Verified Solution • 2025–26 Session moderate

QJEEM24Madd92: JEE Main 2024 Physics - Class 11 JEEM24Madd92 Solution: Let and , where is the origin. If the area of the parallelogram …

Let OA→=2a⃗,OB→=6a⃗+5b⃗\overrightarrow{O A}=2 \vec{a}, \overrightarrow{O B}=6 \vec{a}+5 \vec{b} and OC→=3b⃗\overrightarrow{O C}=3 \vec{b} , where OO is the origin. If the area of the parallelogram with adjacent sides OA→\overrightarrow{O A} and OC→\overrightarrow{O C} is 15 sq. units, then the area (in sq. units) of the quadrilateral OABCO A B C is equal to:

Multiple Choice Options

  • A 32
  • B 38
  • C 35
  • D 40

Step-by-Step Verified Solution

amp;6∣a⃗×b⃗∣=15amp;⇒∣a⃗×b⃗∣=52\begin{aligned}& 6|\vec{a} \times \vec{b}|=15 \\& \Rightarrow|\vec{a} \times \vec{b}|=\frac{5}{2}\end{aligned}

JEE Main 2024 (Online) 9th April Morning Shift Mathematics - Vector Algebra Question 30 English Explanation 1

Area of quadrilateral OABCO A B C

== area of △OAC+\triangle O A C+ area of △ABC\triangle A B C

JEE Main 2024 (Online) 9th April Morning Shift Mathematics - Vector Algebra Question 30 English Explanation 2

amp;=152+12∣(AB→×BC→)∣amp;=152+12∣(4a⃗+5b⃗)×(6a⃗+2b⃗)∣amp;=152+12∣(22a⃗×b⃗)∣amp;=152+11×52amp;=702=35\begin{aligned}& =\frac{15}{2}+\frac{1}{2}|(\overrightarrow{A B} \times \overrightarrow{B C})| \\& =\frac{15}{2}+\frac{1}{2}|(4 \vec{a}+5 \vec{b}) \times(6 \vec{a}+2 \vec{b})| \\& =\frac{15}{2}+\frac{1}{2}|(22 \vec{a} \times \vec{b})| \\& =\frac{15}{2}+11 \times \frac{5}{2} \\& =\frac{70}{2}=35\end{aligned}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring