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Math  /  Geometry

QuestionA 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),(1,1,0)(0,0,0),(1,1,0), and (0,2,3)(0,2,3). By what angle does the tower now deviate from the vertical? □\square radians.

Studdy Solution

STEP 1

1. The original vertical tower was perpendicular to the horizontal plane.
2. The tilt of the ground can be represented by a plane through the given points.
3. We need to find the angle between the original vertical direction and the new normal to the tilted plane.

STEP 2

1. Determine the equation of the plane using the given points.
2. Find the normal vector to the plane.
3. Calculate the angle between the normal vector and the original vertical direction.

STEP 3

To find the equation of the plane, use the three given points: (0,0,0) (0,0,0) , (1,1,0) (1,1,0) , and (0,2,3) (0,2,3) . The general equation of a plane is ax+by+cz=d ax + by + cz = d .

STEP 4

Calculate two vectors on the plane using the given points: - Vector v1=(1,1,0)−(0,0,0)=(1,1,0) \mathbf{v_1} = (1,1,0) - (0,0,0) = (1,1,0) - Vector v2=(0,2,3)−(0,0,0)=(0,2,3) \mathbf{v_2} = (0,2,3) - (0,0,0) = (0,2,3)

STEP 5

Find the normal vector n \mathbf{n} to the plane by taking the cross product of v1 \mathbf{v_1} and v2 \mathbf{v_2} :
n=v1×v2=∣ijk110023∣ \mathbf{n} = \mathbf{v_1} \times \mathbf{v_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 0 \\ 0 & 2 & 3 \end{vmatrix}
n=(1⋅3−0⋅2)i−(1⋅3−0⋅0)j+(1⋅2−1⋅0)k \mathbf{n} = (1 \cdot 3 - 0 \cdot 2) \mathbf{i} - (1 \cdot 3 - 0 \cdot 0) \mathbf{j} + (1 \cdot 2 - 1 \cdot 0) \mathbf{k}
n=(3,−3,2) \mathbf{n} = (3, -3, 2)

STEP 6

The original vertical direction is along the z-axis, represented by the vector v=(0,0,1) \mathbf{v} = (0,0,1) .

STEP 7

Find the angle θ \theta between the normal vector n=(3,−3,2) \mathbf{n} = (3, -3, 2) and the vertical vector v=(0,0,1) \mathbf{v} = (0,0,1) using the dot product formula:
n⋅v=∣n∣∣v∣cos⁡θ \mathbf{n} \cdot \mathbf{v} = | \mathbf{n} | | \mathbf{v} | \cos \theta
n⋅v=3⋅0+(−3)⋅0+2⋅1=2 \mathbf{n} \cdot \mathbf{v} = 3 \cdot 0 + (-3) \cdot 0 + 2 \cdot 1 = 2
∣n∣=32+(−3)2+22=9+9+4=22 | \mathbf{n} | = \sqrt{3^2 + (-3)^2 + 2^2} = \sqrt{9 + 9 + 4} = \sqrt{22}
∣v∣=02+02+12=1 | \mathbf{v} | = \sqrt{0^2 + 0^2 + 1^2} = 1
2=22⋅1⋅cos⁡θ 2 = \sqrt{22} \cdot 1 \cdot \cos \theta
cos⁡θ=222 \cos \theta = \frac{2}{\sqrt{22}}
θ=cos⁡−1(222) \theta = \cos^{-1} \left( \frac{2}{\sqrt{22}} \right)
The angle by which the tower now deviates from the vertical is:
cos⁡−1(222) \boxed{\cos^{-1} \left( \frac{2}{\sqrt{22}} \right)}

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