Match the following. Column I Column II (A) (p) (B) Sides of a triangle are in A.P. and then (q) (C) A line is perpendicular to and passes through The perpendicular distance of this line from the origin is (r) [2006]

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Solution

Q.
Match the following. [2006]

Column I Column II

(A) $\sum_{i = 1}^{\infty} \tan^{- 1} (\frac{1}{2 i^{2}}) = t, then \tan t =$ (p) $1$

(B) Sides $a, b, c$ of a triangle $A B C$ are in A.P. and
$\cos θ_{1} = \frac{a}{b + c}, \cos θ_{2} = \frac{b}{a + c}, \cos θ_{3} = \frac{c}{a + b},$
then $\tan^{2} (\frac{θ_{1}}{2}) + \tan^{2} (\frac{θ_{3}}{2}) =$ (q) $\frac{\sqrt{5}}{3}$

(C) 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 (r) $\frac{2}{3}$

1 (A) → (p); (B) → (r); (C) → (q)

2 (A) → (r); (B) →(p); (C) →(q)

3 (A) → (q); (B) →(p); (C) → (r)

4 (A) → (q); (B) → (r); (C) → (p)

Ans.
(1)

(A) → (p); (B) → (r); (C) → (q)

$(A)$ $t = \sum_{i = 1}^{\infty} \tan^{- 1} (\frac{1}{2 i^{2}}) = \sum_{i = 1}^{\infty} \tan^{- 1} [\frac{(2 i + 1) - (2 i - 1)}{1 + 4 i^{2} - 1}]$

   $= \sum_{i = 1}^{\infty} [\tan^{- 1} (2 i + 1) - \tan^{- 1} (2 i - 1)]$

   $= \tan^{- 1} 3 - \tan^{- 1} 1 + \tan^{- 1} 5 - \tan^{- 1} 3 + \dots + \tan^{- 1} (2 n + 1) - \tan^{- 1} (2 n - 1) + \dots \infty$

$= \lim_{n \to \infty} [\tan^{- 1} (2 n + 1) - \tan^{- 1} 1]$

   $= \lim_{n \to \infty} \tan^{- 1} [\frac{2 n}{1 + (2 n + 1)}] = \lim_{n \to \infty} \tan^{- 1} (\frac{1}{1 + \frac{1}{n}})$

   $= \tan^{- 1} (1) = \frac{π}{4} \Rightarrow \tan t = 1,$ $(A) \to (p)$

$(B)$ $∵$ $a, b, c are in A.P. \Rightarrow 2 b = a + c$

   $Now \cos θ_{1} = \frac{a}{b + c} \Rightarrow \frac{1 - \tan^{2} \frac{θ_{1}}{2}}{1 + \tan^{2} \frac{θ_{1}}{2}} = \frac{a}{b + c}$

   $\Rightarrow \tan^{2} \frac{θ_{1}}{2} = \frac{b + c - a}{b + c + a}$

$Similarly, \tan^{2} \frac{θ_{3}}{2} = \frac{a + b - c}{a + b + c}$

$\Rightarrow \tan^{2} \frac{θ_{1}}{2} + \tan^{2} \frac{θ_{3}}{2} = \frac{2 b}{a + b + c} = \frac{2 b}{3 b} = \frac{2}{3},$ $(B) \to (r)$

$(C)$ $Equation of line through (0, 1, 0) and perpendicular to x + 2 y + 2 z = 0 is$

$\frac{x}{1} = \frac{y - 1}{2} = \frac{z}{2} = λ$

$For some value of λ, the foot of perpendicular from origin to line is (λ, 2 λ + 1, 2 λ)$

$Direction ratios of this perpendicular from origin are λ, 2 λ + 1, 2 λ$

$∴$ $1 \cdot λ + 2 (2 λ + 1) + 2 \cdot 2 λ = 0 \Rightarrow λ = - \frac{2}{9}$

   $∴$ $Foot of perpendicular is (- \frac{2}{9}, \frac{5}{9}, - \frac{4}{9})$

$Hence required distance = \sqrt{\frac{4}{81} + \frac{25}{81} + \frac{16}{81}} = \sqrt{\frac{45}{81}} = \frac{\sqrt{5}}{3},$ $(C) \to (q)$

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