Animated Solution for Mathematics - Vector Algebra: If $A(1,-1,2)$, $B(5,7,-6)$, $C(3,4,-10)$ and $D(-1,-4,-2)$ are the vertices of a quadrilateral $ABCD$, then its area is :

Imagine you are standing in a vast, three-dimensional void. Before you, four points—$A(1,-1,2)$, $B(5,7,-6)$, $C(3,4,-10)$, and $D(-1,-4,-2)$—hang in the air, marking the corners of a quadrilateral.

To the untrained eye, this is just a collection of coordinates. But to you, the physicist and mathematician, this is a challenge of spatial reasoning. How do we measure the area of a shape that isn't sitting flat on a piece of paper? The answer lies in the beautiful, rigid structure of vectors.

When we deal with a quadrilateral in 3D space, we don't need to worry about the complexity of its surface. We can rely on its diagonals.

The area of any quadrilateral is given by the formula:

Think of this as the vector equivalent of the area of a triangle, but scaled up. The cross product of the two diagonals, $\overrightarrow{AC}$ and $\overrightarrow{BD}$, creates a vector whose magnitude is twice the area of the quadrilateral. It is a profound result that turns a potentially messy geometry problem into a clean, algebraic one.

Our first task is to build the diagonal vectors. Remember, a vector is a directed path. To find $\overrightarrow{AC}$, we travel from $A$ to $C$ by subtracting the coordinates of $A$ from $C$:

Now, we do the same for $\overrightarrow{BD}$, traveling from $B$ to $D$:

Take a moment to appreciate the precision here. One small sign error, and the entire calculation collapses. Always double-check your subtraction: $4-(-1)$ is $5$, not $3$. Precision is the soul of physics.

With our diagonals in hand, we enter the heart of the problem: the cross product. We set up our determinant, the gatekeeper of vector orientation:

Expanding this along the first row is where the magic happens. For the $\hat{i}$ component, we have $(5)(4)-(-11)(-12)=20-132=-112$.

For the $\hat{j}$ component, remember the negative sign: $-((2)(4)-(-6)(-12))=-(8-72)=64$. Finally, for the $\hat{k}$ component, we have $(2)(-11)-(-6)(5)=-22+30=8$.

Our resulting vector is $-112\hat{i}+64\hat{j}+8\hat{k}$.

We are almost there. We need the magnitude of this vector. Instead of squaring these large numbers, let's use a pro tip: factor out the common term. Notice that $8$ divides all components:

Now, the magnitude is simply:

Since $261=9\times 29$, we have $8\times 3\sqrt{29}=24\sqrt{29}$. Finally, we apply our area formula:

There it is. The complexity of 3D space, tamed by the elegance of vectors. You didn't just solve a problem; you navigated a geometric landscape. Keep this clarity, and no problem will ever be too daunting. The final answer is $12\sqrt{29}$.