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Question
The diagram shows a parallelogram ABCD.
The coordinates of A, B and D are A(1, 2, 3) , B(6, 4,4 ) and D(2, 5, 5) .
(i) Show that \(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
5\\
2\\
1
\end{array}} \right)\) .
(ii) Find \(\overrightarrow {{\rm{AD}}} \) .
(iii) Hence show that \(\overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}}
6\\
5\\
3
\end{array}} \right)\) .
Find the coordinates of point C.
(i) Find \(\overrightarrow {{\rm{AB}}} \bullet \overrightarrow {{\rm{AD}}} \).
(ii) Hence find angle A.
Hence, or otherwise, find the area of the parallelogram.
Answer/Explanation
Markscheme
(i) evidence of approach M1
e.g. \({\text{B}} – {\text{A}}\) , \(\overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OB}}} \) , \(\left( {\begin{array}{*{20}{c}}
6\\
4\\
4
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
1\\
2\\
3
\end{array}} \right)\)
\(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
5\\
2\\
1
\end{array}} \right)\) AG N0
(ii) evidence of approach (M1)
e.g. \({\text{D}} – {\text{A}}\) , \(\overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OD}}} \) , \(\left( {\begin{array}{*{20}{c}}
2\\
5\\
5
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
1\\
2\\
3
\end{array}} \right)\)
\(\overrightarrow {{\rm{AD}}}= \left( {\begin{array}{*{20}{c}}
1\\
3\\
2
\end{array}} \right)\) A1 N2
(iii) evidence of approach (M1)
e.g. \(\overrightarrow {{\rm{AC}}} = \overrightarrow {{\rm{AB}}} + \overrightarrow {{\rm{AD}}} \)
correct substitution A1
e.g. \(\overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}}
5\\
2\\
1
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
1\\
3\\
2
\end{array}} \right)\)
\(\overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}}
6\\
5\\
3
\end{array}} \right)\) AG N0
[5 marks]
evidence of combining vectors (there are at least 5 ways) (M1)
e.g. \(\overrightarrow {{\rm{OC}}} = \overrightarrow {{\rm{OA}}} + \overrightarrow {{\rm{AC}}} \) , \(\overrightarrow {{\rm{OC}}} = \overrightarrow {{\rm{OB}}} + \overrightarrow {{\rm{AD}}} \), \(\overrightarrow {{\rm{AB}}} = \overrightarrow {{\rm{OC}}} – \overrightarrow {{\rm{OD}}} \)
correct substitution A1
\(\overrightarrow {{\rm{OC}}} = \left( {\begin{array}{*{20}{c}}
1\\
2\\
3
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
6\\
5\\
3
\end{array}} \right)\left( { = \left( {\begin{array}{*{20}{c}}
7\\
7\\
6
\end{array}} \right)} \right)\)
e.g. coordinates of C are \((7{\text{, }}7{\text{, }}6)\) A1 N1
[3 marks]
(i) evidence of using scalar product on \(\overrightarrow {{\rm{AB}}} \) and \(\overrightarrow {{\rm{AD}}} \) (M1)
e.g. \(\overrightarrow {{\rm{AB}}} \bullet \overrightarrow {{\rm{AD}}} = 5(1) + 2(3) + 1(2)\)
\(\overrightarrow {{\rm{AB}}} \bullet \overrightarrow {{\rm{AD}}} = 13\) A1 N2
(ii) \(\left| {\overrightarrow {{\rm{AB}}} } \right| = 5.477 \ldots \) , \(\left| {\overrightarrow {{\rm{AD}}} } \right| = 3.741 \ldots \) (A1)(A1)
evidence of using \(\cos A = \frac{{\overrightarrow {{\rm{AB}}} \bullet \overrightarrow {{\rm{AD}}} }}{{\left| {\overrightarrow {{\rm{AB}}} } \right|\left| {\overrightarrow {{\rm{AD}}} } \right|}}\) (M1)
correct substitution A1
e.g. \(\cos A = \frac{{13}}{{20.493}}\)
\(\widehat A = 0.884\) \((50.6^\circ )\) A1 N3
[7 marks]
METHOD 1
evidence of using \({\rm{area}} = 2\left( {\frac{1}{2}\left| {\overrightarrow {{\rm{AD}}} } \right|\left| {\overrightarrow {{\rm{AB}}} } \right|\sin {\rm{D}}\widehat {\rm{A}}{\rm{B}}} \right)\) (M1)
correct substitution A1
e.g. \({\rm{area}} = 2\left( {\frac{1}{2}(3,741 \ldots )(5.477 \ldots )\sin 0.883 \ldots } \right)\)
\({\rm{area}} = 15.8\) A1 N2
METHOD 2
evidence of using \({\rm{area}} = b \times h\) (M1)
finding height of parallelogram A1
e.g. \(h = 3.741 \ldots \times \sin 0.883 \ldots ( = 2.892 \ldots )\) , \(h = 5.477 \ldots \times \sin 0.883 \ldots ( = 4.234 \ldots )\)
\({\rm{area}} = 15.8\) A1 N2
[3 marks]
Question
Consider the points A(\(5\), \(2\), \(1\)) , B(\(6\), \(5\), \(3\)) , and C(\(7\), \(6\), \(a + 1\)) , \(a \in{\mathbb{R}}\) .
Let \({\rm{q}}\) be the angle between \(\overrightarrow {{\rm{AB}}} \) and \(\overrightarrow {{\rm{AC}}} \) .
Find
(i) \(\overrightarrow {{\rm{AB}}} \) ;
(ii) \(\overrightarrow {{\rm{AC}}} \) .
Find the value of \(a\) for which \({\rm{q}} = \frac{\pi }{2}\) .
i. Show that \(\cos q = \frac{{2a + 14}}{{\sqrt {14{a^2} + 280} }}\) .
ii. Hence, find the value of a for which \({\rm{q}} = 1.2\) .
Hence, find the value of a for which \({\rm{q}} = 1.2\) .
Answer/Explanation
Markscheme
(i) appropriate approach (M1)
eg \(\overrightarrow {{\rm{AO}}} {\rm{ + }}\overrightarrow {{\rm{OB}}} \) , \({\rm{B}} – {\rm{A}}\)
\(\overrightarrow {{\rm{AB}}} = \left( \begin{array}{l}
1\\
3\\
2
\end{array} \right)\) A1 N2
(ii) \(\overrightarrow {{\rm{AC}}} = \left( \begin{array}{l}
2\\
4\\
a
\end{array} \right)\) A1 N1
[3 marks]
valid reasoning (seen anywhere) R1
eg scalar product is zero, \(\cos \frac{\pi }{2} = \frac{{\boldsymbol{u} \bullet \boldsymbol{v}}}{{\left| \boldsymbol{u} \right|\left| \boldsymbol{v} \right|}}\)
correct scalar product of their \(\overrightarrow {{\rm{AB}}} \) and \(\overrightarrow {{\rm{AC}}} \) (may be seen in part (c)) (A1)
eg \(1(2) + 3(4) + 2(a)\)
correct working for their \(\overrightarrow {{\rm{AB}}} \) and \(\overrightarrow {{\rm{AC}}} \) (A1)
eg \(2a + 14\) , \(2a = – 14\)
\(a = – 7\) A1 N3
[4 marks]
correct magnitudes (may be seen in (b)) (A1)(A1)
\(\sqrt {{1^2} + {3^2} + {2^2}} \left( { = \sqrt {14} } \right)\) , \(\sqrt {{2^2} + {4^2} + {a^2}} \left( { = \sqrt {20 + {a^2}} } \right)\)
substitution into formula (M1)
eg \(\cos \theta \frac{{1 \times 2 + 3 \times 4 + 2 \times a}}{{\sqrt {{1^2} + {3^2} + {2^2}} \sqrt {{2^2} + {4^2} + {a^2}} }}\) , \(\frac{{14 + 2a}}{{\sqrt {14} \sqrt {4 + 16 + {a^2}} }}\)
simplification leading to required answer A1
eg \(\cos \theta = \frac{{14 + 2a}}{{\sqrt {14} \sqrt {20 + {a^2}} }}\)
\(\cos \theta = \frac{{2a + 14}}{{\sqrt {14{a^2} + 280} }}\) AG N0
[4 marks]
correct setup (A1)
eg \(\cos 1.2 = \frac{{2a + 14}}{{\sqrt {14{a^2} + 280} }}\)
valid attempt to solve (M1)
eg sketch, \(\frac{{2a + 14}}{{\sqrt {14{a^2} + 280} }} – \cos 1.2 = 0\) , attempt to square
\(a = – 3.25\) A2 N3
[4 marks]
correct setup (A1)
eg \(\cos 1.2 = \frac{{2a + 14}}{{\sqrt {14{a^2} + 280} }}\)
valid attempt to solve (M1)
eg sketch, \(\frac{{2a + 14}}{{\sqrt {14{a^2} + 280} }} – \cos 1.2 = 0\) , attempt to square
\(a = – 3.25\) A2 N3
[4 marks]
Question
Consider the lines \({L_1}\) and \({L_2}\) with equations \({L_1}\) : \(\boldsymbol{r}=\left( \begin{array}{c}11\\8\\2\end{array} \right) + s\left( \begin{array}{c}4\\3\\ – 1\end{array} \right)\) and \({L_2}\) : \(\boldsymbol{r} = \left( \begin{array}{c}1\\1\\ – 7\end{array} \right) + t\left( \begin{array}{c}2\\1\\11\end{array} \right)\).
The lines intersect at point \(\rm{P}\).
Find the coordinates of \({\text{P}}\).
Show that the lines are perpendicular.
The point \({\text{Q}}(7, 5, 3)\) lies on \({L_1}\). The point \({\text{R}}\) is the reflection of \({\text{Q}}\) in the line \({L_2}\).
Find the coordinates of \({\text{R}}\).
Answer/Explanation
Markscheme
appropriate approach (M1)
eg \(\left( \begin{array}{c}11\\8\\2\end{array} \right) + s\left( \begin{array}{c}4\\3\\ – 1\end{array} \right) = \left( \begin{array}{c}1\\1\\ – 7\end{array} \right) + t\left( \begin{array}{c}2\\1\\11\end{array} \right)\), \({L_1} = {L_2}\)
any two correct equations A1A1
eg \(11 + 4s = 1 + 2t,{\text{ }}8 + 3s = 1 + t,{\text{ }}2 – s = – 7 + 11t\)
attempt to solve system of equations (M1)
eg \(10 + 4s = 2(7 + 3s), \left\{ {\begin{array}{*{20}{c}} {4s – 2t = – 10} \\ {3s – t = – 7} \end{array}} \right.\)
one correct parameter A1
eg \(s = – 2,{\text{ }}t = 1\)
\({\text{P}}(3, 2, 4)\) (accept position vector) A1 N3
[6 marks]
choosing correct direction vectors for \({L_1}\) and \({L_2}\) (A1)(A1)
eg \(\left( {\begin{array}{*{20}{c}} 4 \\ 3 \\ { – 1} \end{array}} \right),\left( {\begin{array}{*{20}{c}} 2 \\ 1 \\ {11} \end{array}} \right)\) (or any scalar multiple)
evidence of scalar product (with any vectors) (M1)
eg \(a \cdot b\), \(\left( \begin{array}{c}4\\3\\ – 1\end{array} \right) \bullet \left( \begin{array}{c}2\\1\\11\end{array} \right)\)
correct substitution A1
eg \(4(2) + 3(1) + ( – 1)(11),{\text{ }}8 + 3 – 11\)
calculating \(a \cdot b = 0\) A1
Note: Do not award the final A1 without evidence of calculation.
vectors are perpendicular AG N0
[5 marks]
Note: Candidates may take different approaches, which do not necessarily involve vectors.
In particular, most of the working could be done on a diagram. Award marks in line with the markscheme.
METHOD 1
attempt to find \(\overrightarrow {{\text{QP}}} \) or \(\overrightarrow {{\text{PQ}}} \) (M1)
correct working (may be seen on diagram) A1
eg \(\overrightarrow {{\text{QP}}} \) = \(\left( \begin{array}{c} – 4\\ – 3\\1\end{array} \right)\), \(\overrightarrow {{\text{PQ}}} \) = \(\left( \begin{array}{c}7\\5\\3\end{array} \right) – \left( \begin{array}{c}3\\2\\4\end{array} \right)\)
recognizing \({\text{R}}\) is on \({L_1}\) (seen anywhere) (R1)
eg on diagram
\({\text{Q}}\) and \({\text{R}}\) are equidistant from \({\text{P}}\) (seen anywhere) (R1)
eg \(\overrightarrow {{\text{QP}}} = \overrightarrow {{\text{PR}}} \), marked on diagram
correct working (A1)
eg \(\left( \begin{array}{c}3\\2\\4\end{array} \right) – \left( \begin{array}{c}7\\5\\3\end{array} \right) = \left( \begin{array}{c}x\\y\\z\end{array} \right) – \left( \begin{array}{c}3\\2\\4\end{array} \right),\left( \begin{array}{c} – 4\\ – 3\\1\end{array} \right) + \left( \begin{array}{c}3\\2\\4\end{array} \right)\)
\({\text{R}}(–1, –1, 5)\) (accept position vector) A1 N3
METHOD 2
recognizing \({\text{R}}\) is on \({L_1}\) (seen anywhere) (R1)
eg on diagram
\({\text{Q}}\) and \({\text{R}}\) are equidistant from \({\text{P}}\) (seen anywhere) (R1)
eg \({\text{P}}\) midpoint of \({\text{QR}}\), marked on diagram
valid approach to find one coordinate of mid-point (M1)
eg \({x_p} = \frac{{{x_Q} + {x_R}}}{2},{\text{ }}2{y_p} = {y_Q} + {y_R},{\text{ }}\frac{1}{2}\left( {{z_Q} + {z_R}} \right)\)
one correct substitution A1
eg \({x_R} = 3 + (3 – 7),{\text{ }}2 = \frac{{5 + {y_R}}}{2},{\text{ }}4 = \frac{1}{2}(z + 3)\)
correct working for one coordinate (A1)
eg \({x_R} = 3 – 4,{\text{ }}4 – 5 = {y_R},{\text{ }}8 = (z + 3)\)
\({\text{R}} (-1, -1, 5)\) (accept position vector) A1 N3
[6 marks]
Question
The following diagram shows two perpendicular vectors u and v.
Let \(w = u – v\). Represent \(w\) on the diagram above.
Given that \(u = \left( \begin{array}{c}3\\2\\1\end{array} \right)\) and \(v = \left( \begin{array}{c}5\\n\\3\end{array} \right)\), where \(n \in \mathbb{Z}\), find \(n\).
Answer/Explanation
Markscheme
METHOD 1
A1A1 N2
Note: Award A1 for segment connecting endpoints and A1 for direction (must see arrow).
METHOD 2
A1A1 N2
Notes: Award A1 for segment connecting endpoints and A1 for direction (must see arrow).
Additional lines not required.
[2 marks]
evidence of setting scalar product equal to zero (seen anywhere) R1
eg u \( \cdot \) v \( = 0,{\text{ }}15 + 2n + 3 = 0\)
correct expression for scalar product (A1)
eg \(3 \times 5 + 2 \times n + 1 \times 3,{\text{ }}2n + 18 = 0\)
attempt to solve equation (M1)
eg \(2n = – 18\)
\(n = – 9\) A1 N3
[4 marks]
Question
Let \(u = 6i + 3j + 6k\) and \(v = 2i + 2j + k\).
Find
(i) \(u \bullet v\);
(ii) \(\left| {{u}} \right|\);
(iii) \(\left| {{v}} \right|\).
Find the angle between \({{u}}\) and \({{v}}\).
Answer/Explanation
Markscheme
(i) correct substitution (A1)
eg\(\;\;\;6 \times 2 + 3 \times 2 + 6 \times 1\)
\(u \bullet v = 24\) A1 N2
(ii) correct substitution into magnitude formula for \({{u}}\) or \({{v}}\) (A1)
eg\(\;\;\;\sqrt {{6^2} + {3^2} + {6^2}} ,{\text{ }}\sqrt {{2^2} + {2^2} + {1^2}} \), correct value for \(\left| {{v}} \right|\)
\(\left| {{u}} \right| = 9\) A1 N2
(iii) \(\left| {{v}} \right| = 3\) A1 N1
[5 marks]
correct substitution into angle formula (A1)
eg\(\;\;\;\frac{{24}}{{9 \times 3}},{\text{ }}0.\bar 8\)
\(0.475882,{\text{ }}27.26604^\circ \) A1 N2
\(0.476,{\text{ }}27.3^\circ \)
[2 marks]
Total [7 marks]
Question
The points A and B lie on a line \(L\), and have position vectors \(\left( {\begin{array}{*{20}{c}} { – 3} \\ { – 2} \\ 2 \end{array}} \right)\) and \(\left( {\begin{array}{*{20}{c}} 6 \\ 4 \\ { – 1} \end{array}} \right)\) respectively. Let O be the origin. This is shown on the following diagram.
The point C also lies on \(L\), such that \(\overrightarrow {{\text{AC}}} = 2\overrightarrow {{\text{CB}}} \).
Let \(\theta \) be the angle between \(\overrightarrow {{\text{AB}}} \) and \(\overrightarrow {{\text{OC}}} \).
Let D be a point such that \(\overrightarrow {{\text{OD}}} = k\overrightarrow {{\text{OC}}} \), where \(k > 1\). Let E be a point on \(L\) such that \({\rm{C\hat ED}}\) is a right angle. This is shown on the following diagram.
Find \(\overrightarrow {{\text{AB}}} \).
Show that \(\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 2 \\ 0 \end{array}} \right)\).
Find \(\theta \).
(i) Show that \(\left| {\overrightarrow {{\text{DE}}} } \right| = (k – 1)\left| {\overrightarrow {{\text{OC}}} } \right|\sin \theta \).
(ii) The distance from D to line \(L\) is less than 3 units. Find the possible values of \(k\).
Answer/Explanation
Markscheme
valid approach (addition or subtraction) (M1)
eg\(\,\,\,\,\,\)\({\text{AO}} + {\text{OB}},{\text{ B}} – {\text{A}}\)
\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 9 \\ 6 \\ { – 3} \end{array}} \right)\) A1 N2
[2 marks]
METHOD 1
valid approach using \(\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right)\) (M1)
eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}} {x + 3} \\ {y + 2} \\ {z – 2} \end{array}} \right),{\text{ }}\overrightarrow {{\text{CB}}} = \left( {\begin{array}{*{20}{c}} {6 – x} \\ {4 – y} \\ { – 1 – z} \end{array}} \right)\)
correct working A1
eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} {x + 3} \\ {y + 2} \\ {z – 2} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {12 – 2x} \\ {8 – 2y} \\ { – 2 – 2z} \end{array}} \right)\)
all three equations A1
eg\(\,\,\,\,\,\)\(x + 3 = 12 – 2x,{\text{ }}y + 2 = 8 – 2y,{\text{ }}z – 2 = – 2 – 2z\),
\(\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 2 \\ 0 \end{array}} \right)\) AG N0
METHOD 2
valid approach (M1)
eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{OC}}} – \overrightarrow {{\text{OA}}} = 2\left( {\overrightarrow {{\text{OB}}} – \overrightarrow {{\text{OC}}} } \right)\)
correct working A1
eg\(\,\,\,\,\,\)\(3\overrightarrow {{\text{OC}}} = 2\overrightarrow {{\text{OB}}} + \overrightarrow {{\text{OA}}} \)
correct substitution of \(\overrightarrow {{\text{OB}}} \) and \(\overrightarrow {{\text{OA}}} \) A1
eg\(\,\,\,\,\,\)\(3\overrightarrow {{\text{OC}}} = 2\left( {\begin{array}{*{20}{c}} 6 \\ 4 \\ { – 1} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} { – 3} \\ { – 2} \\ 2 \end{array}} \right),{\text{ }}3\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} 9 \\ 6 \\ 0 \end{array}} \right)\)
\(\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 2 \\ 0 \end{array}} \right)\) AG N0
METHOD 3
valid approach (M1)
eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{AC}}} = \frac{2}{3}\overrightarrow {{\text{AB}}} \), diagram, \(\overrightarrow {{\text{CB}}} = \frac{1}{3}\overrightarrow {{\text{AB}}} \)
correct working A1
eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}} 6 \\ 4 \\ { – 2} \end{array}} \right),{\text{ }}\overrightarrow {{\text{CB}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 2 \\ { – 1} \end{array}} \right)\)
correct working involving \(\overrightarrow {{\text{OC}}} \) A1
eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} { – 3} \\ { – 2} \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 6 \\ 4 \\ { – 2} \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} 6 \\ 4 \\ { – 1} \end{array}} \right) – \left( {\begin{array}{*{20}{c}} 3 \\ 2 \\ { – 1} \end{array}} \right)\)
\(\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 2 \\ 0 \end{array}} \right)\) AG N0
[3 marks]
finding scalar product and magnitudes (A1)(A1)(A1)
scalar product \( = (9 \times 3) + (6 \times 2) + ( – 3 \times 0){\text{ }}( = 39)\)
magnitudes \(\sqrt {81 + 36 + 9} {\text{ }}( = 11.22),{\text{ }}\sqrt {9 + 4} {\text{ }}( = 3.605)\)
substitution into formula M1
eg\(\,\,\,\,\,\)\(\cos \theta = \frac{{(9 \times 3) + 12}}{{\sqrt {126} \times \sqrt {13} }}\)
\(\theta = 0.270549{\text{ }}({\text{accept }}15.50135^\circ )\)
\(\theta = 0.271{\text{ }}({\text{accept }}15.5^\circ )\) A1 N4
[5 marks]
(i) attempt to use a trig ratio M1
eg\(\,\,\,\,\,\)\(\sin \theta = \frac{{{\text{DE}}}}{{{\text{CD}}}},{\text{ }}\left| {\overrightarrow {{\text{CE}}} } \right| = \left| {\overrightarrow {{\text{CD}}} } \right|\cos \theta \)
attempt to express \(\overrightarrow {{\text{CD}}} \) in terms of \(\overrightarrow {{\text{OC}}} \) M1
eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{OC}}} + \overrightarrow {{\text{CD}}} = \overrightarrow {{\text{OD}}} ,{\text{ OC}} + {\text{CD}} = {\text{OD}}\)
correct working A1
eg\(\,\,\,\,\,\)\(\left| {k\overrightarrow {{\text{OC}}} – \overrightarrow {{\text{OC}}} } \right|\sin \theta \)
\(\left| {\overrightarrow {{\text{DE}}} } \right| = (k – 1)\left| {\overrightarrow {{\text{OC}}} } \right|\sin \theta \) AG N0
(ii) valid approach involving the segment DE (M1)
eg\(\,\,\,\,\,\)recognizing \(\left| {\overrightarrow {{\text{DE}}} } \right| < 3,{\text{ DE}} = 3\)
correct working (accept equation) (A1)
eg\(\,\,\,\,\,\)\((k – 1)(\sqrt {13} )\sin 0.271 < 3,{\text{ }}k – 1 = 3.11324\)
\(1 < k < 4.11{\text{ }}({\text{accept }}k < 4.11{\text{ but not }}k = 4.11)\) A1 N2
[6 marks]