Animated Solution for Mathematics - Vector Algebra: The area of the quadrilateral $ABCD$ with vertices $A(2,1,1)$, $B(1,2,5)$, $C(-2,-3,5)$ and $D(1,-6,-7)$ is equal to

When we look at the vertices $A(2,1,1)$, $B(1,2,5)$, $C(-2,-3,5)$, and $D(1,-6,-7)$, we are looking at points suspended in the void. It is easy to feel overwhelmed by the spatial complexity, but remember: every complex problem is just a collection of simple ones waiting to be organized.

The first instinct might be to look for a single, magical formula for the area of a quadrilateral in 3D. But here is the secret: we don't need one. We use the principle of Divide and Conquer.

By drawing a diagonal from $B$ to $D$, we slice this quadrilateral into two manageable triangles: $\Delta ABD$ and $\Delta CBD$. The total area is simply the sum of these two parts:

This is the beauty of vector geometry—it allows us to break down complex shapes into fundamental building blocks.

How do we find the area of a triangle in 3D? We use the cross product. For any triangle with adjacent sides represented by vectors $\vec{u}$ and $\vec{v}$, the area is given by:

This formula is powerful because it does not care about the orientation of the triangle in space; it only cares about the vectors that define it.

Triangle ABD

Let us tackle $\Delta ABD$ first. We define our vectors from vertex $A$:

Now, we compute the cross product $\overrightarrow{AB}\times \overrightarrow{AD}$ using the determinant method. Expanding this, we get $20\hat{i}-12\hat{j}+8\hat{k}$.

The magnitude is:

Thus, $\text{Area}(\Delta ABD)=\frac{1}{2}\times 4\sqrt{38}=2\sqrt{38}$.

Triangle CBD

Now, we repeat this for $\Delta CBD$. Using vertex $C$, we find:

The cross product $\overrightarrow{CB}\times \overrightarrow{CD}$ yields $-60\hat{i}+36\hat{j}-24\hat{k}$. The magnitude is:

Therefore, $\text{Area}(\Delta CBD)=\frac{1}{2}\times 12\sqrt{38}=6\sqrt{38}$.

We have arrived at the finish line. We have the area of the first triangle, $2\sqrt{38}$, and the second, $6\sqrt{38}$. Adding them together, we get:

The final area of the quadrilateral is $8\sqrt{38}$. It is elegant, it is precise, and it is exactly what we set out to find.