What is the difference between $U\wedge(U\wedge V)$ and $U\cdot(U\wedge V)$?

Question

I know the $U\cdot(U\wedge V)$ means:

$$ \begin{pmatrix} i & j & k \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ \end{pmatrix} \implies \begin{pmatrix} (u_2)(v_3) -(u_3)(v_2) \\ (v_1)(u_3) -(v_3)(u_1) \\ (u_1)(v_2) -(u_2)(v_1) \\ \end{pmatrix} $$

but I dont understand $U\wedge(U\wedge V)$.

Answer 1

$U\cdot(U\wedge V)$ should be a scalar, since it is the result of a dot or scalar product. Because $U\wedge V$ is orthogonal to $U$, it will be $0$

$U\wedge(U\wedge V)$ should be a vector, since it is the result of a cross or vector product. Because $U\wedge V$ is orthogonal to $U$, and $U\wedge(U\wedge V)$ is orthogonal to both, it be a multiple of the part of $V$ orthogonal to $U$: it will be a zero vector (not a zero scalar) if $V$ is a multiple of $U$, while it will be a multiple of $V$ if $V$ is orthogonal to $U$

Answer 2

If you understand $U\wedge V$, which is a vector, then there is nothing new with $U\wedge(U\wedge V)$. You can read it as $U\wedge W$ where $W:=U\wedge V$.

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$U\wedge(U\wedge V)$ is and coplanar with $U,V$ and orthogonal to $U$.