1. Let F be the following vector field: F(x, y, z) =< e¯ªsin(y), e¯"sin(z), e¯²sin(x) >.A. Find div FB. Show that there is no vector field G such that curl G = 3yzj+2ri - rz²k.2. Let r be a three-dimensional real vector-valued function such that r" exists.Show that[r(t) x r'(t)] = {r"(t) × −3r(t).

"As per the policy we are allowed to answer only the first question, please repost the other question separately." Given That: Let F be the vector field such that F=e-xsiny , e-ysinz , e-zsinx . To Find: A)Divergence of F. B) Show that there is no vector field G exists such that curl G=2xi^+3yzj^-xz2k^. a) We have to find the divergence of F as follows: div F=∇·F=∂∂xi^+∂∂yj^+∂∂zk^·F=∂∂xi^+∂∂yj^+∂∂zk^·e-xsinyi^ + e-ysinzj^ + e-zsinxk^ =∂∂xe-xsiny +∂∂ye-ysinz +∂∂ze-zsinx =-e-xsiny -e-ysinz -e-zsinx=-e-xsiny + e-ysinz + e-zsinx Therefore, div F=-e-xsiny + e-ysinz + e-zsinx. b) We have to show that there is no vector field G such that curl G=2xi^+3yzj^-xz2k^ by using the following theorem and its contradiction. Theorem: If F=Pi^+Qj^+Rk^ be a vector field and P, Q, and R have continuous second-order partial derivatives, then divcurl F=0. So, by contradiction assume that there is a vector field such that curl G=2xi^+3yzj^-xz2k^ then, div curl G=∇.2xi^+3yzj^-xz2k^=∂∂xi^+∂∂yj^+∂∂zk^.2xi^+3yzj^-xz2k^=∂∂x2xi^.i^+∂∂y3yz j^.j^+∂∂z-xz2k^.k^=2+3z-2xz But this is the contradiction to the theorem that divcurl F=0. Therefore, our assumption is wrong that there is a vector field such that curl G=2xi^+3yzj^-xz2k^. Hence, there is no such vector field exists such that a curl G=2xi^+3yzj^-xz2k^.