$$\require{cancel}$$

# 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

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