A 10 inches long pencil $AB$ with mid point $C$ and a small eraser $P$ are placed on the horizontal top of a table such that $\mathrm{PC}=\sqrt{5}$ inches and $\angle PCB=\tan^{-1}(2)$. The acute angle through which the pencil must be rotated about C so that the perpendicular distance between eraser and pencil becomes exactly 1 inch is:

(D) $\tan^{-1}\left(\frac{3}{4}\right)$

(A) $\tan^{-1}\left(\frac{1}{2}\right)$

(B) $\tan^{-1}\left(\frac{4}{3}\right)$

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Straight Lines

A 10 inches long pencil

A B

with mid point

C

and a small eraser

P

are placed on the horizontal top of a table such that

PC = 5

inches and

∠ P C B = tan^{- 1} (2)

.
The acute angle through which the pencil must be rotated about C so that the perpendicular distance between eraser and pencil becomes exactly 1 inch is:

(A) $tan^{- 1} (\frac{1}{2})$
(B) $tan^{- 1} (\frac{4}{3})$
(C) $tan^{- 1} (1)$
(D) $tan^{- 1} (\frac{3}{4})$ ✓ Correct

Answer: (D)

Solution

The perpendicular distance from the eraser to the line of the pencil is

P C sin θ

, where

θ

is the angle between

C P

and the pencil. Initially

θ = tan^{- 1} 2

, so

sin θ = \frac{2}{5}

and the distance is

5 \cdot \frac{2}{5} = 2

inches.
Rotating the pencil about

C

through an acute angle

α

changes the angle to

θ - α

, and the distance becomes

1

inch when

5 sin (θ - α) = 1 ⟹ sin (θ - α) = \frac{1}{5} ⟹ tan (θ - α) = \frac{1}{2}

With

tan θ = 2

\frac{2 - tan α}{1 + 2 tan α} = \frac{1}{2} ⟹ 4 - 2 tan α = 1 + 2 tan α ⟹ tan α = \frac{3}{4}

The pencil must be rotated through

tan^{- 1} \frac{3}{4}

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Single correct option

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JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Straight Lines

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