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 $\angle ACB$
$\endgroup$ 12 Answers
$\begingroup$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 triangle.
$\endgroup$ $\begingroup$Let's observe picture below :
$h= |\vec{CB}|\sin C\Rightarrow A=\frac{1}{2} |\vec{CA}| |\vec{CB}|\sin C=\frac{1}{2}|CA \times CB|$
