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Step-by-Step Solution
Step 1: Identify the Points and Setup Coordinates
Consider the square floor of side 10 m in the horizontal plane. Let:
A be at coordinates $(0,0,0)$.
B be at $(10,0,0)$.
Since the hall is 10 m wide along one axis, let the point on the adjacent corner be $(0,10,0)$. But to keep consistency with the provided vector notation, we will only explicitly define points A and B and capture others through given vectors.
We are given the corner points G and H in such a way that the height of the hall is $h$. This means the vertical $z$-coordinate for these points is $10$, and their in-plane positions include $h$ along the $y$-axis for both G and H. From the solution reference, one consistent assignment is:
$H = (0,h,10)$
$G = (10,h,10)$
This keeps the square base in the $xy$-plane and assigns $z=10$ at the top edge (interpreting the top diagonals at $z = 10$). Essentially, $A = (0,0,0)$, $B = (10,0,0)$. (Exact labeling choices can vary but must maintain consistency.)
Step 2: Express the Diagonal Vectors
The diagonals in question are:
$\overrightarrow{AG}$: from point A to point G
$\overrightarrow{BH}$: from point B to point H
Using the coordinates:
$\overrightarrow{AG} = (10 - 0)\,\hat{i} + (h - 0)\,\hat{j} + (10 - 0)\,\hat{k} = 10\,\hat{i} + h\,\hat{j} + 10\,\hat{k}$
$\overrightarrow{BH} = (0 - 10)\,\hat{i} + (h - 0)\,\hat{j} + (10 - 0)\,\hat{k} = -10\,\hat{i} + h\,\hat{j} + 10\,\hat{k}$
Step 3: Use the Dot Product to Find the Angle
The cosine of the angle $\theta$ between two vectors $\vec{u}$ and $\vec{v}$ is given by:
$\cos \theta = \frac{\vec{u}\cdot \vec{v}}{\lvert \vec{u}\rvert \;\lvert \vec{v}\rvert}
$
Here, $\theta = \angle GPH$ and $\cos \theta = \frac{1}{5}.$
Step 4: Compute the Dot Product and Magnitudes
First, compute the dot product $\overrightarrow{AG}\cdot \overrightarrow{BH}$:
$\overrightarrow{AG}\cdot \overrightarrow{BH} = (10)(-10) + (h)(h) + (10)(10)$
$= -100 + h^2 + 100 = h^2.$
Now, find the magnitudes:
$\lvert \overrightarrow{AG}\rvert = \sqrt{10^2 + h^2 + 10^2} = \sqrt{h^2 + 200},
\quad
\lvert \overrightarrow{BH}\rvert = \sqrt{(-10)^2 + h^2 + 10^2} = \sqrt{h^2 + 200}.
$
Step 5: Apply the Angle Condition
We are given $\cos \theta = \frac{1}{5}$. Substitute the vectors' dot product and magnitudes:
$\frac{1}{5} = \frac{\overrightarrow{AG}\cdot \overrightarrow{BH}}{\lvert \overrightarrow{AG}\rvert \,\lvert \overrightarrow{BH}\rvert}
= \frac{h^2}{(\sqrt{h^2 + 200})(\sqrt{h^2 + 200})}
= \frac{h^2}{h^2 + 200}.
$
Step 6: Solve for $h$
$\frac{1}{5} = \frac{h^2}{h^2 + 200}
\quad\Longrightarrow\quad
h^2 + 200 = 5\,h^2
\quad\Longrightarrow\quad
4\,h^2 = 200
\quad\Longrightarrow\quad
h^2 = 50
\quad\Longrightarrow\quad
h = 5\sqrt{2}.
$
Final Answer
The height of the hall is $5\sqrt{2}$ meters.
