Animated Solution for Mathematics - Vector Algebra: Match the statements / expressions given in Column-I with the values given in Column-II.

A Multi-Concept Odyssey

Welcome, future engineer. Today, we are not just solving a problem; we are embarking on a journey through four distinct landscapes of mathematics.

The 'Match the Following' format is a classic JEE Advanced staple. It is designed to test your versatility—your ability to switch gears from the rhythmic waves of trigonometry to the discrete jumps of calculus, and finally to the spatial elegance of 3D geometry.

Take a deep breath. We are going to dismantle this problem piece by piece.

The Trigonometric Dance

We begin with the equation $2{\sin }^2\theta +{\sin }^{2}2\theta =2$. When you see a mix of $\theta$ and $2\theta$, your first instinct should always be to unify the arguments.

We know the double-angle identity: $\sin 2\theta =2\sin \theta \cos \theta$. Substituting this, our equation transforms into:

Expanding this, we get $2{\sin }^2\theta +4{\sin }^2\theta {\cos }^2\theta =2$. Divide by 2, and we have ${\sin }^2\theta +2{\sin }^2\theta {\cos }^2\theta =1$.

Now, replace ${\cos }^2\theta$ with $1-{\sin }^2\theta$. This is the turning point. We arrive at:

This is a quadratic in disguise! Factoring it gives $(2{\sin }^2\theta -1)({\sin }^2\theta -1)=0$.

We find that ${\sin }^2\theta =1/2$ or ${\sin }^2\theta =1$. This leads us to $\theta =\pi /4$ and $\theta =\pi /2$. The beauty here lies in the transformation—taking a complex-looking equation and reducing it to a simple quadratic.

The GIF Trap

Next, we encounter the Greatest Integer Function (GIF): $f(x)=[\frac{6x}{\pi }]\cos [\frac{3x}{\pi }]$. Many students fear the GIF, but it is simply a step function.

It is discontinuous whenever the input inside the bracket becomes an integer. Our task is to check the given points: $\pi /6,\pi /4,\pi /3,\pi /2,\pi$.

Let's test $x=\pi /6$. The first term becomes $[6(\pi /6)/\pi ]=[1]=1$. The second term is $\cos [3(\pi /6)/\pi ]=\cos [0.5]=\cos (0)=1$.

But wait, the discontinuity happens at the transition to the integer. If we check $x=\pi /6$, the input $\frac{6x}{\pi }$ hits 1. At $x=\pi /3$, the input $\frac{6x}{\pi }$ hits 2 and $\frac{3x}{\pi }$ hits 1.

Both are integers! This is where the function jumps. By systematically testing each value, we realize that for $\pi /6,\pi /3,\pi /2,$ and $\pi$, at least one of the GIF terms hits an integer.

Only $\pi /4$ leaves us with non-integers ($1.5$ and $0.75$), keeping the function continuous. It is a game of precision.

The Volume of Space

Now, we step into 3D geometry. We need the volume of a parallelepiped defined by $\vec{u}=\hat{i}+\hat{j}$, $\vec{v}=\hat{i}+2\hat{j}$, and $\vec{w}=\hat{i}+\hat{j}+\pi \hat{k}$.

The volume is the absolute value of the scalar triple product, which is the determinant of the matrix formed by these vectors: $V=|\det (\vec{u},\vec{v},\vec{w})|$. Setting up the determinant:

Expanding along the third column is the smartest move here. We get $\pi \times (1\times 2-1\times 1)=\pi (1)=\pi$.

It is elegant, clean, and satisfying. The geometry of space collapses into a single scalar value.

Vector Harmony

Finally, we tackle the vector equation $\vec{a}+\vec{b}+\sqrt{3}\vec{c}=0$. We need the angle between $\vec{a}$ and $\vec{b}$.

The trick is to isolate the vectors we care about: $\vec{a}+\vec{b}=-\sqrt{3}\vec{c}$. Now, square both sides: $|\vec{a}+\vec{b}{|}^2=|-\sqrt{3}\vec{c}{|}^2$.

Expanding the left side gives $|\vec{a}{|}^2+|\vec{b}{|}^2+2\vec{a}\cdot \vec{b}=3|\vec{c}{|}^2$. Since these are unit vectors, their magnitudes are 1.

We get $1+1+2(1)(1)\cos \theta =3(1)$. This simplifies to $2+2\cos \theta =3$, or $2\cos \theta =1$.

Thus, $\cos \theta =1/2$, which means $\theta =\pi /3$.

We have traversed trigonometry, calculus, 3D geometry, and vector algebra. Each section required a different mindset, but they all relied on the same core principles: simplification, identification of critical points, and the power of algebraic manipulation.

You have the tools. Now, go forth and apply them.