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

Question

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

Accepted Answer

$\begin{gathered} 6|\vec{a}	imes \vec{b}|=15 \ \Rightarrow |\vec{a}	imes \vec{b}|=\frac{5}{2} \end{gathered}$

Area of quadrilateral $OABC$

$=$ area of $	riangleOAC+$ area of $	riangleABC$

$\begin{gathered} =\frac{15}{2}+\frac{1}{2}|(\overset{	o \leftarrow }{AB}	imes \overset{	o \leftarrow }{BC})| \ =\frac{15}{2}+\frac{1}{2}|(4\vec{a}+5\vec{b})	imes (6\vec{a}+2\vec{b})| \ =\frac{15}{2}+\frac{1}{2}|(22\vec{a}	imes \vec{b})| \ =\frac{15}{2}+11	imes \frac{5}{2} \ =\frac{70}{2}=35 \end{gathered}$

Solutions · 1