Exam

JEE Main

Year

2024

Date

9 April

Shift

Morning Shift

Subject

Mathematics

Chapter

Vector Algebra

JEE Main 2024 · 9 April · Morning Shift · Mathematics · Vector Algebra

Let O A =2 a , O B =6 a +5 b and O C =3 b , where O is the origin....

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

Step-by-step solution

Area of parallelogram having sides $$\begin{aligned} & \overrightarrow{\mathrm{OA}} & \overrightarrow{\mathrm{OC}}=|\overrightarrow{\mathrm{OA}} \times \overrightarrow{\mathrm{OC}}|=|2 \overrightarrow{\mathrm{a}} \times 3 \overrightarrow{\mathrm{b}}|=15 \\ & 6|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=15 \\ & \Rightarrow|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=\frac{5}{2} \ldots \ldots . .(1) \end{aligned}$$
Area of quadrilateral \begin{aligned} & \mathrm{OABC}=\frac{1}{2}\left|\overrightarrow{\mathrm{d}_1 \times \overrightarrow{\mathrm{d}_2\right| \\ & =\frac{1}{2}|\overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{OB}}|=\frac{1}{2}|(3 \overrightarrow{\mathrm{b}}-2 \overrightarrow{\mathrm{a}}) \times(6 \overrightarrow{\mathrm{a}}+5 \overrightarrow{\mathrm{b}})| \\ & =\frac{1}{2}|18 \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}-10 \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=14|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}| \\ & =14 \times \frac{5}{2}=35 \end{aligned}