A million years ago, an alien species built a vertical tower on a horizontal plane. When they returned they discovered that the ground had tilted so that measurements of 3 points on the ground gave coordinates of (0, 0, 0), (1, 2, 0), and (0, 1, 3). By what angle does the tower now deviate from the vertical?


To solve this problem, we need to determine the angle by which the tower deviates from the vertical due to the tilt of the ground. 

### Step 1: Define the vectors on the ground
Given three points on the ground: 
- \( A(0, 0, 0) \)
- \( B(1, 2, 0) \)
- \( C(0, 1, 3) \)

### Step 2: Find two vectors on the plane
We can find two vectors that lie on the plane:
- Vector \( \mathbf{AB} \) from \( A \) to \( B \):
  \[
  \mathbf{AB} = B - A = (1, 2, 0)
  \]
- Vector \( \mathbf{AC} \) from \( A \) to \( C \):
  \[
  \mathbf{AC} = C - A = (0, 1, 3)
  \]

### Step 3: Find the normal vector to the plane
The normal vector \( \mathbf{n} \) to the plane can be found by taking the cross product of \( \mathbf{AB} \) and \( \mathbf{AC} \):
\[
\mathbf{n} = \mathbf{AB} \times \mathbf{AC}
\]

Calculate the cross product:
\[
\mathbf{n} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 2 & 0 \\
0 & 1 & 3
\end{vmatrix}
= \mathbf{i}(2 \cdot 3 - 0 \cdot 1) - \mathbf{j}(1 \cdot 3 - 0 \cdot 0) + \mathbf{k}(1 \cdot 1 - 2 \cdot 0)
\]
\[
\mathbf{n} = \mathbf{i}(6) - \mathbf{j}(3) + \mathbf{k}(1) = (6, -3, 1)
\]

### Step 4: Calculate the angle between the normal vector and the vertical direction
The vertical direction is represented by the vector \( \mathbf{v} = (0, 0, 1) \).

The angle \( \theta \) between the normal vector \( \mathbf{n} \) and the vertical vector \( \mathbf{v} \) is given by the formula:
\[
\cos \theta = \frac{\mathbf{n} \cdot \mathbf{v}}{|\mathbf{n}| \cdot |\mathbf{v}|}
\]

- The dot product \( \mathbf{n} \cdot \mathbf{v} \) is:
  \[
  \mathbf{n} \cdot \mathbf{v} = 6 \times 0 + (-3) \times 0 + 1 \times 1 = 1
  \]

- The magnitude \( |\mathbf{n}| \) of \( \mathbf{n} \) is:
  \[
  |\mathbf{n}| = \sqrt{6^2 + (-3)^2 + 1^2} = \sqrt{36 + 9 + 1} = \sqrt{46}
  \]

- The magnitude \( |\mathbf{v}| \) of \( \mathbf{v} \) is:
  \[
  |\mathbf{v}| = \sqrt{0^2 + 0^2 + 1^2} = 1
  \]

Therefore,
\[
\cos \theta = \frac{1}{\sqrt{46}}
\]
\[
\theta = \cos^{-1} \left( \frac{1}{\sqrt{46}} \right)
\]

### Final Answer:
The angle \( \theta \) by which the tower deviates from the vertical is:
\[
\theta \approx \cos^{-1} \left( \frac{1}{\sqrt{46}} \right) \approx 81.25^\circ
\]

### Do you want more details or have any questions?
Here are 5 related questions:
1. What is the physical significance of the cross product in this context?
2. How would the solution change if the plane was defined by different points?
3. How can we interpret the vector normal to the plane?
4. How do tilts in different directions affect the angle of deviation?
5. How does the magnitude of the normal vector relate to the plane's properties?

**Tip:** Always ensure that vectors defining a plane are correctly calculated before determining the normal vector.

A million years ago, an alien species built a vertical tower on a horizontal plane. When they returned they discovered that the ground had tilted so that measurements of 3 points on the ground gave coordinates of (0, 0, 0), (1, 2, 0), and (0, 1, 3). By what angle does the tower now deviate from the vertical?

Explore how to calculate the angle by which a tower deviates from the vertical due to ground tilt, using vector geometry and the cross product. This problem involves calculating the angle between a normal vector and the vertical direction, suitable for students in Grades 10-12.

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