingroup$If $F = (-y, x, 1)$, then $\nabla \times F = (0, 0, 2)$ is not orthogonal to $F$. The cross product formalism is a mnemonic for remembering the curl formula, not a literal cross product of vectors: The "components" of $\nabla$ act on the components of $F$ by differentiation, not by multiplication.
$\endgroup$ 1 $\begingroup$No. As an example, consider the vector field $\vec{F} = y \hat{i} - x \hat{j} + \hat{k}$, whose curl is $\vec{\nabla} \times \vec{F} = 2 \hat{k}$. This is obviously not perpendular to $\vec{F}$ itself.
As to why this is not the case, the best answer I can think of is that the value of a function at a particular point and its derivatives at that same point are basically independent of each other, i.e., you can always find a function with a given value and a given set of derivatives at any one point in space. Since the curl of a vector field depends on the field's derivatives, it makes sense that the vector field and its curl could point pretty much any direction relative to each other.
$\endgroup$ 3 $\begingroup$I know this is an old question, but for posterity's sake, something important happens when the curl is orthogonal to the vector field.
If $V$ is a $C^\infty$ vector field and $V\cdot (\nabla \times V) \equiv 0$ on an open set $U$ of $\mathbb{R}^3$, then on $U$, we have that for some nowhere $0$ $C^\infty$ scalar function f, the vector field $fV$ is conservative (see problem 3 of ).
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