Q.Find the area of a triangle having the points $A(1, 1, 1)$, $B(1, 2, 3)$ and $C(2, 3, 1)$ as its vertices.

Key idea: the area of a triangle equals half the magnitude of the cross product of two sides from a common vertex.

Step 1 — Side vectors from $A$.

Step 2 — Cross product.

$\vec{AB}\times\vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 2 \\ 1 & 2 & 0 \end{vmatrix} = \hat{i}(0 - 4) - \hat{j}(0 - 2) + \hat{k}(0 - 1) = -4\hat{i} + 2\hat{j} - \hat{k}.$ …

The magnitude $|\vec{u}\times\vec{v}|$ equals the area of the parallelogram spanned by $\vec{u}$ and $\vec{v}$. A triangle is exactly half of that parallelogram, so its area is $\tfrac{1}{2}|\vec{u}\times\vec{v}|$ where $\vec{u}$ and $\vec{v}$ are two sides sharing a vertex.

1. Choose two sides from the same vertex

Take $A(1,1,1)$ as the common vertex:

2. Cross the two side vectors

Expanding along the top row:

• $\hat{i}$: $1\cdot 0 - 2\cdot 2 = -4$
• $\hat{j}$: $-(0\cdot 0 - 2\cdot 1) = -(-2) = 2$
• $\hat{k}$: $0\cdot 2 - 1\cdot 1 = -1$

So

3. Magnitude, then halve it …

Method: Area of a triangle from its vertices using vectors

The magnitude of a cross product is a parallelogram's area, and a triangle is half of it. So build two side-vectors from one vertex, cross them, and halve.

Steps

Step 1: Make two side vectors from a common vertex

Pick any vertex (say $A$) and form $\overrightarrow{AB} = B - A$ and $\overrightarrow{AC} = C - A$ by subtracting coordinates. Both vectors must start at the same vertex.

Step 2: Cross the two side vectors

$\overrightarrow{AB}\times\overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ (AB)_1 & (AB)_2 & (AB)_3 \\ (AC)_1 & (AC)_2 & (AC)_3 \end{vmatrix}.$ …

Common Mistakes

Mistake 1: Forgetting the factor $\tfrac{1}{2}$.

Why it's wrong: $|\overrightarrow{AB}\times\overrightarrow{AC}|$ is the area of the parallelogram on those sides; the triangle is only half of it. Correct approach: always multiply the cross-product magnitude by $\tfrac{1}{2}$ for a triangle.

Mistake 2: Using two sides that do not share a vertex.

Why it's wrong: vectors like $\overrightarrow{AB}$ and $\overrightarrow{BC}$ (or a mix that doesn't start from one common point) do not span the triangle correctly and give a wrong area. Correct approach: take both side vectors from the same vertex, e.g. $\overrightarrow{AB}$ and $\overrightarrow{AC}$. …