IB Math Analysis & Approaches Questionbank-Topic: SL 3.2 The cosine rule SL Paper 1

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 \(\overrightarrow {{\text{AC}}}  = \left( {\begin{array}{*{20}{c}} 3 \\ 0 \\ 4 \end{array}} \right)\).

Find the coordinates of A.

[2]
a.

Find \(\overrightarrow {{\text{AB}}} \);

[3]
b.i.

Find \(\left| {\overrightarrow {{\text{AB}}} } \right|\).

[2]
b.ii.

Find \(\cos {\rm{B\hat AC}}\).

[5]
c.

Hence or otherwise, find the distance between \({P_1}\) and \({P_2}\) two seconds after they leave A.

[4]
d.
Answer/Explanation

Markscheme

recognizing \(t = 0\) at A     (M1)

A is \((4,{\text{ }} – 1,{\text{ }}3)\)     A1     N2

[2 marks]

a.

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 \(\overrightarrow {{\text{AB}}} \)     (A1)

eg\(\,\,\,\,\,\)\({\text{AO}} + {\text{OB}},{\text{ B}} – {\text{A, }}\left( {\begin{array}{*{20}{c}} 6 \\ 3 \\ { – 1} \end{array}} \right) – \left( {\begin{array}{*{20}{c}} 4 \\ { – 1} \\ 3 \end{array}} \right)\)

\(\overrightarrow {{\text{AB}}}  = \left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ { – 4} \end{array}} \right)\)     A1     N2

METHOD 2

recognizing \(\overrightarrow {{\text{AB}}} \) is two times the direction vector     (M1)

correct working     (A1)

eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{AB}}}  = 2\left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { – 2} \end{array}} \right)\)

\(\overrightarrow {{\text{AB}}}  = \left( {\begin{array}{*{20}{c}} 2 \\ 4 \\ { – 4} \end{array}} \right)\)     A1     N2

[3 marks]

b.i.

correct substitution     (A1)

eg\(\,\,\,\,\,\)\(\left| {\overrightarrow {{\text{AB}}} } \right| = \sqrt {{2^2} + {4^2} + {4^2}} ,{\text{ }}\sqrt {4 + 16 + 16} ,{\text{ }}\sqrt {36} \)

\(\left| {\overrightarrow {{\text{AB}}} } \right| = 6\)     A1     N2

[2 marks]

b.ii.

METHOD 1 (vector approach)

valid approach involving \(\overrightarrow {{\text{AB}}} \) and \(\overrightarrow {{\text{AC}}} \)     (M1)

eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{AB}}}  \bullet \overrightarrow {{\text{AC}}} ,{\text{ }}\frac{{\overrightarrow {{\text{BA}}}  \bullet \overrightarrow {{\text{AC}}} }}{{{\text{AB}} \times {\text{AC}}}}\)

finding scalar product and \(\left| {\overrightarrow {{\text{AC}}} } \right|\)     (A1)(A1)

scalar product \(2(3) + 4(0) – 4(4){\text{ }}( =  – 10)\)

\(\left| {\overrightarrow {{\text{AC}}} } \right| = \sqrt {{3^2} + {0^2} + {4^2}} {\text{ }}( = 5)\)

substitution of their scalar product and magnitudes into cosine formula     (M1)

eg\(\,\,\,\,\,\)\(\cos {\rm{B\hat AC = }}\frac{{6 + 0 – 16}}{{6\sqrt {{3^2} + {4^2}} }}\)

\({\text{cos}}\,B\hat AC =  – \frac{{10}}{{30}}\left( { =  – \frac{1}{3}} \right)\)     A1     N2

 

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     (A1)(A1)

\({\text{AC}} = 5,{\text{ BC}} = 9\)

substitution of their lengths into cosine formula     (M1)

eg\(\,\,\,\,\,\)\(\cos {\rm{B\hat AC}} = \frac{{{5^2} + {6^2} – {9^2}}}{{2 \times 5 \times 6}}\)

\(\cos {\rm{B\hat AC}} =  – \frac{{20}}{{60}}{\text{ }}\left( { =  – \frac{1}{3}} \right)\)     A1     N2

[5 marks]

c.

Note:     Award relevant marks for working seen to find BC in part (c) (if cosine rule used in part (c)).

 

METHOD 1 (using cosine rule)

recognizing need to find BC     (M1)

choosing cosine rule     (M1)

eg\(\,\,\,\,\,\)\({c^2} = {a^2} + {b^2} – 2ab\cos {\text{C}}\)

correct substitution into RHS     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 9     A1     N2

 

METHOD 2 (finding magnitude of \(\overrightarrow {BC} \)

recognizing need to find BC     (M1)

valid approach     (M1)

eg\(\,\,\,\,\,\)attempt to find \(\overrightarrow {{\rm{OB}}} \) or \(\overrightarrow {{\rm{OC}}} \), \(\overrightarrow {{\rm{OB}}}  = \left( {\begin{array}{*{20}{c}} 6 \\ 3 \\ { – 1} \end{array}} \right)\) or \(\overrightarrow {{\rm{OC}}}  = \left( {\begin{array}{*{20}{c}} 7 \\ { – 1} \\ 7 \end{array}} \right),{\text{ }}\overrightarrow {{\rm{BA}}}  + \overrightarrow {{\rm{AC}}} \)

correct working     A1

eg\(\,\,\,\,\,\)\(\overrightarrow {{\rm{BC}}}  = \left( {\begin{array}{*{20}{c}} 1 \\ { – 4} \\ 8 \end{array}} \right),{\text{ }}\overrightarrow {{\rm{CB}}}  = \left( {\begin{array}{*{20}{c}} { – 1} \\ 4 \\ { – 8} \end{array}} \right),{\text{ }}\sqrt {{1^2} + {4^2} + {8^2}}  = \sqrt {81} \)

distance is 9     A1     N2

 

METHOD 3 (finding coordinates and using distance formula)

recognizing need to find BC     (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\(\,\,\,\,\,\)\({\text{BC}} = \sqrt {{{(6 – 7)}^2} + {{\left( {3 – ( – 1)} \right)}^2} + {{( – 1 – 7)}^2}} ,{\text{ }}\sqrt {{1^2} + {4^2} + {8^2}}  = \sqrt {81} \)

distance is 9     A1     N2

[4 marks]

d.

Question

The following diagram shows triangle ABC, with \({\text{AB}} = 3{\text{ cm}}\), \({\text{BC}} = 8{\text{ cm}}\), and \({\rm{A\hat BC = }}\frac{\pi }{3}\).

N17/5/MATME/SP1/ENG/TZ0/04

Show that \({\text{AC}} = 7{\text{ cm}}\).

[4]
a.

The shape in the following diagram is formed by adding a semicircle with diameter [AC] to the triangle.

N17/5/MATME/SP1/ENG/TZ0/04.b

Find the exact perimeter of this shape.

[3]
b.
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)     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\(\,\,\,\,\,\)\({\text{A}}{{\text{C}}^2} = 49,{\text{ }}b = \sqrt {49} \)

\({\text{AC}} = 7{\text{ (cm)}}\)     AG     N0

Note:     Award no marks if the only working seen is \({\text{A}}{{\text{C}}^2} = 49\) or \({\text{AC}} = \sqrt {49} \) (or similar).

[4 marks]

a.

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)     (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]

b.

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