Find the speed of Ryan’s airplane.

Find the height of Ryan’s airplane after two seconds.

The position of Jack’s airplane \(s\) seconds after it takes off is given by r = \(\left( \begin{array}{c} - 39\\44\\0\end{array} \right) + s\left( \begin{array}{c}4\\ - 6\\7\end{array} \right)\).

Show that the paths of the airplanes are perpendicular.

The two airplanes collide at the point \((-23, 20, 28)\).

How long after Ryan’s airplane takes off does Jack’s airplane take off?

valid approach     (M1)

eg     magnitude of direction vector

correct working     (A1)

eg     \(\sqrt {{{( - 4)}^2} + {2^2} + {4^2}} ,{\text{ }}\sqrt { - {4^2} + {2^2} + {4^2}} \)

\(6{\text{ (m}}{{\text{s}}^{ - 1}})\)     A1     N2

[3 marks]

substituting \(2\) for \(t\)     (A1)

eg     \(0 + 2(4)\), r = \(\left( \begin{array}{c}5\\6\\0\end{array} \right) + 2\left( \begin{array}{c} - 4\\2\\4\end{array} \right),\left( \begin{array}{c} - 3\\10\\8\end{array} \right)\), \(y = 10\)

\(8\) (metres)     A1     N2

[2 marks]

METHOD 1

choosing correct direction vectors \(\left( \begin{array}{c} - 4\\2\\4\end{array} \right)\) and \(\left( \begin{array}{c}4\\ - 6\\7\end{array} \right)\)     (A1)(A1)

evidence of scalar product      M1

eg     a \( \cdot \) b

correct substitution into scalar product     (A1)

eg    \(( - 4 \times 4) + (2 \times  - 6) + (4 \times 7)\)

evidence of correct calculation of the scalar product as \(0\)     A1

eg     \( - 16 - 12 + 28 = 0\)

directions are perpendicular     AG     N0

METHOD 2

choosing correct direction vectors \(\left( \begin{array}{c} - 4\\2\\4\end{array} \right)\) and \(\left( \begin{array}{c}4\\ - 6\\7\end{array} \right)\)     (A1)(A1)

attempt to find angle between vectors     M1

correct substitution into numerator     A1

eg     \(\cos \theta  = \frac{{ - 16 - 12 + 28}}{{\left| a \right|\left| b \right|}},{\text{ }}\cos \theta  = 0\)

\(\theta  = 90^\circ \)     A1

directions are perpendicular     AG     N0

[5 marks]

METHOD 1

one correct equation for Ryan's airplane     (A1)

eg     \(5 - 4t =  - 23,{\text{ }}6 + 2t = 20,{\text{ }}0 + 4t = 28\)

\(t = 7\)     A1

one correct equation for Jack's airplane     (A1)

eg     \( - 39 + 4s =  - 23,{\text{ }}44 - 6s = 20,{\text{ }}0 + 7s = 28\)

\(s = 4\)     A1

\(3\) (seconds later)     A1     N2

METHOD 2

valid approach     (M1)

eg     \(\left( \begin{array}{c}5\\6\\0\end{array} \right) + t\left( \begin{array}{c} - 4\\2\\4\end{array} \right) = \left( \begin{array}{c} - 39\\44\\0\end{array} \right) + s\left( \begin{array}{c}4\\ - 6\\7\end{array} \right)\), one correct equation

two correct equations     (A1)

eg     \(5 - 4t =  - 39 + 4s,{\text{ }}6 + 2t = 44 - 6s,{\text{ }}4t = 7s\)

\(t = 7\)     A1

\(s = 4\)    A1

\(3\) (seconds later)     A1     N2

[5 marks]