Let the Arc AC of a Circle Subtend a Right Angle at the Centre O. If B on the Arc AC Divides It So That arc AB : arc BC = 1 : 5, and OC = αOA + βOB, Then α + √2(√3−1)β Is Equal To

JEE Main 2025 — 23 January, Morning Shift. Previous Year Question.

Problem

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 $\dfrac{\text{length of arc }AB}{\text{length of arc }BC}=\dfrac15$, and $\overrightarrow{OC}=\alpha\overrightarrow{OA}+\beta\overrightarrow{OB}$, then $\alpha+\sqrt2(\sqrt3-1)\beta$ is equal to:

Key insight. Since arc lengths are proportional to their subtended angles, the ratio $1:5$ splits the full $90°$ arc into angles of exactly $15°$ (for arc $AB$) and $75°$ (for arc $BC$) — placing $A$, $B$, $C$ at specific angles on the circle then turns the vector equation into two ordinary trigonometric equations, one for each coordinate direction.

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Approach

Split the $90°$ arc $AC$ into $6$ equal parts using the $1:5$ ratio, giving $\angle AOB=15°$ and $\angle BOC=75°$. Place $A$ along one axis and $C$ along the perpendicular axis (since $\angle AOC=90°$), with $B$ at $15°$ from $A$. Write $\overrightarrow{OC}=\alpha\overrightarrow{OA}+\beta\overrightarrow{OB}$ in component form and match coefficients of $\hat i$ and $\hat j$ to solve for $\alpha$ and $\beta$.

Solution

Step 1 — Split the 90° arc using the 1:5 ratio

$$\angle AOB = 90°\times\frac16=15°, \qquad \angle BOC = 90°\times\frac56=75°$$

Step 2 — Place the points using a radius R

Let $\overrightarrow{OA}=R\hat i$ (along the x-axis), $\overrightarrow{OB}=R\cos15°\,\hat i+R\sin15°\,\hat j$, and $\overrightarrow{OC}=R\hat j$ (since $\angle AOC=90°$).

Step 3 — Write the vector equation in component form

$$R\hat j = \alpha(R\hat i) + \beta\left(R\cos15°\,\hat i+R\sin15°\,\hat j\right)$$

Matching $\hat i$: $0=\alpha+\beta\cos15° \implies \alpha=-\beta\cos15°$

Matching $\hat j$: $1=\beta\sin15° \implies \beta=\dfrac{1}{\sin15°}$

Step 4 — Find α

$$\alpha = -\frac{\cos15°}{\sin15°} = -\cot15°$$

Step 5 — Evaluate sin15° and cos15° using the angle-subtraction formulas

$$\sin15° = \sin(45°-30°) = \frac{\sqrt2(\sqrt3-1)}{4}, \qquad \cos15° = \cos(45°-30°) = \frac{\sqrt2(\sqrt3+1)}{4}$$

Step 6 — Compute cot15° and β

$$\cot15° = \frac{\cos15°}{\sin15°} = \frac{\sqrt3+1}{\sqrt3-1} = \frac{(\sqrt3+1)^2}{(\sqrt3-1)(\sqrt3+1)} = \frac{4+2\sqrt3}{2} = 2+\sqrt3$$

So $\alpha = -(2+\sqrt3)$.

$$\beta = \frac{1}{\sin15°} = \frac{4}{\sqrt2(\sqrt3-1)} = \sqrt2(\sqrt3+1)$$

(rationalising: multiply numerator and denominator by $\sqrt3+1$, then simplify $\sqrt2$)

Step 7 — Compute √2(√3−1)β

$$\sqrt2(\sqrt3-1)\times\sqrt2(\sqrt3+1) = 2(\sqrt3-1)(\sqrt3+1) = 2(3-1) = 4$$

Step 8 — Combine

$$\alpha+\sqrt2(\sqrt3-1)\beta = -(2+\sqrt3)+4 = 2-\sqrt3$$

Answer

Common mistakes

• Splitting the arc in the wrong ratio direction. Since arc $AB$ : arc $BC$ = $1:5$ (not $5:1$), $B$ sits much closer to $A$ than to $C$ — this gives $\angle AOB=15°$ (the smaller angle), and mixing up which arc gets which fraction of $90°$ flips the whole calculation.
• Not simplifying $\sqrt2(\sqrt3-1)\beta$ before combining with $\alpha$. Multiplying it out directly (rather than leaving $\beta$ as a standalone messy radical) reveals a clean cancellation to $4$ — trying to combine unsimplified radical expressions risks an arithmetic slip.

Practise next

• Let arc $PR$ of a circle subtend a right angle at centre $O$, with point $Q$ on the arc dividing it so that arc $PQ$:arc $QR$=$1:2$. If $\overrightarrow{OR}=\alpha\overrightarrow{OP}+\beta\overrightarrow{OQ}$, find $\alpha+\beta$, using the same placement-and-component-matching approach.