Moving cross product in equation

Question

Is $\vec{n} \times(\vec{H_{t1}}-\vec{H_{t2}}) = \vec{K} $ the same as $(\vec{H_{t1}}-\vec{H_{t2}}) = \vec{K} \times-\vec{n} $?

Answer 1

You just have to disprove that by multiplying both sides of $\vec{n} \times(\vec{H_{t1}}-\vec{H_{t2}}) = \vec{K} $ from left by $-\vec{n}$, $$ (\vec{n} \times(\vec{H_{t1}}-\vec{H_{t2}})) \times (-\vec{n}) = \vec{K} \times (-\vec{n}) $$ and by using identity $\vec{A} \times (\vec{B} \times \vec{C}) = \vec{B}(\vec{A} \cdot \vec{C}) - \vec{C}(\vec{A} \cdot \vec{B}) $, denote $\vec{H} = (\vec{H_{t1}}-\vec{H_{t2}})$,

you'll have $\vec{K} \times (-\vec{n}) = \vec{n} (\vec{n} \cdot \vec{H}) - \vec{H} (\vec{n} \cdot \vec{n})$, which is not always equal to $(\vec{H_{t1}}-\vec{H_{t2}})$.

Answer 2

Unless you have some reason to assume that $\langle H_{t_1}-H_{t_2}; n\rangle=0$, no, because of the identity $$(a\times b)\times(-a)=-\langle a;a\rangle b+\langle b;a\rangle a$$