JEE MAIN - Mathematics (2024 - 9th April Morning Shift - No. 20)
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 $$\overrightarrow{O A}$$ and $$\overrightarrow{O C}$$ is 15 sq. units, then the area (in sq. units) of the quadrilateral $$O A B C$$ is equal to:
32
38
35
40
Explanation
$$\begin{aligned} & 6|\vec{a} \times \vec{b}|=15 \\ & \Rightarrow|\vec{a} \times \vec{b}|=\frac{5}{2} \end{aligned}$$
_9th_April_Morning_Shift_en_20_1.png)
Area of quadrilateral $$O A B C$$
$$=$$ area of $$\triangle O A C+$$ area of $$\triangle A B C$$
_9th_April_Morning_Shift_en_20_2.png)
$$\begin{aligned} & =\frac{15}{2}+\frac{1}{2}|(\overrightarrow{A B} \times \overrightarrow{B C})| \\ & =\frac{15}{2}+\frac{1}{2}|(4 \vec{a}+5 \vec{b}) \times(6 \vec{a}+2 \vec{b})| \\ & =\frac{15}{2}+\frac{1}{2}|(22 \vec{a} \times \vec{b})| \\ & =\frac{15}{2}+11 \times \frac{5}{2} \\ & =\frac{70}{2}=35 \end{aligned}$$
Comments (0)
