13.2: Triangles

I shall start with a geometric theorem involving triangles, which will be useful as we progress towards our aim of computing orbital elements.

$\text{FIGURE XIII.1}$

Figure $\text{XIII.1}$ shows three coplanar vectors. It is clearly possible to express $\textbf{r}_2$ as a linear combination of the other two. That is to say, it should be possible to find coefficients such that

$$\textbf{r}_2 = a_1 \textbf{r}_1 + a_3 \textbf{r}_3 . \label{13.2.1}$$

The notation I am going to use is as follows:

• The area of the triangle formed by joining the tips of $\textbf{r}_2$ and $\textbf{r}_3$ is $A_1$.
• The area of the triangle formed by joining the tips of $\textbf{r}_3$ and $\textbf{r}_1$ is $A_2$.
• The area of the triangle formed by joining the tips of $\textbf{r}_1$ and $\textbf{r}_2$ is $A_3$.

To find the coefficients in Equation $\ref{13.2.1}$, multiply both sides by $\textbf{r}_1 \times$:

$$\textbf{r}_1 \times \textbf{r}_2 = a_3 \textbf{r}_1 \times \textbf{r}_3 . \label{13.2.2}$$

The two vector products are parallel vectors (they are each perpendicular to the plane of the paper), of magnitudes $2A_3$ and $2A_2$ respectively. ($2A_3$ is the area of the parallelogram of which the vectors $\textbf{r}_1$ and $\textbf{r}_2$ form two sides.)

$$\therefore a_3 = A_3/A_2 . \label{13.2.3}$$

Similarly by multiplying both sides of Equation $\ref{13.2.1}$ by $\textbf{r}_3 \times$ it will be found that

$$a_1 = A_1/ A_2 . \label{13.2.4}$$

Hence we find that

$$A_2 \textbf{r}_2 = A_1 \textbf{r}_1 + A_3 \textbf{r}_3 . \label{13.2.5}$$