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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
[N/A]
[N/A]
Question
OACB is a parallelogram with \(\overrightarrow {{\text{OA}}} = \) a and \(\overrightarrow {{\text{OB}}} = \) b, where a and b are non-zero vectors.
Show that
(i) \({\left| {\overrightarrow {{\text{OC}}} } \right|^2} = |\)a\({|^2} + 2\)a \( \bullet \) b \( + |\)b\({|^2}\);
(ii) \({\left| {\overrightarrow {{\text{AB}}} } \right|^2} = |\)a\({|^2} – 2\)a \( \bullet \) b \( + |\)b\({|^2}\).
Given that \(\left| {\overrightarrow {{\text{OC}}} } \right| = \left| {\overrightarrow {{\text{AB}}} } \right|\), prove that OACB is a rectangle.
Answer/Explanation
Markscheme
METHOD 1
\({\left| {\overrightarrow {{\text{OC}}} } \right|^2} = \overrightarrow {{\text{OC}}} \bullet \overrightarrow {{\text{OC}}} \)
= (a + b) \( \bullet \) (a + b) A1
= a \( \bullet \) a + a \( \bullet \) b + b \( \bullet \) a + b \( \bullet \) b A1
= \(|\)a\({|^2}\) + 2a \( \bullet \) b + \(|\)b\({|^2}\) AG
METHOD 2
\({\left| {\overrightarrow {{\text{OC}}} } \right|^2} = {\left| {\overrightarrow {{\text{OA}}} } \right|^2} + {\left| {\overrightarrow {{\text{OB}}} } \right|^2} – 2\left| {\overrightarrow {{\text{OA}}} } \right|\left| {\overrightarrow {{\text{OB}}} } \right|\cos ({\rm{O\hat AC}})\) A1
\(\left| {\overrightarrow {{\text{OA}}} } \right|\left| {\overrightarrow {{\text{OB}}} } \right|\cos ({\rm{O\hat AC}}) = – \)(a \( \bullet \) b) A1
\({\left| {\overrightarrow {{\text{OC}}} } \right|^2} = |\)a\({|^2}\) + 2a \( \bullet \) b + \(|\)b\({|^2}\) AG
(ii) METHOD 1
\({\left| {\overrightarrow {{\text{AB}}} } \right|^2} = \overrightarrow {{\text{AB}}} \bullet \overrightarrow {{\text{AB}}} \)
= (b − a) \( \bullet \) (b − a) A1
= b \( \bullet \) b − b \( \bullet \) a − a \( \bullet \) b + a \( \bullet \) a A1
= \(|\)a\({|^2}\) – 2a \( \bullet \) b + \(|\)b\({|^2}\) AG
METHOD 2
\({\left| {\overrightarrow {{\text{AB}}} } \right|^2} = {\left| {\overrightarrow {{\text{AC}}} } \right|^2} + {\left| {\overrightarrow {{\text{BC}}} } \right|^2} – 2\left| {\overrightarrow {{\text{AC}}} } \right|\left| {\overrightarrow {{\text{BC}}} } \right|\cos ({\rm{A\hat CB}})\) A1
\(\left| {\overrightarrow {{\text{AC}}} } \right|\left| {\overrightarrow {{\text{BC}}} } \right|\cos ({\rm{A\hat CB}}) = \) a \( \bullet \) b A1
\({\left| {\overrightarrow {{\text{AB}}} } \right|^2} = |\)a \({|^2} – \) 2a \( \bullet \) b + \(|\)b\({|^2}\) AG
[4 marks]
\(\left| {\overrightarrow {{\text{OC}}} } \right| = \left| {\overrightarrow {{\text{AB}}} } \right| \Rightarrow {\left| {\overrightarrow {{\text{OC}}} } \right|^2} = {\left| {\overrightarrow {{\text{AB}}} } \right|^2} \Rightarrow |\)a\({|^2} + {\text{2}}\)a \( \bullet \) b \( + |\)b\({|^2} = |\)a\({|^2} – 2\)a \( \bullet \) b \( + |\)b\({|^2}\) R1(M1)
Note: Award R1 for \(\left| {\overrightarrow {{\text{OC}}} } \right| = \left| {\overrightarrow {{\text{AB}}} } \right| \Rightarrow {\left| {\overrightarrow {{\text{OC}}} } \right|^2} = {\left| {\overrightarrow {{\text{AB}}} } \right|^2}\) and (M1) for \(|\)a\({|^2} + {\text{2}}\)a \( \bullet \) b \( + |\)b\({|^2} = |\)a\({|^2} – 2\)a \( \bullet \) b \( + |\)b\({|^2}\).
a \( \bullet \) b \( = 0\) A1
hence OACB is a rectangle (a and b both non-zero)
with adjacent sides at right angles R1
Note: Award R1(M1)A0R1 if the dot product has not been used.
[4 marks]
Examiners report
In part (a), a significant number of candidates either did not use correct vector notation or simply did not use vector notation at all. A large number of candidates who appeared to adopt a scalar product approach, did not use the scalar product ‘dot’ and represented a \( \bullet \) b as ab. A few candidates successfully used the cosine rule with correct vector notation. A small number of candidates expressed a and b in general component form. In part (a) (ii), quite a number of candidates expressed \({\overrightarrow {{\text{AB}}} }\) as a – b rather than as b – a.
In part (b), some very well structured proofs were offered by a small number of candidates.
Question
OACB is a parallelogram with \(\overrightarrow {{\text{OA}}} = \) a and \(\overrightarrow {{\text{OB}}} = \) b, where a and b are non-zero vectors.
Show that
(i) \({\left| {\overrightarrow {{\text{OC}}} } \right|^2} = |\)a\({|^2} + 2\)a \( \bullet \) b \( + |\)b\({|^2}\);
(ii) \({\left| {\overrightarrow {{\text{AB}}} } \right|^2} = |\)a\({|^2} – 2\)a \( \bullet \) b \( + |\)b\({|^2}\).
Given that \(\left| {\overrightarrow {{\text{OC}}} } \right| = \left| {\overrightarrow {{\text{AB}}} } \right|\), prove that OACB is a rectangle.
Answer/Explanation
Markscheme
METHOD 1
\({\left| {\overrightarrow {{\text{OC}}} } \right|^2} = \overrightarrow {{\text{OC}}} \bullet \overrightarrow {{\text{OC}}} \)
= (a + b) \( \bullet \) (a + b) A1
= a \( \bullet \) a + a \( \bullet \) b + b \( \bullet \) a + b \( \bullet \) b A1
= \(|\)a\({|^2}\) + 2a \( \bullet \) b + \(|\)b\({|^2}\) AG
METHOD 2
\({\left| {\overrightarrow {{\text{OC}}} } \right|^2} = {\left| {\overrightarrow {{\text{OA}}} } \right|^2} + {\left| {\overrightarrow {{\text{OB}}} } \right|^2} – 2\left| {\overrightarrow {{\text{OA}}} } \right|\left| {\overrightarrow {{\text{OB}}} } \right|\cos ({\rm{O\hat AC}})\) A1
\(\left| {\overrightarrow {{\text{OA}}} } \right|\left| {\overrightarrow {{\text{OB}}} } \right|\cos ({\rm{O\hat AC}}) = – \)(a \( \bullet \) b) A1
\({\left| {\overrightarrow {{\text{OC}}} } \right|^2} = |\)a\({|^2}\) + 2a \( \bullet \) b + \(|\)b\({|^2}\) AG
(ii) METHOD 1
\({\left| {\overrightarrow {{\text{AB}}} } \right|^2} = \overrightarrow {{\text{AB}}} \bullet \overrightarrow {{\text{AB}}} \)
= (b − a) \( \bullet \) (b − a) A1
= b \( \bullet \) b − b \( \bullet \) a − a \( \bullet \) b + a \( \bullet \) a A1
= \(|\)a\({|^2}\) – 2a \( \bullet \) b + \(|\)b\({|^2}\) AG
METHOD 2
\({\left| {\overrightarrow {{\text{AB}}} } \right|^2} = {\left| {\overrightarrow {{\text{AC}}} } \right|^2} + {\left| {\overrightarrow {{\text{BC}}} } \right|^2} – 2\left| {\overrightarrow {{\text{AC}}} } \right|\left| {\overrightarrow {{\text{BC}}} } \right|\cos ({\rm{A\hat CB}})\) A1
\(\left| {\overrightarrow {{\text{AC}}} } \right|\left| {\overrightarrow {{\text{BC}}} } \right|\cos ({\rm{A\hat CB}}) = \) a \( \bullet \) b A1
\({\left| {\overrightarrow {{\text{AB}}} } \right|^2} = |\)a \({|^2} – \) 2a \( \bullet \) b + \(|\)b\({|^2}\) AG
[4 marks]
\(\left| {\overrightarrow {{\text{OC}}} } \right| = \left| {\overrightarrow {{\text{AB}}} } \right| \Rightarrow {\left| {\overrightarrow {{\text{OC}}} } \right|^2} = {\left| {\overrightarrow {{\text{AB}}} } \right|^2} \Rightarrow |\)a\({|^2} + {\text{2}}\)a \( \bullet \) b \( + |\)b\({|^2} = |\)a\({|^2} – 2\)a \( \bullet \) b \( + |\)b\({|^2}\) R1(M1)
Note: Award R1 for \(\left| {\overrightarrow {{\text{OC}}} } \right| = \left| {\overrightarrow {{\text{AB}}} } \right| \Rightarrow {\left| {\overrightarrow {{\text{OC}}} } \right|^2} = {\left| {\overrightarrow {{\text{AB}}} } \right|^2}\) and (M1) for \(|\)a\({|^2} + {\text{2}}\)a \( \bullet \) b \( + |\)b\({|^2} = |\)a\({|^2} – 2\)a \( \bullet \) b \( + |\)b\({|^2}\).
a \( \bullet \) b \( = 0\) A1
hence OACB is a rectangle (a and b both non-zero)
with adjacent sides at right angles R1
Note: Award R1(M1)A0R1 if the dot product has not been used.
[4 marks]
Examiners report
In part (a), a significant number of candidates either did not use correct vector notation or simply did not use vector notation at all. A large number of candidates who appeared to adopt a scalar product approach, did not use the scalar product ‘dot’ and represented a \( \bullet \) b as ab. A few candidates successfully used the cosine rule with correct vector notation. A small number of candidates expressed a and b in general component form. In part (a) (ii), quite a number of candidates expressed \({\overrightarrow {{\text{AB}}} }\) as a – b rather than as b – a.
In part (b), some very well structured proofs were offered by a small number of candidates.