In this question, the distance is measured in metres and time is in seconds.
A radio-controlled airplane takes off from ground level and is assumed to be travelling at a steady speed in a straight line. The position vector of the airplane \(t\) seconds after it takes off is given by \(\mathbf{r}=\mathbf{a}+t\mathbf{b}\), where \(\mathbf{a}\) refers to the position vector of the point where it departs and \(\mathbf{b}\) is known as its velocity vector.
Paul stands at the point \(C\) with coordinates \(\left( -7,\,\,5,\,\,2 \right)\) to observe the airplane.
While the airplane is in the air, a drone is seen flying at a steady speed in a straight line with equation
\(\frac{2x-1}{-6}=\frac{y+7}{4}=\frac{z-10}{k}\).
At \(P\), the airplane suddenly changes its speed and direction. The position vector of the airplane \(s\) seconds after it leaves \(P\) is given by
\[\mathbf{r} = \begin{pmatrix} -9 \\ 4 \\ 6 \end{pmatrix} + s \begin{pmatrix} 3 \\ 2 \\ -1 \end{pmatrix}, \text{ where}\hspace{0.5em} s \in \mathbb{R}, s \geq 0.\]
It travels at a steady speed in a straight line towards an inclined slope, which is assumed to be a plane with equation
\(x-y+7z=2\).
TGM correction to part (iii): the symmetric equation describes a full line, so use \(\lambda\in\mathbb R\). The source paper says “non-negative constant”, but gives no starting point or direction that would restrict the drone to a ray. The parameter is not assumed to be elapsed time.
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