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Part 1:

Represent the information as three coordinates (x, y, z) and let (0, 0, 0) be the position of the first drill at ground level.
Positive x represents east, and negative x represents west.
Positive y represents north, and negative y represents south.
Positive z represents up, and negative z represents down.

First: (0, 0, -1.1)
Second: (3, 83, -2.4)
Third: (-75 + 1.9cos(85°), 0, -1.9sin(85°)) or (-74.8, 0, -1.89)

Now, we can solve for the equation of the plane.
In 2 dimensions, a possible equation for the line would be [math]y = mx + y_0[/math].
In 3 dimensions, a possible equation for the plane would be [math]z = m_xx + m_yy + z_0[/math].
Plugging in our values, we get these three equations:
[math]-1.1 = m_x\cdot 0 + m_y\cdot 0 + z_0[/math]
[math]-2.4 = m_x\cdot 3 + m_y\cdot 83 + z_0[/math]
[math]-1.89 = m_x\cdot -74.8 + m_y\cdot 0 + z_0[/math]

Part 2:

Now, we have three linear equations in three variables. Solve this however you like! Here, I'll use Cramer's rule (best used with a calculator that supports matrices), but you can use any method you want:

[math]m_x = \frac{\begin{vmatrix}-1.1 & 0 & 1 \\ -2.4 & 83 & 1 \\ -1.89 & 0 & 1\end{vmatrix}}{\begin{vmatrix}0 & 0 & 1 \\ 3 & 83 & 1 \\ -74.8 & 0 & 1\end{vmatrix}} = 0.0106[/math]

[math]m_y = \frac{\begin{vmatrix}0 & -1.1 & 1 \\ 3 & -2.4 & 1 \\ -74.8 & -1.89 & 1\end{vmatrix}}{\begin{vmatrix}0 & 0 & 1 \\ 3 & 83 & 1 \\ -74.8 & 0 & 1\end{vmatrix}} = -0.160[/math]

[math]z_0 = \frac{\begin{vmatrix}0 & 0 & -1.1 \\ 3 & 83 & -2.4 \\ -74.8 & 0 & -1.89\end{vmatrix}}{\begin{vmatrix}0 & 0 & 1 \\ 3 & 83 & 1 \\ -74.8 & 0 & 1\end{vmatrix}} = -1.1[/math]

So we get the equation [math]z = 0.0106x - 0.160y - 1.1[/math]

Part 3:

The strike (measured clockwise from north) is [math]\arctan\frac{0.160}{0.0106} = 86.2\degree[/math]

(For those that understand calculus notation, [math]\frac{0.160}{0.0106} = \frac{\partial x}{\partial y}[/math])

So, the dip direction is either 176.2° or -1.8°. In this case, it is -1.8°. Using the apparent dip equation, the dip is [math]\arctan\frac{\tan{AD}}{\sin(90\degree - 1.8\degree)} = \arctan\frac{0.160}{\sin{88.2\degree}} = 9.09\degree\textrm{NW}[/math]

([math]\tan{AD} = \frac{-\partial z}{\partial y}[/math])

The thickness is [math]0.1\cos{9.09\degree} = 0.0987\ \textrm{km}[/math]

1) high
2) Sedimentary
3) High
4) Micas, High