If the magnitudes of vectors $\mathbf{a}$ and $\mathbf{b}$ are $5$ and $12$, respectively, then the magnitude of vector $(\mathbf{b-a})$ could NOT be (A) 5 (B) 7 (C) 10 (D) 12 (E) 17
The triangle inequality states that $7 < ||(\mathbf{b-a})||<17$ (i.e. the sums cannot be equal to that third length!). If it was permissible for them to be equal, then it would be (A) 5. Is it like a special exception for vectors because we are dealing with magnitudes?
When dealing with literal triangles in the plane, the inequality is strict so as to prevent degenerate triangles. To get a magnitude of $7$, draw $\textbf{b}$ so that it points east and draw $\textbf{a}$ parallel (so that it also points east). Then their difference $\textbf{b} - \textbf{a}$ is a vector of magnitude $7$ that points east. Note that the vectors don't really form a triangle (in the intuitive two-dimensional sense); it's all one-dimensional, along the same line.
Likewise, to get a magnitude of $17$, do the same thing except draw $\textbf{a}$ antiparallel (so that it points west).