Three-point problem: A 600 m N, C 400 m W (400, 100, 300)

Method: the three-point problem. Three points at known altitudes define a plane. Find a point on the line AB with the same altitude as C; the line joining it to C is a strike line (a horizontal line on the bed).

Given: B is the origin. A is 600 m north of B (altitude 400 m). C is 400 m west of B (altitude 300 m). B is at 100 m.

        A (400 m)
        |
        | 600 m
        |
  C (300 m) -------- B (100 m)   [C is 400 m west]

Step 1: Strike line

Along AB the altitude rises uniformly from 100 m (B) to 400 m (A), i.e. 300 m in 600 m (0.5 m per m). The point D on AB with altitude 300 m (equal to C) is at a distance from B of

$$BD = \frac{300-100}{400-100} \times 600 = 400\ \text{m north of B}$$

Line DC joins two points at 300 m, so it is the strike line. It runs from D (0, 400) to C (−400, 0).

Step 2: Strike direction

Taking east as +x and north as +y, the northward rise is $b = (400-100)/600 = 0.5$ and the eastward rise is $a = (300-100)/(-400) = -0.5$ (the bed rises towards the west).

From D to C the line runs 400 m west and 400 m south. The acute angle it makes with the north-south line is

$$\theta = \arctan\left(\frac{400}{400}\right) = 45°$$

D to C runs towards S45°W, so the strike is N45°E (the same line as S45°W).

Cross-check: the strike line is perpendicular to the direction of rise (a, b) = (−0.5, 0.5), which points to N45°W, giving a strike azimuth of 45° (or 225°).

Step 3: Dip direction and dip amount

The bed rises towards the north-west (A and C are higher than B), so it dips the opposite way (down the slope), at 90° to the strike. Dip direction azimuth = 135°, which is S45°E.

$$\tan\delta = \sqrt{a^2 + b^2} = \sqrt{0.5^2 + 0.5^2} = 0.7071 \Rightarrow \delta = 35.26°$$

Answer: Strike N45°E; dip direction S45°E; dip amount 35.3°.