If your group is simple the adjoint representation is irreducible, so a (nonzero) orbit always spans $\mathfrak g$ and hence will not be contained in $B^\perp$ for any (nonzero) $B$. If $G$ is only semi-simple, you get $G(A)\subset B^\perp$ by taking $A$ and $B$ in different simple factors.

(For your extended question: let's again take $G$ simple and $A\ne0$. As the orbit $G(A)$ spans $\mathfrak g$, you can draw on it a curve $g_t(A)$ that still spans, hence is not in $B^\perp$ for any nonzero $B$. On the other hand, you may of course first fix some $B\ne0$ and then draw, in $B^\perp\cap G(A)$, a curve having your desired property.)

If your group is simple the adjoint representation is irreducible, so a (nonzero) orbit always spans $\mathfrak g$ and hence will not be contained in $B^\perp$ for any (nonzero) $B$. If $G$ is only semi-simple, you get $G(A)\subset B^\perp$ by taking $A$ and $B$ in different simple factors.

(For your extended question: let's again take $G$ simple and $A\ne0$. As the orbit $G(A)$ spans $\mathfrak g$, you can draw on it a curve $g_t(A)$ that still spans, hence is not in $B^\perp$ for any nonzero $B$. On the other hand, you may of course first fix some $B\in A^\perp$ and then draw, in $B^\perp\cap G(A)$, a curve having your desired property.)

If your group is simple the adjoint representation is irreducible, so a (nonzero) orbit always spans $\mathfrak g$ and hence will not be contained in $B^\perp$ for any (nonzero) $B$. If $G$ is only semi-simple, you get $G(A)\subset B^\perp$ by taking $A$ and $B$ in different simple factors.

For your extended question, let's again take $G$ simple and $A\ne0$. As the orbit $G(A)$ spans $\mathfrak g$, you can draw on it a curve $g_t(A)$ that still spans, hence is not in $B^\perp$ for any nonzero $B$. On the other hand, you may of course first fix some $B\ne0$ and then draw, in $B^\perp\cap G(A)$, a curve having your desired property.

If your group is simple the adjoint representation is irreducible, so a (nonzero) orbit always spans $\mathfrak g$ and hence will not be contained in $B^\perp$ for any (nonzero) $B$. If $G$ is only semi-simple, you get $G(A)\subset B^\perp$ by taking $A$ and $B$ in different simple factors.

(For your extended question: let's again take $G$ simple and $A\ne0$. As the orbit $G(A)$ spans $\mathfrak g$, you can draw on it a curve $g_t(A)$ that still spans, hence is not in $B^\perp$ for any nonzero $B$. On the other hand, you may of course first fix some $B\ne0$ and then draw, in $B^\perp\cap G(A)$, a curve having your desired property.)