Match the following Column I (A) Σ_ i=1...

MathematicsEquation of a Line in SpaceJEE Advanced 2006Moderate

Match the following

List-I

(P)

\sum_{i = 1}^{\infty} tan^{- 1} (\frac{1}{2 i ^{2}}) = t

, then

tan t =

(Q)

Sides

a, b, c

of a triangle

A B C

are in AP and

cos θ_{1} = \frac{a}{b + c}

cos θ_{2} = \frac{b}{a + c}

cos θ_{3} = \frac{c}{a + b}

, then

tan^{2} (\frac{θ _{1}}{2}) + tan^{2} (\frac{θ _{3}}{2}) =

(R)

A line is perpendicular to

x + 2 y + 2 z = 0

and passes through

(0, 1, 0)

. The perpendicular distance of this line from the origin is

List-II

(1)

(2)

\frac{5}{3}

(3)

\frac{2}{3}

Answer:P → 1, Q → 3, R → 2

Visualized Solution (English)

Problem Overview

Part A: Calculate $t = \sum_{i = 1}^{\infty} tan^{- 1} (\frac{1}{2 i ^{2}})$ and find $tan t$ .
Part B: Given sides $a, b, c$ in AP, find $tan^{2} (\frac{θ _{1}}{2}) + tan^{2} (\frac{θ _{3}}{2})$ .
Part C: Find the perpendicular distance of a specific line from the origin.

Part A: Telescoping the Series

General term: $T_{i} = tan^{- 1} (\frac{1}{2 i ^{2}})$
Multiply numerator and denominator by $2$ : $T_{i} = tan^{- 1} (\frac{2}{4 i ^{2}})$
Rewrite as: $T_{i} = tan^{- 1} (\frac{( 2 i + 1 ) - ( 2 i - 1 )}{1 + ( 2 i + 1 ) ( 2 i - 1 )})$

Part A: Summing to Infinity

Using identity: $tan^{- 1} x - tan^{- 1} y = tan^{- 1} (\frac{x - y}{1 + x y})$
Sum $t = \sum_{i = 1}^{\infty} [tan^{- 1} (2 i + 1) - tan^{- 1} (2 i - 1)]$
Limit as $n \to \infty$ : $t = tan^{- 1} (\infty) - tan^{- 1} (1) = \frac{π}{2} - \frac{π}{4} = \frac{π}{4}$
Therefore, $tan t = tan (\frac{π}{4}) = 1$

Part B: Triangle and AP

Sides $a, b, c$ in AP $⟹ 2 b = a + c$
Half-angle formula: $tan^{2} (\frac{θ}{2}) = \frac{1 - c o s θ}{1 + c o s θ}$
Given $cos θ_{1} = \frac{a}{b + c}$ and $cos θ_{3} = \frac{c}{a + b}$

Part B: Substitution and Algebra

$tan^{2} (\frac{θ _{1}}{2}) = \frac{1 - \frac{a}{b + c}}{1 + \frac{a}{b + c}} = \frac{b + c - a}{b + c + a}$
$tan^{2} (\frac{θ _{3}}{2}) = \frac{1 - \frac{c}{a + b}}{1 + \frac{c}{a + b}} = \frac{a + b - c}{a + b + c}$

Part B: Final Result

Sum $= \frac{b + c - a + a + b - c}{a + b + c} = \frac{2 b}{a + b + c}$
Since $a + c = 2 b$ , denominator $= 2 b + b = 3 b$
Sum $= \frac{2 b}{3 b} = \frac{2}{3}$

Part C: 3D Geometry Setup

Line passes through $P (0, 1, 0)$ and is $⊥$ to $x + 2 y + 2 z = 0$ .
Direction of line = Normal to plane $= (1, 2, 2)$ .
Equation of line: $\frac{x}{1} = \frac{y - 1}{2} = \frac{z}{2} = λ$

Part C: Finding the Foot

General point on line: $Q (λ, 2 λ + 1, 2 λ)$
Vector $O Q = (λ, 2 λ + 1, 2 λ)$
$O Q \cdot (1, 2, 2) = 0 ⟹ 1 (λ) + 2 (2 λ + 1) + 2 (2 λ) = 0$
Solve: $9 λ + 2 = 0 ⟹ λ = - \frac{2}{9}$

Part C: Calculating Distance

Foot of perpendicular: $Q (- \frac{2}{9}, \frac{5}{9}, - \frac{4}{9})$
Distance $O Q = (- \frac{2}{9})^{2} + (\frac{5}{9})^{2} + (- \frac{4}{9})^{2}$
$O Q = \frac{4 + 25 + 16}{81} = \frac{45}{81} = \frac{5}{3}$

Final Matching

Final Match:
(A) $\to$ (p): $tan t = 1$
(B) $\to$ (r): Sum $= \frac{2}{3}$
(C) $\to$ (q): Distance $= \frac{5}{3}$

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Conceptually Similar Problems

MathematicsFundamental Theorem & Properties of Definite IntegralsJEE Advanced 2010Difficult

View in:English Hindi

Match the statement in Column-I with the values in Column-II

List-I

(P)

A line from the origin meets the lines

\frac{x - 2}{1} = \frac{y - 1}{- 2} = \frac{z + 1}{1}

and

\frac{x - \frac{8}{3}}{2} = \frac{y + 3}{- 1} = \frac{z - 1}{1}

P

and

Q

respectively. If length

P Q = d

, then

d^{2}

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The values of

x

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tan^{- 1} (x + 3) - tan^{- 1} (x - 3) = sin^{- 1} (\frac{3}{5})

are

(R)

Non-zero vectors

a, b

and

c

satisfy

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(b - a) \cdot (b + c) = 0

and

2∣ b + c ∣ = ∣ b - a ∣

. If

a = μ b + 4 c

, then the possible values of

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(S)

Let

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[- π, π]

given by

f (0) = 9

and

f (x) = \frac{s i n ( \frac{9 x}{2} )}{s i n ( \frac{x}{2} )}

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(1)

-4

(2)

(3)

(4)

(5)

Answer:P → 5, P → 3, Q → 4, S → 3

MathematicsComponents of a VectorJEE Advanced 2015Difficult

View in:English Hindi

Match the following:

List-I

(P)

In a triangle

Δ X Y Z

, let

a, b

, and

c

be the lengths of the sides opposite to the angles

X, Y

and

Z

, respectively. If

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In a triangle

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a, b

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c

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X, Y

, and

Z

respectively. If

1 + cos 2 X - 2 cos 2 Y = 2 sin X sin Y

, then possible value(s) of

\frac{a}{b}

is (are)

(R)

R^{2}

, let

3 \hat{i} + \hat{j}, \hat{i} + 3 \hat{j}

and

β \hat{i} + (1 - β) \hat{j}

be the position vectors of

X, Y

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Z

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O

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Z

from the bisector of the acute angle of

O X

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O Y

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, then possible value(s) of

∣ β ∣

is (are)

(S)

Suppose that

F (α)

denotes the area of the region bounded by

x = 0, x = 2, y^{2} = 4 x

and

y = ∣ α x - 1∣ + ∣ α x - 2∣ + α x

, where

α \in {0, 1}

. Then the value(s) of

F (α) + \frac{8}{3} 2

, when

α = 0

and

α = 1

, is (are)

List-II

(1)

(2)

(3)

(4)

(5)

Answer:P → NaN, Q → 1, P → 2, S → 5

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Match the following :

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(P)

Two rays

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Point

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. Let

a = α \hat{i} + β \hat{j} + γ \hat{k}

\hat{k} \times (\hat{k} \times a) = 0

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sin A sin B sin C + cos A cos B = 1

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(1)

(2)

4/3

(3)

\int_{0}^{1} 1 - x d x + \int_{- 1}^{0} 1 + x d x

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Match List I with List II:

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(P)

Let

y (x) = cos (3 cos^{- 1} x), x \in [- 1, 1], x \neq = \pm \frac{3}{2}

. Then

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Number of positive solutions satisfying the equation

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(1)

(2)

(3)

(4)

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Match List I with List II:

List-I

(P)

[\frac{1}{y ^{2}} (\frac{c o s ( t a n ^{- 1} y ) + y s i n ( t a n ^{- 1} y )}{c o t ( s i n ^{- 1} y ) + t a n ( s i n ^{- 1} y )})^{2} + y^{4}]^{1/2}

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(Q)

cos x + cos y + cos z = 0 = sin x + sin y + sin z

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cos \frac{x - y}{2}

(R)

cos (\frac{π}{4} - x) cos 2 x + sin x sin 2 x sec x = cos x sin 2 x sec x + cos (\frac{π}{4} + x) cos 2 x

then possible value of

sec x

(S)

cot (sin^{- 1} 1 - x^{2}) = sin (tan^{- 1} (x 6))

x \neq = 0

, then possible value of

x

List-II

(1)

\frac{1}{2} \frac{5}{3}

(2)

2

(3)

\frac{1}{2}

(4)

Answer:P → 4, Q → 3, R → 2, S → 1

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Let $x, y$ and $z$ be three vectors each of magnitude $2$ and the angle between each pair of them is $\frac{π}{3}$ . If $a$ is a non-zero vector perpendicular to $x$ and $y \times z$ and $b$ is a non-zero vector perpendicular to $y$ and $z \times x$ , then

* Multiple Correct Options

(A)

b = (b \cdot z) (z - x)

(B)

a = (a \cdot y) (y - z)

(C)

a \cdot b = - (a \cdot y) (b \cdot z)

(D)

a = - (a \cdot y) (z - y)

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Let $a = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ , $b = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$ and $c = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k}$ be three non-zero vectors such that $c$ is a unit vector perpendicular to both the vectors $a$ and $b$ . If the angle between $a$ and $b$ is $\frac{π}{6}$ , then $a_{1} b_{1} c_{1} a_{2} b_{2} c_{2} a_{3} b_{3} c_{3}^{2}$ is equal to

(A)

(B)

(C)

\frac{1}{4} (a_{1}^{2} + a_{2}^{2} + a_{3}^{2}) (b_{1}^{2} + b_{2}^{2} + b_{3}^{2})

(D)

\frac{3}{4} (a_{1}^{2} + a_{2}^{2} + a_{3}^{2}) (b_{1}^{2} + b_{2}^{2} + b_{3}^{2}) (c_{1}^{2} + c_{2}^{2} + c_{3}^{2})

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MathematicsScalar Triple ProductJEE Advanced 2009Moderate

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Match the statements / expressions given in Column-I with the values given in Column-II.

List-I

(P)

Root(s) of the equation

2 sin^{2} θ + sin^{2} 2 θ = 2

(Q)

Points of discontinuity of the function

f (x) = [\frac{6 x}{π}] cos [\frac{3 x}{π}]

, where

[y]

denotes the largest integer less than or equal to

y

(R)

Volume of the parallelopiped with its edges represented by the vectors

\hat{i} + \hat{j}, \hat{i} + 2 \hat{j}

and

\hat{i} + \hat{j} + π \hat{k}

(S)

Angle between vector

a

and

b

where

a, b

and

c

are unit vectors satisfying

a + b + 3 c = 0

List-II

(1)

π /6

(2)

π /4

(3)

π /3

(4)

π /2

(5)

π

Answer:Q → 4, Q → NaN, R → 5, S → 3

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Match the statements given in Column-I with the values given in Column-II.

List-I

(P)

a = \hat{j} + 3 \hat{k}

b = - \hat{j} + 3 \hat{k}

and

c = 23 \hat{k}

form a triangle, then the internal angle of the triangle between

a

and

b

(Q)

\int_{a}^{b} (f (x) - 3 x) d x = a^{2} - b^{2}

, then the value of

f (\frac{π}{6})

(R)

The value of

\frac{π ^{2}}{l n 3} \int_{7/6}^{5/6} sec (π x) d x

(S)

The maximum value of

Arg (\frac{1}{1 - z})

for

∣ z ∣ = 1, z \neq = 1

is given by

List-II

(1)

\frac{π}{6}

(2)

\frac{2 π}{3}

(3)

\frac{π}{3}

(4)

π

(5)

\frac{π}{2}

Answer:P → 2, Q → 1, R → 4, S → 5

MathematicsShortest Distance Between Two Skew LinesJEE Advanced 2008Moderate

View in:English Hindi

Comprehension Passage

Consider the lines

L_{1} : \frac{x + 1}{3} = \frac{y + 2}{1} = \frac{z + 1}{2}

and

L_{2} : \frac{x - 2}{1} = \frac{y + 2}{2} = \frac{z - 3}{3}

Question 1:

The unit vector perpendicular to both $L_{1}$ and $L_{2}$ is

(A)

\frac{- i ^ + 7 j ^ + 7 k ^}{99}

(B)

\frac{- i ^ - 7 j ^ + 5 k ^}{5 3}

(C)

\frac{- i ^ + 7 j ^ + 5 k ^}{5 3}

(D)

\frac{7 i ^ - 7 j ^ - k ^}{99}

Answer:B

Question 2:

The shortest distance between $L_{1}$ and $L_{2}$ is

(A)

(B)

\frac{17}{3}

(C)

\frac{41}{5 3}

(D)

\frac{17}{5 3}

Answer:D

Question 3:

The distance of the point (1, 1, 1) from the plane passing through the point (-1, -2, -1) and whose normal is perpendicular to both the lines $L_{1}$ and $L_{2}$ is

(A)

\frac{2}{75}

(B)

\frac{7}{75}

(C)

\frac{13}{75}

(D)

\frac{23}{75}

Answer:C

Match the following

List-I

List-II

Visualized Solution (English)

Problem Overview

Part A: Telescoping the Series

Part A: Summing to Infinity

Part B: Triangle and AP

Part B: Substitution and Algebra

Part B: Final Result

Part C: 3D Geometry Setup

Part C: Finding the Foot

Part C: Calculating Distance

Final Matching

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Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The unit vector perpendicular to both L1​ and L2​ is

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The unit vector perpendicular to both L1​ and L2​ is

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