Logo
Jeeify
MCQMediumJEE 2025Basics of Vectors

JEE Mathematics 2025 Question with Solution

Let the arc AC of a circle subtend a right angle at the center O. If the point B on the arc AC divides the arc AC such that: length of arc ABlength of arc BC=15\frac{\text{length of arc AB}}{\text{length of arc BC}} = \frac{1}{5} and OC→=αOA→+βOB→,\overrightarrow{OC} = \alpha \overrightarrow{OA} + \beta \overrightarrow{OB}, then α=2(3−1)β\alpha = \sqrt{2} (\sqrt{3}-1) \beta is equal to:

  • A

    2−32 - \sqrt{3}

  • B

    232 \sqrt{3}

  • C

    535 \sqrt{3}

  • D

    2+32 + \sqrt{3}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Arc ACAC subtends a right angle at the center, so ∠AOC=90∘\angle AOC = 90^\circ. Also,

length of arc ABlength of arc BC=15\frac{\text{length of arc AB}}{\text{length of arc BC}} = \frac{1}{5}

and

OC→=αOA→+βOB→\overrightarrow{OC} = \alpha \overrightarrow{OA} + \beta \overrightarrow{OB}

Find: The value asked in the options.

Since arc lengths are proportional to the corresponding central angles,

∠AOB:∠BOC=1:5\angle AOB : \angle BOC = 1:5

With

∠AOC=90∘\angle AOC = 90^\circ

we get

∠AOB=16×90∘=15∘,∠BOC=75∘\angle AOB = \frac{1}{6}\times 90^\circ = 15^\circ, \qquad \angle BOC = 75^\circ

Coordinate Vector Method

Take the circle to be of unit radius and choose coordinates so that

OA→=(1,0),OB→=(cos⁡15∘,sin⁡15∘),OC→=(0,1)\overrightarrow{OA} = (1,0), \qquad \overrightarrow{OB} = (\cos 15^\circ, \sin 15^\circ), \qquad \overrightarrow{OC} = (0,1)

Then from

OC→=αOA→+βOB→\overrightarrow{OC} = \alpha \overrightarrow{OA} + \beta \overrightarrow{OB}

we obtain

(0,1)=α(1,0)+β(cos⁡15∘,sin⁡15∘)(0,1) = \alpha (1,0) + \beta (\cos 15^\circ, \sin 15^\circ)

Common mistakes

  • Using chord ratio instead of arc-length ratio is incorrect because the question compares arc AB and arc BC, not the straight line segments. Convert the arc ratio directly into the central angle ratio.

  • Taking BB to divide the 90∘90^\circ angle in the ratio 1:51:5 but then assigning the larger angle to ∠AOB\angle AOB is wrong. Since arc ABarc BC=15\frac{\text{arc }AB}{\text{arc }BC}=\frac{1}{5}, the smaller angle is 15∘15^\circ and the larger one is 75∘75^\circ.

  • Equating vector coefficients without resolving components is incorrect because OA→\overrightarrow{OA} and OB→\overrightarrow{OB} are not perpendicular basis vectors. Write the vectors in coordinates or compare components carefully.

Practice more Basics of Vectors questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Basics of Vectors

Related questions