Question

A hall has a square floor of dimension 10 m \(\times\) 10 m (see the figure) 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 : 

 

• A. \(5\sqrt{2}\)
• B. \(5\sqrt{3}\)
• C. 5
• D. \(2\sqrt{10}\)

Hint:

Vector algebra is a very powerful tool for solving 3D geometry problems. Setting up a coordinate system is often the best first step. Remember the dot product formula for the angle between two vectors, as it's frequently tested.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The hall is a cuboid with a square base. We are given the dimensions of the base and the angle between two of its space diagonals. We need to find the height of the hall.
Step 2: Key Formula or Approach:
We can use vector algebra to solve this problem. Let's set up a coordinate system with one corner of the hall at the origin. Let the corner C be the origin (0, 0, 0). Since the floor is a square of side 10 m, the coordinates of the vertices on the floor are:
C = (0, 0, 0)
D = (10, 0, 0)
B = (0, 10, 0)
A = (10, 10, 0)
Let the height of the hall be `h`. The coordinates of the vertices on the ceiling are:
G = (0, 0, h)
H = (10, 0, h)
F = (0, 10, h)
E = (10, 10, h)
Now we can find the vectors representing the diagonals AG and BH.
Vector AG = \( \vec{G} - \vec{A} = (0, 0, h) - (10, 10, 0) = \langle -10, -10, h \rangle \)
Vector BH = \( \vec{H} - \vec{B} = (10, 0, h) - (0, 10, 0) = \langle 10, -10, h \rangle \)
The angle \(\theta\) between two vectors \( \vec{u} \) and \( \vec{v} \) is given by the dot product formula:
\[ \cos\theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|} \] Step 3: Detailed Explanation:
We are given that the angle \(\theta\) between AG and BH is \( \cos^{-1}\frac{1}{5} \), so \( \cos\theta = \frac{1}{5} \).
Let's calculate the components for the dot product formula:
1. Dot product AG \( \cdot \) BH:
\[ \langle -10, -10, h \rangle \cdot \langle 10, -10, h \rangle = (-10)(10) + (-10)(-10) + (h)(h) \] \[ = -100 + 100 + h^2 = h^2 \] 2. Magnitude of AG:
\[ |\text{AG}| = \sqrt{(-10)^2 + (-10)^2 + h^2} = \sqrt{100 + 100 + h^2} = \sqrt{200 + h^2} \] 3. Magnitude of BH:
\[ |\text{BH}| = \sqrt{(10)^2 + (-10)^2 + h^2} = \sqrt{100 + 100 + h^2} = \sqrt{200 + h^2} \] Now substitute these into the cosine formula:
\[ \cos\theta = \frac{h^2}{(\sqrt{200 + h^2})(\sqrt{200 + h^2})} = \frac{h^2}{200 + h^2} \] We are given \( \cos\theta = \frac{1}{5} \). So, we can set up the equation:
\[ \frac{1}{5} = \frac{h^2}{200 + h^2} \] Cross-multiply to solve for h:
\[ 1 \cdot (200 + h^2) = 5 \cdot h^2 \] \[ 200 + h^2 = 5h^2 \] \[ 200 = 4h^2 \] \[ h^2 = \frac{200}{4} = 50 \] \[ h = \sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2} \] Step 4: Final Answer:
The height of the hall is \(5\sqrt{2}\) meters. This corresponds to option (A).