Change in area of a triangle formed by position vectors

[MissMCHP]

Homework Statement

Consider two particles located at positions $$\vec{r_{1}}(t)$$ and $$\vec{r_{2}}(t)$$, respectively, with corresponding velocities given respectively by $$\vec{v_{1}}(t)$$ and $$\vec{v_{2}}(t)$$. Consider the triangle formed by the origin and the position vectors of the two particles. Show that the rate of change of the area A(t) of this triangle is given by the expression
$$\frac{dA}{dt} = \frac{\left|r_{2}\right|^{2}(v_{1}\bullet r_{1})+\left|r_{1}\right|^{2}(v_{2}\bullet r_{2})-(r_{2}\bullet r_{1})(v_{1}\bullet r_{2}+v_{2}\bullet r_{1})}{2\left|r_{2}\times r_{1}\right|}$$

Homework Equations

$$\left|a\times b\right|=|a||b|sin\theta$$ maybe?

The Attempt at a Solution

Originally, I thought I might get the formula for the area of the triangle, then simply differentiate it in respect to t. However, I only got as far as $$A=|r_{1}\times r_{2}|/2$$before realizing I don't know how to differentiate the magnitude of a cross product. However, I can't think of any other way to express the area of the triangle in terms of the two vectors...

Any hints would be greatly appreciated.

 

[lanedance]

before jumping into the cross product, trying differntiating a magnitude of a vector may help:

magnitude of a vector
$$|\textbf{u}|= (\textbf{u} \bullet \textbf{u})^{\frac{1}{2}}$$

differentiating
$$\frac{d}{dt}|\textbf{u}|<br /> = \frac{d}{dt}|(\textbf{u} \bullet \textbf{u})^{\frac{1}{2}} <br /> = \frac{ \frac{d}{dt}(\textbf{u} \bullet \textbf{u})}{2|\textbf{u}|} <br /> = \frac{ (\textbf{u}' \bullet \textbf{u}+\textbf{u} \bullet \textbf{u}')}{2|\textbf{u}|} <br /> = \frac{ \textbf{u}' \bullet \textbf{u} }{|\textbf{u}|} <br /> = \textbf{u}' \bullet \widehat{\textbf{u}}$$