I have a tetrahedron formed by some points $P_1, P_2, P_3, P_4$.
Form (column) vectors $ \vec{v}_1 = P_1 - P_4, \vec{v}_2 = P_2 - P_4, \vec{v}_3 = P_3 - P_4$ that represent three edges of the tetrahedron.
Form a matrix $A = [\vec{v}_1, \vec{v}_2, \vec{v}_3]$.
What can be said about its operator norm if the length of each vector $\vec{v}_1, \vec{v}_2, \vec{v}_3$ is not greater than one? Is there any special bound?
In general, given two normed vector space $U$ and $V$, the operator norm for a linear transform $T : U \to V$ depends on the choice of norms on $U$ and $V$.
For the $3 \times 3$ matrices at hand, we will assume the corresponding $U$ and $V$ are the ordinary $\mathbb{R}^3$ equipped with Euclidean norm.
By (one of the many equivalent) definition, $$\|A\| = \sup_{u \in S^2} \|Au\| = \sup_{(\lambda_1,\lambda_2,\lambda_3)\in S^2} \| \lambda_1 v_1 + \lambda_2 v_2 + \lambda_3 v_3\| $$ where $\displaystyle\;S^2 = \left\{ u = (\lambda_1, \lambda_2, \lambda_3) \in \mathbb{R}^3 : \|u\| = \sqrt{\lambda_1^2+\lambda_2^2+\lambda_3^2} = 1 \right\}$.
Notice for $(\lambda_1,\lambda_2,\lambda_3) \in S^2$, we have $$\begin{align} \| \lambda_1 v_1 + \lambda_2 v_2 + \lambda_3 v_3\| &\le |\lambda_1|\|v_1\| + |\lambda_2|\|v_2\| + |\lambda_3| \|v_3\|\\ &\stackrel{C.S}\le \sqrt{\lambda_1^2 + \lambda_2^2 + \lambda_3^2}\sqrt{\|v_1\|^2 + \|v_2\|^2 + \|v_3\|^2}\\ &= \sqrt{\|v_1\|^2 + \|v_2\|^2 + \|v_3\|^2} \le \sqrt{3} \end{align} $$ This implies the operator norm $\|A\|$ is bounded from above by $\sqrt{3}$.