Let vector {OA} = 2 vector a, vector {OB} = 6 vector a + 5 vector b and vector {OC} = 3 vector b, where O is the origin.

Question

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

(1) 38

(2) 40

(3) 32

(4) 35

Answer 1

Correct option is : (4) 35 

Area of parallelogram having sides

\(\overrightarrow{\mathrm{OA}} \ \&\ \overrightarrow{\mathrm{OC}}=|\overrightarrow{\mathrm{OA}} \times \overrightarrow{\mathrm{OC}}|=|2 \overrightarrow{\mathrm{a}} \times 3 \overrightarrow{\mathrm{b}}|=15\)

\(6|\vec{a} \times \vec{b}|=15\)

\(\Rightarrow|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=\frac{5}{2}\) ........(1)

Area of quadrilateral

\(\mathrm{OABC}=\frac{1}{2}\left|\overrightarrow{\mathrm{d}}_{1} \times \overrightarrow{\mathrm{d}}_{2}\right|\)

\(=\frac{1}{2}|\overrightarrow{\mathrm{AC}} \times \overrightarrow{\mathrm{OB}}|=\frac{1}{2}|(3 \overrightarrow{\mathrm{b}}-2 \overrightarrow{\mathrm{a}}) \times(6 \overrightarrow{\mathrm{a}}+5 \overrightarrow{\mathrm{b}})|\)

\(=\frac{1}{2}|18 \vec{b} \times \vec{a}-10 \vec{a} \times \vec{b}|=14|\vec{a} \times \vec{b}|\)

\(=14 \times \frac{5}{2}=35\)