# 13.2: Triangles

- Page ID
- 6869

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

\[\textbf{r}_2 = a_1 \textbf{r}_1 + a_3 \textbf{r}_3 . \label{13.2.1} \tag{13.2.1}\]

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 \(\textbf{r}_1 \times\):

\[\textbf{r}_1 \times \textbf{r}_2 = a_3 \textbf{r}_1 \times \textbf{r}_3 . \label{13.2.2} \tag{13.2.2}\]

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.)

\[\therefore a_3 = A_3/A_2 . \label{13.2.3} \tag{13.2.3}\]

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

\[a_1 = A_1/ A_2 . \label{13.2.4} \tag{13.2.4}\]

Hence we find that

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