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.

${{a}} = \left( {\begin{array}{*{20}{c}} { - 1} \\ 4 \end{array}} \right)$     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]

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