Given:
Dip separation vector magnitude: \(d = 6 \text{ m}\)
Dip direction: \(60^\circ\) → \(040^\circ\) (dip angle)
Step 1 — Understand the relation:
For a normal fault, the heave is the horizontal component of the dip separation vector. If \(\delta\) is the dip angle: \[ \text{Heave} = d \cdot \sin(\delta) \]
Step 2 — Plug in values:
\[ \text{Heave} = 6 \cdot \sin 30^\circ \quad (\text{assume dip } \delta = 30^\circ \text{ from vertical}) \] \[ \sin 30^\circ = 0.5 \] \[ \text{Heave} = 6 \cdot 0.5 = 3 \text{ m} \] Answer: \(\boxed{3 \text{ m}}\)
