Flights as vectors - Where do they intersect? We have a flight description and we want to write this as vectors.
Fligh1 : When we go from $A$ (which is on the origin) to $B$ which is $100$ km east and $30$ km south of $A$ then do we have the vector for this flight $(100,-30)-(0,0)=(100,-30)$ ?
Flight2 : When we go from $C$ to $D$, where $C$ is $60$ km east and $90$ km north to origin and $D$ is $40$ km east and $50$ km south to origin. Then do we have the vector for this flight $(40,-50)-(60,90)=(-20,-140)$ ?
Is this correct to write these vectors in 2D ?
Let's suppose that we flight on height $h$. I want to find the coordinates of the point where the two flight routes intersect.
I thought the following but this is not correct :
Since both flights are on height $h$ we consider the 3D vectors $\vec{F1}=\begin{pmatrix}100 \\ -30 \\ h \end{pmatrix}$ and $\vec{F2}=\begin{pmatrix}-20 \\ -140 \\ h\end{pmatrix}$.
Let $P=\begin{pmatrix}x\\ y\\ h\end{pmatrix}$ be the intersection point.
The vectors of the flights from their starting point to that point are :  $\vec{F1_{AP}}=\begin{pmatrix}x \\ y \\ h \end{pmatrix}-\begin{pmatrix}0 \\ 0  \\ h\end{pmatrix}=\begin{pmatrix}x \\ y \\ 0 \end{pmatrix}$ and $\vec{F2_{CP}}=\begin{pmatrix}x \\ y \\ h  \end{pmatrix}-\begin{pmatrix}65 \\ 89 \\ h\end{pmatrix}=\begin{pmatrix}x-60 \\ y-90 \\ 0 \end{pmatrix}$.
We have that $\vec{F2_{CP}}-\vec{F1_{AP}}=\vec{AC}$, where $\vec{AC}$ is the vector from $A$ to $C$, so $\vec{AC}=\begin{pmatrix}60 \\ 90 \\ h \end{pmatrix}-\begin{pmatrix}0 \\ 0 \\ h \end{pmatrix}=\begin{pmatrix}60 \\ 90 \\ h \end{pmatrix}$.
Then $\vec{F2_{CP}}-\vec{F1_{AP}}=\vec{AC}$ i.e. $\begin{pmatrix}x-60 \\ y-90 \\ 0 \end{pmatrix}-\begin{pmatrix}x \\ y \\ 0 \end{pmatrix}=\begin{pmatrix}60 \\ 90 \\ h \end{pmatrix}$.
But this is trivial. So my idea is wrong.
Could you give me a hint?
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EDIT :
I have also an other question : Is the flight plane using vectors the following ? $$E: \ \vec{x}=\vec{A}+r\cdot \vec{F1}+s\cdot \vec{F2}=\begin{pmatrix}0 \\ 0 \\ h \end{pmatrix}+r\cdot \begin{pmatrix}100 \\ -30 \\ 0 \end{pmatrix}+s\cdot \begin{pmatrix}-20 \\ -140 \\ 0\end{pmatrix}$$
 A: The current point $M$ of line $AB$ is such that :
$$\vec{AM}=t \vec{AB}\tag{1}$$
for a certain $t$. Relationship (1) can be written:
$$M-A=t \vec{AB} \ \iff \ M=A+t\vec{AB}$$
In coordinates (you have done these calculations):
$$\pmatrix{x\\y}=t\pmatrix{100\\-30}$$
Said otherwise:
$$\begin{cases}x&=&t(100)\\y&=&t(-30)\end{cases}\tag{2}$$
Now, the second line $CD$. Its current point $N$ of  is such that :
$$\vec{CM}=u \vec{CD}\tag{3}$$
for a certain $u$. Relationship (3) can be written:
$$M-C=u \vec{CD} \ \iff \ M=C+u\vec{CD}$$
In coordinates (you have done these calculations):
$$\pmatrix{x\\y}=\pmatrix{60\\90}+u\pmatrix{-20\\-140}$$
which is equivalent to:
$$\begin{cases}x&=&60+u(-20)\\y&=&90+u(-140)\end{cases}\tag{4}$$
Now express, using (2) and (4), that for a certain $t$ and a certain $u$, points $M$ and $N$ are in fact a same point with the same $x$ and $y$ coordinates :
$$\begin{cases}t(100)&=&60+u(-20)\\t(-30)&=&90+u(-140)\end{cases},\tag{5}$$
which is a system of 2 linear equations with two unknowns that I let you solve.
Don't forget afterwards to check your result on a graphical representation of the "scene".
Remark: the third coordinate isn't necessary in this issue.
