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Question
Ed walks in a straight line from point \({\text{P}}( – 1,{\text{ }}4)\) to point \({\text{Q}}(4,{\text{ }}16)\) with constant speed.
Ed starts from point \(P\) at time \(t = 0\) and arrives at point \(Q\) at time \(t = 3\), where \(t\) is measured in hours.
Given that, at time \(t\), Ed’s position vector, relative to the origin, can be given in the form, \({{r}} = {{a}} + t{{b}}\),
find the vectors \({{a}}\) and \({{b}}\).
Roderick is at a point \({\text{C}}(11,{\text{ }}9)\). During Ed’s walk from \(P\) to \(Q\) Roderick wishes to signal to Ed. He decides to signal when Ed is at the closest point to \(C\).
Find the time when Roderick signals to Ed.
Answer/Explanation
Markscheme
\({{a}} = \left( {\begin{array}{*{20}{c}} { – 1} \\ 4 \end{array}} \right)\) A1
\({{b}} = \frac{1}{3}\left( {\left( {\begin{array}{*{20}{c}} 4 \\ {16} \end{array}} \right) – \left( {\begin{array}{*{20}{c}} { – 1} \\ 4 \end{array}} \right)} \right) = \left( {\begin{array}{*{20}{c}} {\frac{5}{3}} \\ 4 \end{array}} \right)\) (M1)A1
[3 marks]
METHOD 1
Roderick must signal in a direction vector perpendicular to Ed’s path. (M1)
the equation of the signal is \({\mathbf{s}} = \left( {\begin{array}{*{20}{c}} {11} \\ 9 \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} { – 12} \\ 5 \end{array}} \right)\;\;\;\)(or equivalent) A1
\(\left( {\begin{array}{*{20}{c}} { – 1} \\ 4 \end{array}} \right) + \frac{t}{3}\left( {\begin{array}{*{20}{c}} 5 \\ {12} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {11} \\ 9 \end{array}} \right) + \lambda \left( {\begin{array}{*{20}{c}} { – 12} \\ 5 \end{array}} \right)\) M1
\(\frac{5}{3}t + 12\lambda = 12\) and \(4t – 5\lambda = 5\) M1
\(t = 2.13\;\;\;\left( { = \frac{{360}}{{169}}} \right)\) A1
METHOD 2
\(\left( {\begin{array}{*{20}{c}} 5 \\ {12} \end{array}} \right) \bullet \left( {\left( {\begin{array}{*{20}{c}} {11} \\ 9 \end{array}} \right) – \left( {\begin{array}{*{20}{c}} { – 1 + \frac{5}{3}t} \\ {4 + 4t} \end{array}} \right)} \right) = 0\;\;\;\)(or equivalent) M1A1A1
Note: Award the M1 for an attempt at a scalar product equated to zero, A1 for the first factor and A1 for the complete second factor.
attempting to solve for \(t\) (M1)
\(t = 2.13\;\;\;\left( {\frac{{360}}{{169}}} \right)\) A1
METHOD 3
\(x = \sqrt {{{\left( {12 – \frac{{5t}}{3}} \right)}^2} + {{(5 – 4t)}^2}} \;\;\;\)(or equivalent)\(\;\;\;\left( {{x^2} = {{\left( {12 – \frac{{5t}}{3}} \right)}^2} + {{(5 – 4t)}^2}} \right)\) M1A1A1
Note: Award M1 for use of Pythagoras’ theorem, A1 for \({\left( {12 – \frac{{5t}}{3}} \right)^2}\) and A1 for \({(5 – 4t)^2}\).
attempting (graphically or analytically) to find \(t\) such that \(\frac{{{\text{d}}x}}{{{\text{d}}t}} = 0\left( {\frac{{{\text{d}}({x^2})}}{{{\text{d}}t}} = 0} \right)\) (M1)
\(t = 2.13\;\;\;\left( { = \frac{{360}}{{169}}} \right)\) A1
METHOD 4
\(\cos \theta = \frac{{\left( {\begin{array}{*{20}{c}} {12} \\ 5 \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 5 \\ {12} \end{array}} \right)}}{{\left| {\left( {\begin{array}{*{20}{c}} {12} \\ 5 \end{array}} \right)} \right|\left| {\left( {\begin{array}{*{20}{c}} 5 \\ {12} \end{array}} \right)} \right|}} = \frac{{120}}{{169}}\) M1A1
Note: Award M1 for attempting to calculate the scalar product.
\(\frac{{120}}{{13}} = \frac{t}{3}\left| {\left( {\begin{array}{*{20}{c}} 5 \\ {12} \end{array}} \right)} \right|\;\;\;\)(or equivalent) (A1)
attempting to solve for \(t\) (M1)
\(t = 2.13\;\;\;\left( { = \frac{{360}}{{169}}} \right)\) A1
[5 marks]
Total [8 marks]
Examiners report
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