Proving the Equivalence of the Algebraic and Geometric Forms of the Determinant in 2D

Question

How can we prove that the following two formulas are equivalent:

$$||u||\cdot||v||\sin(\theta) = u_1​v_2​-u_2​v_1​$$

Answer 1

Let's write the vectors in polar form as $u = (r\cos(\alpha),r\sin(\alpha))$ and $v = (s\cos(\beta),s\sin(\beta))$. Then $||u||=r$, $||v||=s$, and the angle between the vectors is $\beta-\alpha$.

We have \begin{align*} u_1v_2 - u_2v_1 &= r\cos(\alpha)s\sin(\beta)-r\sin(\alpha)s\cos(\beta)\\ &=rs(\cos(\alpha)\sin(\beta)-\sin(\alpha)\cos(\beta)) \\ &= ||u||\cdot||v||\sin(\beta-\alpha)\\ &= ||u||\cdot||v||\sin(\theta) \end{align*}