Match the following
Column I
(A) Σ_{i=1}^∞ tan⁻¹(1/(2i²)) = t, then tan t =
(B) Sides a, b, c of a t

Q: Wait, why does the telescoping series sum to $\frac{\pi}{2} - \frac{\pi}{4}$? Where did the $\frac{\pi}{2}$ come from?

Great question! As $n \to \infty$, the term $\tan^{-1}(2n+1)$ approaches $\tan^{-1}(\infty)$, which is $\frac{\pi}{2}$. The telescoping nature leaves us with the limit of the upper bound minus the first lower bound, $\tan^{-1}(1)$, which is $\frac{\pi}{4}$. It is the beauty of limits in action!

Q: Can I use the Sine Rule for Part B instead of the half-angle formula?

You absolutely could, but it would be a much longer path! The half-angle formula is the 'surgical' tool here. Because you are given $\cos\theta$ directly, using the identity $\tan^2(\theta/2) = \frac{1-\cos\theta}{1+\cos\theta}$ allows the algebra to collapse beautifully. Always look for the path of least resistance in JEE problems.

Q: Why is the direction vector of the line the same as the normal of the plane?

Think about the geometry: a line perpendicular to a plane must run parallel to the plane's normal vector. Since the normal vector $\vec{n} = \langle 1, 2, 2 \rangle$ defines the orientation of the plane, any line perpendicular to that plane must share that same orientation. It is a fundamental property of 3D space!

Question

Match the following

List-I:

P. $\sum_{i=1}^{\infty} 	an^{-1}\left(\frac{1}{2i^2}ight) = t$, then $	an t =$

Q. Sides $a, b, c$ of a triangle $ABC$ are in AP and $\cos 	heta_1 = \frac{a}{b+c}$, $\cos 	heta_2 = \frac{b}{a+c}$, $\cos 	heta_3 = \frac{c}{a+b}$, then $	an^2\left(\frac{	heta_1}{2}ight) + 	an^2\left(\frac{	heta_3}{2}ight) =$

R. A line is perpendicular to $x + 2y + 2z = 0$ and passes through $(0, 1, 0)$. The perpendicular distance of this line from the origin is

List-II:

1. 1

2. $\frac{\sqrt{5}}{3}$

3. $\frac{2}{3}$

Sigma Percentile · Accepted Answer

Welcome, future engineer! Today, we are not just solving a problem; we are dissecting a masterpiece of JEE Advanced mathematics. This 'Match the Following' question is a beautiful triad—it tests your ability to handle infinite series, your command over trigonometric identities in geometry, and your spatial intuition in 3D.

### Part A: The Telescoping Dance

We begin with the infinite sum $t = \sum_{i=1}^{\infty} 	an^{-1} \left( \frac{1}{2i^2} ight)$. At first glance, this looks intimidating. The secret lies in the 'telescoping' technique.

We need to transform the argument $\frac{1}{2i^2}$ into the form $\frac{x-y}{1+xy}$. By multiplying the numerator and denominator by $2$, we obtain:

$$ \frac{2}{4i^2} = \frac{2}{1 + (4i^2 - 1)} = \frac{(2i+1) - (2i-1)}{1 + (2i+1)(2i-1)} $$

This is the 'Aha!' moment. Our term becomes:

$$ 	an^{-1} \left( \frac{(2i+1)-(2i-1)}{1+(2i+1)(2i-1)} ight) = 	an^{-1}(2i+1) - 	an^{-1}(2i-1) $$

Using the identity $	an^{-1}x - 	an^{-1}y = 	an^{-1} \left( \frac{x-y}{1+xy} ight)$, the sum collapses. As $i 	o \infty$, the terms cancel out, leaving only:

$$ \lim_{n 	o \infty} 	an^{-1}(2n+1) - 	an^{-1}(1) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} $$

Thus, $	an t = 1$.

### Part B: The Geometry of Arithmetic Progressions

Next, we step into the world of triangles. We are given sides $a, b, c$ in A.P., meaning $2b = a+c$. We need to evaluate $	an^2(	heta_1/2) + 	an^2(	heta_3/2)$.

The half-angle identity $	an^2(	heta/2) = \frac{1-\cos	heta}{1+\cos	heta}$ is our best friend here. Substituting $\cos	heta_1 = \frac{a}{b+c}$, we get:

$$ 	an^2(	heta_1/2) = \frac{1 - \frac{a}{b+c}}{1 + \frac{a}{b+c}} = \frac{b+c-a}{a+b+c} $$

Similarly, for $	heta_3$, we obtain $\frac{a+b-c}{a+b+c}$. Adding these, the numerators sum to:

$$ (b+c-a) + (a+b-c) = 2b $$

Since $a+c=2b$, the denominator $a+b+c$ becomes $3b$. The final result is:

$$ \frac{2b}{3b} = \frac{2}{3} $$

### Part C: The 3D Vector Odyssey

Finally, we tackle the 3D geometry. We have a line perpendicular to $x + 2y + 2z = 0$ passing through $(0, 1, 0)$. The normal vector $\vec{n} = \langle 1, 2, 2 angle$ is the direction of our line.

The line equation is $\frac{x}{1} = \frac{y-1}{2} = \frac{z}{2} = \lambda$. A general point on the line is $Q(\lambda, 2\lambda+1, 2\lambda)$.

To find the perpendicular distance from the origin, we find the foot of the perpendicular where the vector $\vec{OQ}$ is orthogonal to the line's direction. The dot product $\vec{OQ} \cdot \vec{n} = 0$ leads to:

$$ \lambda(1) + (2\lambda+1)(2) + (2\lambda)(2) = 0 \Rightarrow 9\lambda + 2 = 0 \Rightarrow \lambda = -2/9 $$

Calculating the distance from the origin to $Q(-2/9, 5/9, -4/9)$ yields:

$$ \sqrt{\left(-\frac{2}{9}ight)^2 + \left(\frac{5}{9}ight)^2 + \left(-\frac{4}{9}ight)^2} = \sqrt{\frac{4+25+16}{81}} = \sqrt{\frac{45}{81}} = \frac{\sqrt{5}}{3} $$

We have conquered the mountain!

Animated Solution for Mathematics - Three Dimensional Geometry: Match the following

List-I

List-II

Select Matching Pairs:

Visualized Solution

Problem Overview

Part A: Telescoping the Series

Part A: Summing to Infinity

Part B: Triangle and A.P.

Part B: Substitution and Algebra

Part B: Final Result

Part C: 3D Geometry Setup

Part C: Finding the Foot of Perpendicular

Part C: Solving for $λ$

Part C: Calculating Distance

Final Matching

The Sigma Insight: Equation of a Line in Space

Part A

Part B

Part C

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