Animated Solution for Mathematics - Vector Algebra: A hall has a square floor of dimension $10\text{ m}\times 10\text{ m}$ (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 Journey into 3D Thinking

My dear student, welcome to the world of 3D geometry. When you look at a problem like this, it is easy to feel overwhelmed by the lines, the vertices, and the unknown height $h$.

But I want you to take a deep breath. In the JEE Advanced, the most complex problems are often just simple concepts dressed in fancy clothes. Today, we are going to strip away the complexity and reveal the elegant, beating heart of this problem.

Establishing Our Reference Frame

Imagine you are standing in the corner of this hall, at point $D$. We need a language to describe where everything is, and in physics and mathematics, that language is the coordinate system.

By setting $D$ as our origin $(0,0,0)$, we anchor ourselves. The floor lies flat on the $xy$-plane, meaning every point on the floor has a $z$-coordinate of $0$.

Since the floor is a $10\text{ m}\times 10\text{ m}$ square, we can map the corners with absolute precision: - $D=(0,0,0)$ - $C=(10,0,0)$ - $A=(0,10,0)$ - $B=(10,10,0)$

Now, the roof. The roof is simply the floor, shifted upwards by a height $h$. So, any point on the roof is just the corresponding floor point with a $z$-coordinate of $h$.

Thus, $H=(0,0,h)$ and $G=(10,0,h)$. We have successfully translated the physical hall into a mathematical map. We are no longer guessing; we are calculating.

The Vector Bridge

Now, let us look at the diagonals. The problem asks us to consider the angle between the lines $AG$ and $BH$. In 3D space, the most robust way to handle the angle between two lines is to treat them as vectors.

Let us define vector $\overrightarrow{AG}$. To find it, we subtract the position of $A$ from the position of $G$:

Next, let us define vector $\overrightarrow{BH}$ by subtracting the position of $B$ from the position of $H$:

Do you see the elegance here? We have turned a geometric visualization into pure algebraic components. We are ready to use the dot product.

The Dot Product Engine

The dot product is the secret key to finding angles. We know that for any two vectors $\vec{u}$ and $\vec{v}$, the cosine of the angle $\theta$ between them is given by:

Let us calculate the numerator, the dot product $\overrightarrow{AG}\cdot \overrightarrow{BH}$:

Look at that! The $-100$ and $+100$ cancel out perfectly. This is the moment in the exam where you smile, because you know you are on the right track.

Now, for the denominator, we need the magnitudes. The magnitude of $\overrightarrow{AG}$ is:

And because of the symmetry of our square floor, the magnitude of $\overrightarrow{BH}$ is identical:

The Final Calculation

We are almost there. We substitute our findings back into the cosine formula:

This simplifies beautifully to:

Now, we cross-multiply to solve for $h$:

Taking the square root of both sides, we find:

And there it is. The height of the hall is $5\sqrt{2}$ meters.

I want you to pause and appreciate what you just did. You took a 3D structure, mapped it into a coordinate system, used vector algebra to define the lines, and employed the dot product to find the relationship between the height and the angle.

This is not just solving a problem; this is mastering the art of spatial reasoning. Keep this confidence. When you face the next problem, remember: define your coordinates, find your vectors, and trust the algebra. You are capable of solving anything.