JEE Main 2025 · MathsMultiple choiceSingle correctMediumCalculation

JEE Main 7 April 2025, Shift 1, Maths Q15

Question 15 of 75 in this shift, Maths question 15 of 25, Section A.

Let the angle

\theta

0 < \theta < \frac{\pi}{2}

between two unit vectors

\hat{a}

and

\hat{b}

\sin^{-1}\left(\frac{\sqrt{65}}{9}\right)

. If the vector

\vec{c} = 3\hat{a} + 6\hat{b} + 9(\hat{a} \times \hat{b})

, then the value of

9(\vec{c} \cdot \hat{a}) - 3(\vec{c} \cdot \hat{b})

(1) $24$
(2) $27$
(3) $29$ Official answer
(4) $31$

Official answer

Option 3

NTA final key.

Chapter: Vector Algebra
Topic: Cross Product
Idea tested: Orthogonality Dot Product Trap

Same topic in other shifts

All Cross Product questions

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Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.