Answer
Theorem 2 (iii): ${\bf{v}} \times {\bf{w}} = {\bf{0}}$ if and only if ${\bf{w}} = \lambda {\bf{v}}$ for some scalar $\lambda$ or ${\bf{v}} = 0$.
Work Step by Step
According to Theorem 1: ${\bf{v}} \times {\bf{w}}$ has length $||{\bf{v}}||||{\bf{w}}||\sin \theta $, where $\theta$ is the angle between ${\bf{v}}$ and ${\bf{w}}$ for $0 \le \theta \le \pi $.
If ${\bf{v}} \times {\bf{w}} = {\bf{0}}$ this implies that the length of ${\bf{v}} \times {\bf{w}}$ is zero. If ${\bf{v}}$ and ${\bf{w}}$ are nonzero vectors, then they must be parallel such that $\theta=0$. Thus, ${\bf{w}} = \lambda {\bf{v}}$ for some scalar $\lambda$. Or in the case ${\bf{v}} = 0$, then the length of ${\bf{v}} \times {\bf{w}}$ is zero.
Conversely, if ${\bf{v}}$ and ${\bf{w}}$ are parallel and nonzero vectors, then ${\bf{w}} = \lambda {\bf{v}}$ for some scalar $\lambda$ such that ${\bf{v}} \times {\bf{w}} = {\bf{0}}$. If ${\bf{v}} = 0$, then we have ${\bf{v}} \times {\bf{w}} = {\bf{0}}$.
Hence, it proves Theorem 2 (iii): ${\bf{v}} \times {\bf{w}} = {\bf{0}}$ if and only if ${\bf{w}} = \lambda {\bf{v}}$ for some scalar $\lambda$ or ${\bf{v}} = 0$.
