Animated Solution for Mathematics - Vector Algebra: Let $\overrightarrow{OA}=\vec{a}$, $\overrightarrow{OB}=12\vec{a}+4\vec{b}$ and $\overrightarrow{OC}=\vec{b}$, where $O$ is the origin. If $S$ is the parallelogram with adjacent sides $OA$ and $OC$, then $\frac{\text{area of the quadrilateral OABC}}{\text{area of S}}$ is equal to

Welcome, future engineer! Today, we are going to peel back the layers of a classic vector geometry problem. It might look like a simple ratio calculation at first glance, but it is actually a beautiful demonstration of how vectors allow us to manipulate shapes without ever needing to draw them on a Cartesian grid.

We are given two fundamental vectors, $\overrightarrow{OA}=\vec{a}$ and $\overrightarrow{OC}=\vec{b}$, originating from the origin $O$. These two vectors define our universe and form the adjacent sides of a parallelogram $S$.

The area of this parallelogram is a fundamental constant in our problem, defined by the magnitude of their cross product:

Now, we introduce a third vector, $\overrightarrow{OB}=12\vec{a}+4\vec{b}$. This vector defines point $B$ and, consequently, the quadrilateral $OABC$.

Calculating the area of an irregular quadrilateral directly is often a nightmare. But here is the secret: geometry is about finding the right perspective. By drawing a diagonal from $O$ to $B$, we slice this quadrilateral into two manageable triangles: $\Delta OAB$ and $\Delta OBC$.

Let us tackle $\Delta OAB$ first. The area of a triangle formed by vectors $\vec{u}$ and $\vec{v}$ is $\frac{1}{2}|\vec{u}\times \vec{v}|$. For $\Delta OAB$, our sides are $\overrightarrow{OA}$ and $\overrightarrow{OB}$.

Substituting our values, we get:

Applying the distributive property of the cross product, we get $12(\vec{a}\times \vec{a})+4(\vec{a}\times \vec{b})$. Since $\vec{a}\times \vec{a}=0$, the expression simplifies beautifully:

Now, let us do the same for $\Delta OBC$. Its sides are $\overrightarrow{OB}$ and $\overrightarrow{OC}$.

Distributing the cross product, we get $12(\vec{a}\times \vec{b})+4(\vec{b}\times \vec{b})$. Again, $\vec{b}\times \vec{b}$ vanishes, leaving us with:

We have the two pieces of our puzzle. The total area of the quadrilateral $OABC$ is the sum of these two triangles:

Finally, the question asks for the ratio of the area of the quadrilateral to the area of the parallelogram $S$:

The magnitude terms cancel out, leaving us with a clean, satisfying result of 8. This is the power of vector algebra—it turns a complex geometric problem into a simple, elegant cancellation.