Use a cross-product magnitude for triangle area

Use a cross-product magnitude for triangle area

Choose two sides from the same vertex and calculate their cross product. Take its magnitude for a parallelogram area, then halve it for a triangle. Reversing the factor order changes orientation but leaves the area unchanged.

For a nondegenerate triangle $ABC$, choose two side displacements with the same starting vertex. With $\theta=\angle BAC$, the complete source proof is

$$\begin{aligned} \operatorname{Area}(ABC)&=\frac12(AB)(AC)\sin\theta\\ &=\frac12|\overrightarrow{AB}||\overrightarrow{AC}|\sin\theta\\ &=\frac12|\overrightarrow{AB}||\overrightarrow{AC}|\sin\theta|\widehat{\mathbf n}|\\ &=\frac12|\overrightarrow{AB}\times\overrightarrow{AC}|. \end{aligned}$$

Here $\widehat{\mathbf n}$ is a unit normal to the two side directions. The result is a nonnegative scalar area, not the cross-product vector. The source also gives the equivalent choices

$$\operatorname{Area}(ABC)=\frac12|\overrightarrow{BA}\times\overrightarrow{BC}|=\frac12|\overrightarrow{CA}\times\overrightarrow{CB}|.$$

Switching vertices changes the two displacements together; swapping their order changes the normal sign but not its magnitude. The formula also yields zero for a degenerate triangle, without requiring a unit normal there.

Extend from a triangle to a parallelogram

A parallelogram is twice either triangle made by one diagonal. Thus the same two adjacent side vectors give

$$\begin{aligned} \operatorname{Area}(ABCD)&=2\operatorname{Area}(ABD)\\ &=2\left(\frac12|\overrightarrow{AB}\times\overrightarrow{AD}|\right)\\ &=|\overrightarrow{AB}\times\overrightarrow{AD}|. \end{aligned}$$

The length of the cross-product vector is the parallelogram area. If one side is scaled by $t$, the cross-product vector scales by $t$ and its length scales by $|t|$. This explains the Teacher’s suggested comparison between changing vector length and changing area. Example 18 computes both areas from the same cross product; its diagram is schematic.

Form sides from position vectors

For three position vectors, first subtract to form two sides. Example 19 expands $(\mathbf b-\mathbf a)\times(\mathbf c-\mathbf a)$ and obtains

$$\operatorname{Area}(ABC)=\frac12|\mathbf a\times\mathbf b+\mathbf b\times\mathbf c+\mathbf c\times\mathbf a|.$$

Every term inside the norm is a vector: in particular, $\mathbf a\times\mathbf a=\mathbf0$, the zero vector. Reversing the triangle orientation negates the vector sum and leaves its magnitude unchanged. Translating the origin leaves both side differences unchanged, so the area is independent of that choice.

Separate orientation from area

Two sides from the same vertex are $\mathbf u=(2,0,0)$ and $\mathbf v=(1,3,0)$. Find the triangle area. Predict what happens to the cross product and to the area if you swap the sides, then if you replace $\mathbf v$ by $2\mathbf v$. Explain using a base and height as well as a product.

After your attempt: compare the reasoning

Here $\mathbf u\times\mathbf v=(0,0,6)$, so the triangle area is $\tfrac12(6)=3$. The base length is 2 and the perpendicular height is 3, confirming this area. Swapping the sides reverses the cross product to $(0,0,-6)$ but leaves its magnitude and the area unchanged. Doubling the second side gives cross product $(0,0,12)$ and area 6: the height doubles. The cross-product magnitude gives a parallelogram area; a triangle needs the factor $1/2$.