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Friday, 9 October

JEE Main 2024 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 9 April 2024, Shift 1, Maths Q17

Question 17 of 90 in this shift, Maths question 17 of 30, Section A.

Let OA→=2a⃗\overrightarrow{OA}=2\vec{a}, OB→=6a⃗+5b⃗\overrightarrow{OB}=6\vec{a}+5\vec{b} and OC→=3b⃗\overrightarrow{OC}=3\vec{b}, where O is the origin. If the area of the parallelogram with adjacent sides OA→\overrightarrow{OA} and OC→\overrightarrow{OC} is 1515 sq. units, then the area (in sq. units) of the quadrilateral OABC is equal to :
  1. (1)3232
  2. (2)3535Official answer
  3. (3)3838
  4. (4)4040

Official answer

Option 2

NTA final key.

Chapter
Vector Algebra
Topic
Cross Product
Idea tested
Area of Quadrilateral via Diagonals

Same idea in other shifts

Asked 3× in all
  1. 9 Apr 2024, Shift 2 · Q18Let a⃗=2i^+αj^+k^, b⃗=−i^+k^, c⃗=βj^−k^\vec{a}=2\hat{i}+\alpha\hat{j}+\hat{k},\ \vec{b}=-\hat{i}+\hat{k},\ \vec{c}=\beta\hat{j}-\hat{k}, where α\alpha and β\beta are…MediumSingle correct
  2. 8 Apr 2023, Shift 2 · Q15The area of the quadrilateral ABCDABCD with vertices A(2,1,1)A(2, 1, 1), B(1,2,5)B(1, 2, 5), C(−2,−3,5)C(-2, -3, 5) and D(1,−6,−7)D(1, -6, -7) is equal toMediumSingle correct

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.