Using vectors, find the area of the triangle ABC with vertices A(1, 2, 3), B(2, -1, 4) and C(4, 5, -1).

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A(1, 2, 3)

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Concept understanding — Cross Product Area

Area from the Cross Product

The cross product $\vec{a}\times\vec{b}$ of two vectors in 3D is itself a vector, and the most useful thing about its magnitude is that it measures area.

Place the two vectors tail-to-tail. They span a parallelogram. The magnitude of their cross product is exactly the area of that parallelogram:

\text{Area of parallelogram} = |\vec{a}\times\vec{b}| = |\vec{a}|\,|\vec{b}|\,\sin\theta

where $\theta$ is the angle between them.

Why sine, not cosine

The area of a parallelogram is base $\times$ height. Take $|\vec{a}|$ as the base. The height is the part of $\vec{b}$ perpendicular to $\vec{a}$ , namely $|\vec{b}|\sin\theta$ . Multiplying gives $|\vec{a}||\vec{b}|\sin\theta$ — precisely $|\vec{a}\times\vec{b}|$ . The dot product uses $\cos\theta$ (overlap along); the cross product uses $\sin\theta$ (spread across), and "across" is what builds area.

Area of a triangle

A triangle with adjacent sides $\vec{a}$ and $\vec{b}$ is half that parallelogram:

\text{Area of triangle} = \tfrac{1}{2}\,|\vec{a}\times\vec{b}|

For a triangle with vertices $A, B, C$ , take $\vec{a}=\overrightarrow{AB}$ and $\vec{b}=\overrightarrow{AC}$ .

A quick example

For $\vec{a}=\hat{i}+2\hat{j}$ and $\vec{b}=3\hat{i}+\hat{j}$ ,

$\vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&2&0\\3&1&0\end{vmatrix} = (1\cdot 1 - 2\cdot 3)\hat{k} = -5\hat{k}.$ …

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The area of triangle ABC is half the magnitude of the cross product of vectors AB and AC. Using the coordinates, the area is $\frac{\sqrt{274}}{2}$ square units.

The key insight: the area of a triangle formed by three points in space is exactly half the area of the parallelogram spanned by two of its sides. That parallelogram's area is given by the magnitude of the cross product of the two side vectors. So we never need to find heights or angles — just compute two vectors, take their cross product, and halve the magnitude.

Let's work through it.

Choose two sides from the common vertex A. We take vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ :

\overrightarrow{AB} = B - A = (2-1,\ -1-2,\ 4-3) = (1,\ -3,\ 1)

\overrightarrow{AC} = C - A = (4-1,\ 5-2,\ -1-3) = (3,\ 3,\ -4)

Compute the cross product $\overrightarrow{AB} \times \overrightarrow{AC}$ . Using the determinant formula:

\overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -3 & 1 \\ 3 & 3 & -4 \end{vmatrix}

Expand:

= \mathbf{i}\big((-3)(-4) - (1)(3)\big) - \mathbf{j}\big((1)(-4) - (1)(3)\big) + \mathbf{k}\big((1)(3) - (-3)(3)\big)

= \mathbf{i}(12 - 3) - \mathbf{j}(-4 - 3) + \mathbf{k}(3 + 9)

= \mathbf{i}(9) - \mathbf{j}(-7) + \mathbf{k}(12)

$= (9,\ 7,\ 12)$ …

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Q.Using vectors, find the area of the triangle ABC with vertices $A(1, 2, 3)$ , $B(2, -1, 4)$ and $C(4, 5, -1)$ .

Concept understanding — Cross Product Area

Area from the Cross Product

Why sine, not cosine

Area of a triangle

A quick example

Method: Area of a triangle from vertices using the cross product

Steps

Common Mistakes

Q.Using vectors, find the area of the triangle ABC with vertices A(1,2,3)A(1, 2, 3)A(1,2,3), B(2,−1,4)B(2, -1, 4)B(2,−1,4) and C(4,5,−1)C(4, 5, -1)C(4,5,−1).

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Q.Using vectors, find the area of the triangle ABC with vertices $A(1, 2, 3)$ , $B(2, -1, 4)$ and $C(4, 5, -1)$ .

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