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 $\frac{\text{length of arc } AB}{\text{length of arc } BC}=\frac{1}{5}$ and $\vec{OC}=\alpha\vec{OA}+\beta\vec{OB},$ then $\alpha+\sqrt{2}(\sqrt{3}-1)\beta$ is equal to Options: (A) $2\sqrt{3}$ (B) $2-\sqrt{3}$ (C) $5\sqrt{3}$ (D) $2+\sqrt{3}$

### Analyzing the Geometric Setup We begin with a circle centered at $O$. We define the position vectors of points $A$ and $C$ as $\vec{OA} = \vec{a}$ and $\vec{OC} = \vec{c}$. Since these are radii, their magnitudes are equal: $|\vec{a}| = |\vec{c}| = R$. The arc $AC$ subtends a right angle of $90^\circ$ at the center. Point $B$ lies on this arc such that the ratio of arc lengths $AB:BC = 1:5$. Because arc length is proportional to the central angle, the ratio of the angles $\angle AOB$ to $\angle BOC$ is also $1:5$. Let $\angle AOB = x$ and $\angle BOC = 5x$. Solving $x + 5x = 90^\circ$ yields $6x = 90^\circ$, so $x = 15^\circ$. Thus, we have: $$\angle AOB = 15^\circ, \quad \angle BOC = 75^\circ$$ ### The Vector Equation We are given the linear combination: $$\vec{c} = \alpha \vec{a} + \beta \vec{b}$$ To isolate the scalars $\alpha$ and $\beta$, we utilize the dot product. First, we take the dot product of the equation with $\vec{c}$: $$\vec{c} \cdot \vec{c} = \alpha (\vec{a} \cdot \vec{c}) + \beta (\vec{b} \cdot \vec{c})$$ Since $\vec{a} \perp \vec{c}$, we have $\vec{a} \cdot \vec{c} = 0$. Given $|\vec{b}| = |\vec{c}| = R$ and the angle between them is $75^\circ$, the equation becomes: $$R^2 = 0 + \beta (R^2 \cos 75^\circ)$$ Solving for $\beta$: $$\beta = \frac{1}{\cos 75^\circ}$$ ### Solving for Alpha Next, we take the dot product of the original equation with $\vec{a}$: $$\vec{a} \cdot \vec{c} = \alpha (\vec{a} \cdot \vec{a}) + \beta (\vec{a} \cdot \vec{b})$$ Substituting $\vec{a} \cdot \vec{c} = 0$, $|\vec{a}|^2 = R^2$, and $\vec{a} \cdot \vec{b} = R^2 \cos 15^\circ$, we get: $$0 = \alpha R^2 + \beta (R^2 \cos 15^\circ)$$ This simplifies to: $$\alpha = -\beta \cos 15^\circ = -\frac{\cos 15^\circ}{\cos 75^\circ} = -\cot 15^\circ$$ Using the identity $\cot 15^\circ = 2 + \sqrt{3}$, we find: $$\alpha = -(2 + \sqrt{3})$$ ### Final Calculation We now evaluate the expression $\alpha + \sqrt{2}(\sqrt{3}-1)\beta$. Substituting our values for $\alpha$ and $\beta$: $$\text{Result} = -(2 + \sqrt{3}) + \sqrt{2}(\sqrt{3}-1) \cdot \frac{1}{\cos 75^\circ}$$ Recalling that $\cos 75^\circ = \sin 15^\circ = \frac{\sqrt{6}-\sqrt{2}}{4}$, the expression simplifies through the properties of surds. Upon completing the algebraic reduction, we arrive at the final value: $$\mathbf{2 - \sqrt{3}}$$

JEE Main 2025 (January)

LEVELJEE Advanced

Animated Solution for Mathematics - Vector Algebra: 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 $\frac{length of arc A B}{length of arc B C} = \frac{1}{5}$ and $O C = α O A + β O B,$ then $α + 2 (3 - 1) β$ is equal to

Select Answer:

Visualized Solution

Visualizing the Circle and Arcs

Let the center of the circle be the origin $O (0, 0)$ .
The arc $A C$ subtends $∠ A O C = 9 0^{\circ}$ at the center.
Let $O A = a$ and $O C = c$ . Since they are radii, $∣ a ∣ = ∣ c ∣ = R$ .

Introducing Point $B$

Point $B$ lies on the arc $A C$ , dividing it into two parts.
Let $O B = b$ . Its magnitude is also $R$ .
We are given the ratio of arc lengths: $\frac{arc A B}{arc B C} = \frac{1}{5}$ .

Relating Arc Lengths to Angles

Recall the formula: $Arc Length = R θ$ .
This means the ratio of the central angles is equal to the ratio of the arc lengths.
Therefore, $\frac{∠ A O B}{∠ B O C} = \frac{1}{5}$ .

Calculating the Central Angles

Let $∠ A O B = x$ and $∠ B O C = 5 x$ .
The total angle is $x + 5 x = 9 0^{\circ} ⟹ 6 x = 9 0^{\circ} ⟹ x = 1 5^{\circ}$ .
Thus, $∠ A O B = 1 5^{\circ}$ and $∠ B O C = 7 5^{\circ}$ .

Setting up the Vector Equation

We are given the relation: $c = α a + β b$ .
We know the magnitudes: $∣ a ∣ = ∣ b ∣ = ∣ c ∣ = R$ .
We also know the angles between them: $(a, c) = 9 0^{\circ}$ , $(a, b) = 1 5^{\circ}$ , $(b, c) = 7 5^{\circ}$ .

Dot Product with $c$

To find $β$ , let's take the dot product of the entire equation with $c$ .
$c \cdot c = α (a \cdot c) + β (b \cdot c)$ .
Substituting the dot product formula: $R^{2} = α (R^{2} cos 9 0^{\circ}) + β (R^{2} cos 7 5^{\circ})$ .

Simplifying the First Dot Product

Since $cos 9 0^{\circ} = 0$ , the term with $α$ vanishes.
We can cancel $R^{2}$ from both sides.
This leaves us with: $1 = 0 + β cos 7 5^{\circ}$ .

Finding the Value of $β$

Rearranging gives $β = \frac{1}{c o s 7 5 ^{\circ}}$ .
We know $cos 7 5^{\circ} = sin 1 5^{\circ} = \frac{3 - 1}{2 2}$ .
Therefore, $β = \frac{2 2}{3 - 1}$ .

Dot Product with $a$

Now, to find $α$ , let's take the dot product of the original equation with $a$ .
$a \cdot c = α (a \cdot a) + β (a \cdot b)$ .
Substituting the values: $R^{2} cos 9 0^{\circ} = α R^{2} + β (R^{2} cos 1 5^{\circ})$ .

Simplifying the Second Dot Product

Again, $cos 9 0^{\circ} = 0$ , so the left side is $0$ .
Canceling $R^{2}$ gives: $0 = α + β cos 1 5^{\circ}$ .
Rearranging for $α$ : $α = - β cos 1 5^{\circ}$ .

Finding the Value of $α$

Substitute $β = \frac{1}{c o s 7 5 ^{\circ}}$ into the equation.
$α = - \frac{c o s 1 5 ^{\circ}}{c o s 7 5 ^{\circ}} = - cot 1 5^{\circ}$ .
Since $cot 1 5^{\circ} = 2 + 3$ , we get $α = - (2 + 3)$ .

Setting up the Final Expression

The question asks for the value of: $α + 2 (3 - 1) β$ .
Let's substitute the exact values of $α$ and $β$ we just found.
Expression: $- (2 + 3) + 2 (3 - 1) (\frac{2 2}{3 - 1})$ .

Evaluating the Final Expression

Simplify the second term: the $(3 - 1)$ cancels out.
We are left with $2 \cdot 22 = 4$ .
So the expression becomes: $- (2 + 3) + 4$ .
Final result: $2 - 3$ .

00:00 / 00:00

The Sigma Insight: Scalar (Dot) Product

Solution Diagram

Analyzing the Geometric Setup

We begin with a circle centered at

O

. We define the position vectors of points

A

and

C

O A = a

and

O C = c

. Since these are radii, their magnitudes are equal:

∣ a ∣ = ∣ c ∣ = R

The arc

A C

subtends a right angle of

9 0^{\circ}

at the center. Point

B

lies on this arc such that the ratio of arc lengths

A B : B C = 1 : 5

Because arc length is proportional to the central angle, the ratio of the angles

∠ A O B

∠ B O C

is also

1 : 5

. Let

∠ A O B = x

and

∠ B O C = 5 x

Solving

x + 5 x = 9 0^{\circ}

yields

6 x = 9 0^{\circ}

, so

x = 1 5^{\circ}

. Thus, we have:

∠ A O B = 1 5^{\circ}, ∠ B O C = 7 5^{\circ}

The Vector Equation

We are given the linear combination:

c = α a + β b

To isolate the scalars

α

and

β

, we utilize the dot product. First, we take the dot product of the equation with

c

c \cdot c = α (a \cdot c) + β (b \cdot c)

Since

a ⊥ c

, we have

a \cdot c = 0

. Given

∣ b ∣ = ∣ c ∣ = R

and the angle between them is

7 5^{\circ}

, the equation becomes:

R^{2} = 0 + β (R^{2} cos 7 5^{\circ})

Solving for

β

β = \frac{1}{cos 7 5 ^{\circ}}

Solving for Alpha

Next, we take the dot product of the original equation with

a

a \cdot c = α (a \cdot a) + β (a \cdot b)

Substituting

a \cdot c = 0

∣ a ∣^{2} = R^{2}

, and

a \cdot b = R^{2} cos 1 5^{\circ}

, we get:

0 = α R^{2} + β (R^{2} cos 1 5^{\circ})

This simplifies to:

α = - β cos 1 5^{\circ} = - \frac{cos 1 5 ^{\circ}}{cos 7 5 ^{\circ}} = - cot 1 5^{\circ}

Using the identity

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

, we find:

α = - (2 + 3)

Final Calculation

We now evaluate the expression

α + 2 (3 - 1) β

. Substituting our values for

α

and

β

Result = - (2 + 3) + 2 (3 - 1) \cdot \frac{1}{cos 7 5 ^{\circ}}

Recalling that

cos 7 5^{\circ} = sin 1 5^{\circ} = \frac{6 - 2}{4}

, the expression simplifies through the properties of surds. Upon completing the algebraic reduction, we arrive at the final value:

2 - 3

Common Traps & FAQs

Why do we use the dot product instead of just resolving the vectors into x and y components?

That is a fantastic question. While you absolutely could use components, it would force you to define a coordinate system and calculate the coordinates of point B, which involves extra trigonometric steps. The dot product is coordinate-independent and elegant. By dotting with \vec{c} and \vec{a}, we exploit the orthogonality of the vectors to isolate the coefficients \alpha and \beta directly. It is the 'JEE way'—finding the path of least resistance through the geometry.

Wait, why does the \alpha term vanish when we dot with \vec{c}?

Remember that the problem states the arc AC subtends a right angle at the center O. This means the angle between \vec{OA} (which is \vec{a}) and \vec{OC} (which is \vec{c}) is 90^\circ. By definition, the dot product of two perpendicular vectors is zero. So, \vec{a} \cdot \vec{c} = |\vec{a}||\vec{c}| \cos(90^\circ) = 0. This is the 'hidden trap' that makes the algebra collapse beautifully!

How do I handle the \cos(75^\circ) and \cos(15^\circ) values without a calculator?

You must master your compound angle identities! Remember that 75^\circ = 45^\circ + 30^\circ and 15^\circ = 45^\circ - 30^\circ. Using the expansion \cos(A \pm B) = \cos A \cos B \mp \sin A \sin B, you can derive these values easily. For JEE Advanced, these are standard tools you should be able to recall instantly.

Select Answer:

Visualized Solution

Visualizing the Circle and Arcs

Introducing Point $B$

Relating Arc Lengths to Angles

Calculating the Central Angles

Setting up the Vector Equation

Dot Product with $c$

Simplifying the First Dot Product

Finding the Value of $β$

Dot Product with $a$

Simplifying the Second Dot Product

Finding the Value of $α$

Setting up the Final Expression

Evaluating the Final Expression

The Sigma Insight: Scalar (Dot) Product

Analyzing the Geometric Setup

The Vector Equation

Solving for Alpha

Final Calculation

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Select Answer:

Visualized Solution

Visualizing the Circle and Arcs

Introducing Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Relating Arc Lengths to Angles

Calculating the Central Angles

Setting up the Vector Equation

Dot Product with .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

Simplifying the First Dot Product

Finding the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

Dot Product with .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Simplifying the Second Dot Product

Finding the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Setting up the Final Expression

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The Sigma Insight: Scalar (Dot) Product

Analyzing the Geometric Setup

The Vector Equation

Solving for Alpha

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