Let OA=a, OB=12a+4b OC=b, where O is the origin.

Question

Let $\overset{	o \leftarrow }{OA}=\vec{a},\overset{	o \leftarrow }{OB}=12\vec{a}+4\vec{b}\overset{	o \leftarrow }{OC}=\vec{b}$, where O is the origin. If S is the parallelogram with adjacent sides OA and OC, then $\frac{area\, of\, the\, quadrilateral\, OA\, BC}{area\, of\, S}$ is equal to \_\_\_\_\_\_\_\_\_.

Accepted Answer

Area of parallelogram, $S=|\vec{a}	imes \vec{b}|$

Area of quadrilateral $=Area(	riangleOAB)+Area(	riangleOBC)$

$\begin{gathered} =\frac{1}{2}{|\vec{a}	imes (12\vec{a}+4\vec{b})|+|\vec{b}	imes (12\vec{a}+4\vec{b})|} \ =8|(\vec{a}	imes \vec{b})| \end{gathered}$

$=\frac{8|(\vec{a}	imes \vec{b})|}{|(\vec{a}	imes \vec{b})|}=8$

Solutions · 1