Q.The area of the quadrilateral $ABCD$, where $A(0, 4, 1)$, $B(2, 3, -1)$, $C(4, 5, 0)$ and $D(2, 6, 2)$, is equal to
(A) $9$ sq. units
(B) $18$ sq. units
(C) $27$ sq. units
(D) $81$ sq. units

The key idea is that quadrilateral $ABCD$ is a parallelogram (opposite sides are equal and parallel), so its area equals the magnitude of the cross product of two adjacent vectors.

Step 1: Find vectors for two adjacent sides.

$\overrightarrow{AB} = (2-0, 3-4, -1-1) = (2, -1, -2)$

$\overrightarrow{AD} = (2-0, 6-4, 2-1) = (2, 2, 1)$

Step 2: Compute the cross product $\overrightarrow{AB} \times \overrightarrow{AD}$:

$\overrightarrow{AB} \times \overrightarrow{AD} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & -2 \\ 2 & 2 & 1 \end{vmatrix} = \mathbf{i}((-1)(1) - (-2)(2)) - \mathbf{j}((2)(1) - (-2)(2)) + \mathbf{k}((2)(2) - (-1)(2))$ …

The key insight: when you have four points in 3D, the quadrilateral might not be planar, but here it is — and more importantly, it’s a parallelogram. That means we don’t need to split it into two triangles; we can use the cross product directly.

Why cross product? The magnitude of the cross product of two vectors gives the area of the parallelogram they span. If the quadrilateral is a parallelogram, its area is simply $|\vec{AB} \times \vec{AD}|$ (or any adjacent pair).

Let’s verify the shape first.

1. Find the side vectors

   $\vec{AB} = B - A = (2-0, 3-4, -1-1) = (2, -1, -2)$

   $\vec{BC} = C - B = (4-2, 5-3, 0-(-1)) = (2, 2, 1)$

   $\vec{CD} = D - C = (2-4, 6-5, 2-0) = (-2, 1, 2)$

   $\vec{DA} = A - D = (0-2, 4-6, 1-2) = (-2, -2, -1)$

   Notice: $\vec{CD} = -\vec{AB}$ and $\vec{DA} = -\vec{BC}$. That’s the hallmark of a parallelogram — opposite sides are parallel and equal in length.

2. Pick two adjacent sides — say $\vec{AB}$ and $\vec{AD}$.

   $\vec{AD} = D - A = (2-0, 6-4, 2-1) = (2, 2, 1)$

   (Check: this is exactly $\vec{BC}$, as expected.)

3. Compute the cross product

   $\vec{AB} \times \vec{AD} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & -2 \\ 2 & 2 & 1 \end{vmatrix}$

   Expand:

   $\mathbf{i}[(-1)(1) - (-2)(2)] - \mathbf{j}[(2)(1) - (-2)(2)] + \mathbf{k}[(2)(2) - (-1)(2)]$

   $= \mathbf{i}[-1 + 4] - \mathbf{j}[2 + 4] + \mathbf{k}[4 + 2]$ …

Method: Area of a Triangle or Parallelogram from the Cross Product

The magnitude of a cross product measures the area two vectors span — the tool for area problems given vertices in space.

Steps

Step 1: Build vectors for two adjacent sides from a common vertex.

From vertex $A$, form $\overrightarrow{AB}$ and $\overrightarrow{AD}$ by subtracting coordinates. Use adjacent sides sharing a corner — never a diagonal.

Step 2 (for a quadrilateral): confirm the shape first.

If $\overrightarrow{AB}=\overrightarrow{DC}$ (opposite sides equal and parallel) it is a parallelogram and one cross product suffices; otherwise split it into two triangles.

Step 3: Take the cross product and its magnitude. …

Common Mistakes

Mistake 1: Using a diagonal instead of two adjacent sides.

Why it's wrong: the cross product gives the area its two vectors span, so vectors like $\overrightarrow{AB}$ and $\overrightarrow{AC}$ (a diagonal) don't span the parallelogram's sides. Correct approach: use two adjacent side vectors from one vertex, e.g. $\overrightarrow{AB}$ and $\overrightarrow{AD}$.

Mistake 2: Halving the parallelogram's area. …