= Solution
{c}
Use the <vector triple product identity>
$$
\mathbf a\times(\mathbf b\times\mathbf c)=\mathbf b(\mathbf a\cdot\mathbf c)-\mathbf c(\mathbf a\cdot\mathbf b).
$$
Since $\mathbf n$ is a <unit vector>, put
$$
\mathbf w=\mathbf n\times(\mathbf v\times\mathbf n)=\mathbf v-(\mathbf v\cdot\mathbf n)\mathbf n,\qquad \mathbf q=\mathbf n\times\mathbf v.
$$
Thus $\mathbf w$ is the component of $\mathbf v$ in the <orthogonal complement> of $\mathbf n$, and $\mathbf n\cdot\mathbf w=0$. A <cross product> is <orthogonal> to both its factors, so $\mathbf n\cdot\mathbf q=0$ and $\mathbf v\cdot\mathbf q=0$. Consequently
$$
\mathbf q\cdot\mathbf w=\mathbf q\cdot\mathbf v-(\mathbf v\cdot\mathbf n)(\mathbf q\cdot\mathbf n)=0.
$$
\b[All three pairs are <orthogonal>.] This remains true when $\mathbf v$ is parallel to $\mathbf n$: then $\mathbf q$ and $\mathbf w$ are <zero vectors>. Pairwise <orthogonality> does not assert <linear independence> in this case.
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