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Date May 2014 Marks available 5 Reference code 14M.1.sl.TZ2.9
Level SL only Paper 1 Time zone TZ2
Command term Show that Question number 9 Adapted from N/A

Question

Distances in this question are in metres.

Ryan and Jack have model airplanes, which take off from level ground. Jack’s airplane takes off after Ryan’s.

The position of Ryan’s airplane \(t\) seconds after it takes off is given by \(\boldsymbol{r}=\left( \begin{array}{c}5\\6\\0\end{array} \right) + t\left( \begin{array}{c} - 4\\2\\4\end{array} \right)\).

Find the speed of Ryan’s airplane.

[3]
a.

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

[2]
b.

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.

[5]
c.

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

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

[5]
d.

Markscheme

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]

a.

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]

b.

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]

c.

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]

d.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.

Syllabus sections

Topic 4 - Vectors » 4.2 » The scalar product of two vectors.
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