A hall has a square floor of dimension 10 , m | vedclass

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(D) Let the vertices of the floor be $A(0,0,0)$,$B(10,0,0)$,$C(10,10,0)$,and $D(0,10,0)$. Let the height of the hall be $h$. Then the vertices of the

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$5 \sqrt{2}$

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(D) Let the vertices of the floor be $A(0,0,0)$,$B(10,0,0)$,$C(10,10,0)$,and $D(0,10,0)$. Let the height of the hall be $h$. Then the vertices of the ceiling are $E(0,0,h)$,$F(10,0,h)$,$G(10,10,h)$,and $H(0,10,h)$.The vector $\overrightarrow{AG} = (10-0)\hat{i} + (10-0)\hat{j} + (h-0)\hat{k} = 10\hat{i} + 10\hat{j} + h\hat{k}$.The vector $\overrightarrow{BH} = (0-10)\hat{i} + (10-0)\hat{j} + (h-0)\hat{k} = -10\hat{i} + 10\hat{j} + h\hat{k}$.The angle $	heta$ between $\overrightarrow{AG}$ and $\overrightarrow{BH}$ is given by $\cos 	heta = \frac{\overrightarrow{AG} \cdot \overrightarrow{BH}}{|\overrightarrow{AG}| |\overrightarrow{BH}|}$.Given $\cos 	heta = \frac{1}{5}$,we have:$\frac{1}{5} = \frac{(10\hat{i} + 10\hat{j} + h\hat{k}) \cdot (-10\hat{i} + 10\hat{j} + h\hat{k})}{\sqrt{10^2 + 10^2 + h^2} \sqrt{(-10)^2 + 10^2 + h^2}}$$\frac{1}{5} = \frac{-100 + 100 + h^2}{200 + h^2} = \frac{h^2}{200 + h^2}$.$200 + h^2 = 5h^2 \Rightarrow 4h^2 = 200 \Rightarrow h^2 = 50 \Rightarrow h = \sqrt{50} = 5\sqrt{2}$ meters.

$A$ hall has a square floor of dimension $10 \, m \times 10 \, m$ and vertical walls. If the angle $GPH$ between the diagonals $AG$ and $BH$ is $\cos^{-1} \frac{1}{5}$,then the height of the hall (in $meters$) is:

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