Question
Note: In this question, distance is in metres and time is in seconds.
Two particles \({P_1}\) and \({P_2}\) start moving from a point A at the same time, along different straight lines.
After \(t\) seconds, the position of \({P_1}\) is given by r = \(\left( {\begin{array}{*{20}{c}} 4 \\ { – 1} \\ 3 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { – 1} \end{array}} \right)\).
Two seconds after leaving A, \({P_1}\) is at point B.
Two seconds after leaving A, \({P_2}\) is at point C, where \(\overrightarrow {{\text{AC}}}Â = \left( {\begin{array}{*{20}{c}} 3 \\ 0 \\ 4 \end{array}} \right)\).
Find the coordinates of A.
Find \(\overrightarrow {{\text{AB}}} \);
Find \(\left| {\overrightarrow {{\text{AB}}} } \right|\).
Find \(\cos {\rm{B\hat AC}}\).
Hence or otherwise, find the distance between \({P_1}\) and \({P_2}\) two seconds after they leave A.
Answer/Explanation
Markscheme
recognizing \(t = 0\) at A Â Â (M1)
A is \((4,{\text{ }} – 1,{\text{ }}3)\) Â Â A1 Â Â N2
[2 marks]
METHOD 1
valid approach   (M1)
eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} 4 \\ { – 1} \\ 3 \end{array}} \right) + 2\left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { – 2} \end{array}} \right),{\text{ }}(6,{\text{ }}3,{\text{ }} – 1)\)
correct approach to find \(\overrightarrow {{\text{AB}}} \) Â Â (A1)
eg\(\,\,\,\,\,\)\({\text{AO}} + {\text{OB}},{\text{ B}} – {\text{A, }}\left( {\begin{array}{*{20}{c}} 6 \\ 3 \\ { – 1} \end{array}} \right) – \left( {\begin{array}{*{20}{c}} 4 \\ { – 1} \\ 3 \end{array}} \right)\)
\(\overrightarrow {{\text{AB}}}Â = \left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ { – 4} \end{array}} \right)\) Â Â A1 Â Â N2
METHOD 2
recognizing \(\overrightarrow {{\text{AB}}} \) is two times the direction vector   (M1)
correct working   (A1)
eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{AB}}}Â = 2\left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { – 2} \end{array}} \right)\)
\(\overrightarrow {{\text{AB}}}Â = \left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ { – 4} \end{array}} \right)\) Â Â A1 Â Â N2
[3 marks]
correct substitution   (A1)
eg\(\,\,\,\,\,\)\(\left| {\overrightarrow {{\text{AB}}} } \right| = \sqrt {{2^2} + {4^2} + {4^2}} ,{\text{ }}\sqrt {4 + 16 + 16} ,{\text{ }}\sqrt {36} \)
\(\left| {\overrightarrow {{\text{AB}}} } \right| = 6\) Â Â A1 Â Â N2
[2 marks]
METHOD 1 (vector approach)
valid approach involving \(\overrightarrow {{\text{AB}}} \) and \(\overrightarrow {{\text{AC}}} \) Â Â (M1)
eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{AB}}}Â \bullet \overrightarrow {{\text{AC}}} ,{\text{ }}\frac{{\overrightarrow {{\text{BA}}}Â \bullet \overrightarrow {{\text{AC}}} }}{{{\text{AB}} \times {\text{AC}}}}\)
finding scalar product and \(\left| {\overrightarrow {{\text{AC}}} } \right|\) Â Â (A1)(A1)
scalar product \(2(3) + 4(0) – 4(4){\text{ }}( =Â – 10)\)
\(\left| {\overrightarrow {{\text{AC}}} } \right| = \sqrt {{3^2} + {0^2} + {4^2}} {\text{ }}( = 5)\)
substitution of their scalar product and magnitudes into cosine formula   (M1)
eg\(\,\,\,\,\,\)\(\cos {\rm{B\hat AC = }}\frac{{6 + 0 – 16}}{{6\sqrt {{3^2} + {4^2}} }}\)
\({\text{cos}}\,B\hat AC = Â – \frac{{10}}{{30}}\left( { = Â – \frac{1}{3}} \right)\) Â Â Â A1 Â Â N2
Â
METHOD 2 (triangle approach)
valid approach involving cosine rule   (M1)
eg\(\,\,\,\,\,\)\(\cos {\rm{B\hat AC = }}\frac{{{\text{A}}{{\text{B}}^2} + {\text{A}}{{\text{C}}^2} – {\text{B}}{{\text{C}}^2}}}{{2 \times {\text{AB}} \times {\text{AC}}}}\)
finding lengths AC and BC Â Â (A1)(A1)
\({\text{AC}} = 5,{\text{ BC}} = 9\)
substitution of their lengths into cosine formula   (M1)
eg\(\,\,\,\,\,\)\(\cos {\rm{B\hat AC}} = \frac{{{5^2} + {6^2} – {9^2}}}{{2 \times 5 \times 6}}\)
\(\cos {\rm{B\hat AC}} =Â – \frac{{20}}{{60}}{\text{ }}\left( { =Â – \frac{1}{3}} \right)\) Â Â A1 Â Â N2
[5 marks]
Note: Â Â Award relevant marks for working seen to find BC in part (c) (if cosine rule used in part (c)).
Â
METHOD 1 (using cosine rule)
recognizing need to find BC Â Â (M1)
choosing cosine rule   (M1)
eg\(\,\,\,\,\,\)\({c^2} = {a^2} + {b^2} – 2ab\cos {\text{C}}\)
correct substitution into RHS Â Â A1
eg\(\,\,\,\,\,\)\({\text{B}}{{\text{C}}^2} = {(6)^2} + {(5)^2} – 2(6)(5)\left( { – \frac{1}{3}} \right),{\text{ }}36 + 25 + 20\)
distance is 9Â Â Â A1 Â Â N2
Â
METHOD 2 (finding magnitude of \(\overrightarrow {BC} \))Â
recognizing need to find BC Â Â (M1)
valid approach   (M1)
eg\(\,\,\,\,\,\)attempt to find \(\overrightarrow {{\rm{OB}}} \) or \(\overrightarrow {{\rm{OC}}} \), \(\overrightarrow {{\rm{OB}}}Â = \left( {\begin{array}{*{20}{c}} 6 \\ 3 \\ { – 1} \end{array}} \right)\) or \(\overrightarrow {{\rm{OC}}}Â = \left( {\begin{array}{*{20}{c}} 7 \\ { – 1} \\ 7 \end{array}} \right),{\text{ }}\overrightarrow {{\rm{BA}}}Â + \overrightarrow {{\rm{AC}}} \)
correct working   A1
eg\(\,\,\,\,\,\)\(\overrightarrow {{\rm{BC}}}Â = \left( {\begin{array}{*{20}{c}} 1 \\ { – 4} \\ 8 \end{array}} \right),{\text{ }}\overrightarrow {{\rm{CB}}}Â = \left( {\begin{array}{*{20}{c}} { – 1} \\ 4 \\ { – 8} \end{array}} \right),{\text{ }}\sqrt {{1^2} + {4^2} + {8^2}}Â = \sqrt {81} \)
distance is 9 Â Â Â A1 Â Â N2
Â
METHOD 3 (finding coordinates and using distance formula)
recognizing need to find BC Â Â (M1)
valid approach   (M1)
eg\(\,\,\,\,\,\)attempt to find coordinates of B or C, \({\text{B}}(6,{\text{ }}3,{\text{ }} – 1)\) or \({\text{C}}(7,{\text{ }} – 1,{\text{ }}7)\)
correct substitution into distance formula   A1
eg\(\,\,\,\,\,\)\({\text{BC}} = \sqrt {{{(6 – 7)}^2} + {{\left( {3 – ( – 1)} \right)}^2} + {{( – 1 – 7)}^2}} ,{\text{ }}\sqrt {{1^2} + {4^2} + {8^2}}Â = \sqrt {81} \)
distance is 9 Â Â A1 Â Â N2
[4 marks]
Question
The following diagram shows triangle ABC, with \({\text{AB}} = 3{\text{ cm}}\), \({\text{BC}} = 8{\text{ cm}}\), and \({\rm{A\hat BC = }}\frac{\pi }{3}\).
Show that \({\text{AC}} = 7{\text{ cm}}\).
The shape in the following diagram is formed by adding a semicircle with diameter [AC] to the triangle.
Find the exact perimeter of this shape.
Answer/Explanation
Markscheme
evidence of choosing the cosine rule   (M1)
eg\(\,\,\,\,\,\)\({c^2} = {a^2} + {b^2} – ab\cos C\)
correct substitution into RHS of cosine rule   (A1)
eg\(\,\,\,\,\,\)\({3^2} + {8^2} – 2 \times 3 \times 8 \times \cos \frac{\pi }{3}\)
evidence of correct value for \(\cos \frac{\pi }{3}\) (may be seen anywhere, including in cosine rule) Â Â A1
eg\(\,\,\,\,\,\)\(\cos \frac{\pi }{3} = \frac{1}{2},{\text{ A}}{{\text{C}}^2} = 9 + 64 – \left( {48 \times \frac{1}{2}} \right),{\text{ }}9 + 64 – 24\)
correct working clearly leading to answer   A1
eg\(\,\,\,\,\,\)\({\text{A}}{{\text{C}}^2} = 49,{\text{ }}b = \sqrt {49} \)
\({\text{AC}} = 7{\text{ (cm)}}\) Â Â AG Â Â N0
Note: Â Â Award no marks if the only working seen is \({\text{A}}{{\text{C}}^2} = 49\) or \({\text{AC}} = \sqrt {49} \) (or similar).
[4 marks]
correct substitution for semicircle   (A1)
eg\(\,\,\,\,\,\)\({\text{semicircle}} = \frac{1}{2}(2\pi \times 3.5),{\text{ }}\frac{1}{2} \times \pi \times 7,{\text{ }}3.5\pi \)
valid approach (seen anywhere) Â Â (M1)
eg\(\,\,\,\,\,\)\({\text{perimeter}} = {\text{AB}} + {\text{BC}} + {\text{semicircle, }}3 + 8 + \left( {\frac{1}{2} \times 2 \times \pi \times \frac{7}{2}} \right),{\text{ }}8 + 3 + 3.5\pi \)
\(11 + \frac{7}{2}\pi {\text{ }}( = 3.5\pi + 11){\text{ (cm)}}\)   A1   N2
[3 marks]