# Area of Triangle via Vectors

I don't understand the formula given by a book that

$$\text{Area of }\triangle ABC = \frac{1}{2}|CA \times CB|$$

How is this derived?

The explanation given by the book was:

$$\text{Area of } \triangle ABC = \frac{1}{2}ab \sin{C} = \frac{1}{2}|CA||CB| \sin{C}$$

I think C refers to $\angle ACB$

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 vector product – pedja Nov 8 '11 at 14:42

## 2 Answers

One can show (through various means) that $|{\bf a} \times {\bf b}|^2 = ({\bf a} \times {\bf b}) {\bf\cdot} ({\bf a} \times {\bf b}) = |{\bf a}|^2|{\bf b}|^2 - ({\bf a}{\bf\cdot} {\bf b})^2$.

Keeping in mind that ${\bf a}{\bf\cdot} {\bf b} = |{\bf a}||{\bf b}|\cos(\theta)$ where $\theta$ is the angle between ${\bf a }$ and ${\bf b}$, we have that $|{\bf a}|^2|{\bf b}|^2 - |{\bf a}|^2|{\bf b}|^2\cos^2(\theta) = |{\bf a}|^2|{\bf b}|^2(1-\cos^2(\theta))$ $= |{\bf a}|^2|{\bf b}|^2\sin^2(\theta)$. Taking square roots and keeping in mind that $0 \leq \theta \leq \pi$ so that $\sin(\theta)\geq 0$, we have $|{\bf a} \times {\bf b}| = |{\bf a}||{\bf b}|\sin(\theta)$.

Looking at the (poorly drawn) picture, notice that the orange triangle's opposite side has length $|{\bf b}|\sin(\theta)$. Thus the area of the parallelogram is base $\times$ height = $|{\bf a}||{\bf b}|\sin(\theta)=|{\bf a} \times {\bf b}|$.

To find the area of the triangle (in red) we simply need to chop the parallelogram in half. Thus $\frac{1}{2}|{\bf a} \times {\bf b}|$ gives the area of the parallelogram.

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Let's observe picture below :

$h= |\vec{CB}|\sin C\Rightarrow A=\frac{1}{2} |\vec{CA}| |\vec{CB}|\sin C=\frac{1}{2}|CA \times CB|$

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