Let the arc AC of a circle subtend a right angle at the centre O . If the point B on the $\operatorname{arc} \mathrm{AC}$, divides the arc AC such that $\frac{\text { length of } \operatorname{arc} \mathrm{AB}}{\text { length of } \operatorname{arc} \mathrm{BC}}=\frac{1}{5}$, and $\overrightarrow{\mathrm{OC}}=\alpha \overrightarrow{\mathrm{OA}}+\beta \overrightarrow{\mathrm{OB}}$, then $\alpha=\sqrt{2}(\sqrt{3}-1) \beta$ is equal to

Vector Algebra — JEE Main · 2025 · 23 Jan 2025 · Shift 1 Question with Solution

Let the arc AC of a circle subtend a right angle at the centre O . If the point B on the

arc AC

, divides the arc AC such that

\frac{length of arc AB}{length of arc BC} = \frac{1}{5}

, and

OC = α OA + β OB

, then

α = 2 (3 - 1) β

is equal to

Answer: (A)

Place

O

at the origin with

∣ O A ∣ = ∣ O B ∣ = ∣ O C ∣ = r

. Since arc

A B : B C = 1 : 5

and arc

A C

subtends

9 0^{\circ}

∠ A O B = 1 5^{\circ}

and

∠ A O C = 9 0^{\circ}

.

Write

O C = α O A + β O B

. Taking the dot product with

O A

r^{2} cos 9 0^{\circ} = α r^{2} + β r^{2} cos 1 5^{\circ} ⟹ α = - β cos 1 5^{\circ}

Taking the dot product with

O B

r^{2} cos 7 5^{\circ} = α r^{2} cos 1 5^{\circ} + β r^{2} ⟹ cos 7 5^{\circ} = - β cos^{2} 1 5^{\circ} + β = β sin^{2} 1 5^{\circ}

Since

cos 7 5^{\circ} = sin 1 5^{\circ}

β = \frac{sin 1 5 ^{\circ}}{sin ^{2} 1 5 ^{\circ}} = \frac{1}{sin 1 5 ^{\circ}}, α = - \frac{cos 1 5 ^{\circ}}{sin 1 5 ^{\circ}} = - cot 1 5^{\circ}

Using

cot 1 5^{\circ} = 2 + 3

and

\frac{1}{sin 1 5 ^{\circ}} = 6 + 2

α + 2 (3 - 1) β = - (2 + 3) + 2 (3 - 1) (6 + 2)

2 (3 - 1) (6 + 2) = 2 (18 + 6 - 6 - 2) = 2 (32 - 2) = 22 \cdot 2 = 4

α + 2 (3 - 1) β = - (2 + 3) + 4 = 2 - 3

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Single correct option