IB DP Maths Topic 4.2 The scalar product of two vectors SL Paper 2

 

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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 \(\overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}}
6\\
5\\
3
\end{array}} \right)\) .

[5]
a(i), (ii) and (iii).

Find the coordinates of point C.

[3]
b.

(i)     Find \(\overrightarrow {{\rm{AB}}} \bullet \overrightarrow {{\rm{AD}}} \).

(ii)    Hence find angle A.

[7]
c(i) and (ii).

Hence, or otherwise, find the area of the parallelogram.

[3]
d.
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]

a(i), (ii) and (iii).

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]

b.

(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]

c(i) and (ii).

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]

d.

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)     \(\overrightarrow {{\rm{AC}}} \) .

[3]
a.

Find the value of \(a\) for which \({\rm{q}} = \frac{\pi }{2}\) .

[4]
b.

i. Show that \(\cos q = \frac{{2a + 14}}{{\sqrt {14{a^2} + 280} }}\) .

ii. Hence, find the value of a for which \({\rm{q}} = 1.2\) .

[8]
c.

Hence, find the value of a for which \({\rm{q}} = 1.2\) .

[4]
c.ii.
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]

a.

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]

b.

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]

c.

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]

c.ii.

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 \({\text{P}}\).

[6]
a.

Show that the lines are perpendicular.

[5]
b.

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 \({\text{R}}\).

[6]
c.
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]

a.

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]

b.

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]

c.

Question

The following diagram shows two perpendicular vectors u and v.

Let \(w = u – v\). Represent \(w\) on the diagram above.

[2]
a.

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 \(n\).

[4]
b.
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]

a.

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]

b.

Question

Let \(u = 6i + 3j + 6k\) and \(v = 2i + 2j + k\).

Find

(i)     \(u \bullet v\);

(ii)     \(\left| {{u}} \right|\);

(iii)     \(\left| {{v}} \right|\).

[5]
a.

Find the angle between \({{u}}\) and \({{v}}\).

[2]
b.
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]

a.

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]

b.

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.

M16/5/MATME/SP2/ENG/TZ1/10

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.

M16/5/MATME/SP2/ENG/TZ1/10.d

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

[2]
a.

Show that \(\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 2 \\ 0 \end{array}} \right)\).

[[N/A]]
b.

Find \(\theta \).

[5]
c.

(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 \(k\).

[6]
d.
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]

a.

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, \(\overrightarrow {{\text{CB}}}  = \frac{1}{3}\overrightarrow {{\text{AB}}} \)

M16/5/MATME/SP2/ENG/TZ1/10.b/M

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]

b.

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]

c.

(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]

d.
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