Animated Solution for Mathematics - Vector Algebra: Let $\vec{a},\vec{b},\vec{c}$ be three mutually perpendicular vectors of the same magnitude and equally inclined at an angle $\theta$, with the vector $\vec{a}+\vec{b}+\vec{c}$. Then $36{\cos }^{2}2\theta$ is equal to ___

A Vector Odyssey

Imagine you are standing in the corner of a room. Look at the three edges meeting at that corner—one along the floor, one along the wall, and one rising vertically. These are your vectors $\vec{a},\vec{b},$ and $\vec{c}$.

They are mutually perpendicular, meaning they form the basis of a 3D coordinate system. The problem asks us to consider their sum, $\vec{v}=\vec{a}+\vec{b}+\vec{c}$.

If you visualize these as edges of a cube, $\vec{v}$ is the space diagonal cutting through the center of that cube. This is the geometric heart of our problem.

We are given that $|\vec{a}|=|\vec{b}|=|\vec{c}|=k$. The beauty of this problem lies in the condition that these vectors are mutually perpendicular.

In the language of linear algebra, this means their dot products are zero:

This is our most powerful tool. Whenever you see 'mutually perpendicular' in a JEE problem, think of it as a green light to eliminate cross-terms in your expansions.

To find the angle $\theta$, we first need the magnitude of the resultant vector $\vec{v}$. We square it to make the math easier:

Expanding this trinomial, we get:

Because of the orthogonality we discussed, the pairwise dot products vanish. We are left with $|\vec{v}{|}^2=k^2+k^2+k^2=3k^2$, which means $|\vec{v}|=\sqrt{3}k$.

Connecting Vectors to Trigonometry

The problem defines $\theta$ as the angle between $\vec{a}$ and $\vec{v}$. We use the dot product definition:

Substituting $\vec{v}=\vec{a}+\vec{b}+\vec{c}$, the numerator becomes:

Again, the perpendicularity saves us: $\vec{a}\cdot \vec{b}=0$ and $\vec{a}\cdot \vec{c}=0$. The numerator simplifies to $|\vec{a}{|}^2=k^2$. Thus:

The Double Angle Identity

We have found $\cos \theta =\frac{1}{\sqrt{3}}$. The question asks for $36{\cos }^{2}2\theta$. We use the double angle identity:

Substituting our value:

Finally, we calculate:

The elegance of this result—a clean, integer answer—is the hallmark of a well-crafted JEE problem. You have successfully navigated the 3D space, utilized the power of orthogonality, and bridged the gap to trigonometry. The final answer is 4.