IB DP Maths Topic 4.2 The angle between two vectors SL Paper 2

 

https://play.google.com/store/apps/details?id=com.iitianDownload IITian Academy  App for accessing Online Mock Tests and More..

Question

The point O has coordinates (0 , 0 , 0) , point A has coordinates (1 , – 2 , 3) and point B has coordinates (– 3 , 4 , 2) .

(i)     Show that \(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
{ – 4}\\
6\\
{ – 1}
\end{array}} \right)\) .

(ii)    Find \({\rm{B}}\widehat {\rm{A}}{\rm{O}}\) .

[8]
a(i) and (ii).

The line \({L_1}\) has equation \(\left( {\begin{array}{*{20}{c}}
x\\
y\\
z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{ – 3}\\
4\\
2
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
{ – 4}\\
6\\
{ – 1}
\end{array}} \right)\) .

Write down the coordinates of two points on \({L_1}\) .

[2]
b.

The line \({L_2}\) passes through A and is parallel to \(\overrightarrow {{\rm{OB}}} \) .

(i)     Find a vector equation for \({L_2}\) , giving your answer in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) . 

(ii)    Point \(C(k, – k,5)\) is on  \({L_2}\) . Find the coordinates of C.

[6]
c(i) and (ii).

The line \({L_3}\) has equation \(\left( {\begin{array}{*{20}{c}}
x\\
y\\
z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
3\\
{ – 8}\\
0
\end{array}} \right) + p\left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
{ – 1}
\end{array}} \right)\) and passes through the point C. 

 Find the value of p at C.

[2]
d.
Answer/Explanation

Markscheme

(i) evidence of approach     M1

e.g. \(\overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OB}}} = \overrightarrow {{\rm{AB}}} \)

\(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
{ – 4}\\
6\\
{ – 1}
\end{array}} \right)\)    AG     N0

(ii) for choosing correct vectors, (\(\overrightarrow {{\rm{AO}}} \) with \(\overrightarrow {{\rm{AB}}} \) or \(\overrightarrow {{\rm{OA}}} \) with \(\overrightarrow {{\rm{BA}}} \) )     (A1)(A1)

Note: Using \(\overrightarrow {{\rm{AO}}} \) with \(\overrightarrow {{\rm{BA}}} \) will lead to \(\pi  – 0.799\) . If they then say \({\rm{B}}\widehat {\rm{A}}{\rm{O}} = 0.799\) , this is a correct solution.

calculating \(\overrightarrow {{\rm{AO}}} \bullet \overrightarrow {{\rm{AB}}} \) , \(\left| {\overrightarrow {{\rm{AO}}} } \right|\) , \(\left| {\overrightarrow {{\rm{AB}}} } \right|\)     (A1)(A1)(A1)

e.g. \({d_1} \bullet {d_2} = ( – 1)( – 4) + (2)(6) + ( – 3)( – 1)( = 19)\)

\(\left| {{d_1}} \right| = \sqrt {{{( – 1)}^2} + {2^2} + {{( – 3)}^2}} ( = \sqrt {14} )\) , \(\left| {{d_2}} \right| = \sqrt {{{( – 4)}^2} + {6^2} + {{( – 1)}^2}} ( = \sqrt {53} )\)

evidence of using the formula to find the angle     M1

e.g. \(\cos \theta = \frac{{( – 1)( – 4) + (2)(6) + ( – 3)( – 1)}}{{\sqrt {{{( – 1)}^2} + {2^2} + {{( – 3)}^2}} \sqrt {{{( – 4)}^2} + {6^2} + {{( – 1)}^2}} }}\) , \(\frac{{19}}{{\sqrt {14} \sqrt {53} }}\) , \(0.69751 \ldots \)

\({\rm{B}}\widehat {\rm{A}}{\rm{O}} = 0.799\) radians (accept \(45.8^\circ \))     A1     N3

[8 marks]

a(i) and (ii).

two correct answers     A1A1

e.g. (1, \( – 2\), 3) , (\( – 3\), 4, 2) , (\( – 7\), 10, 1), (\( – 11\), 16, 0)     N2

[2 marks]

b.

(i) \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
3
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
{ – 3}\\
4\\
2
\end{array}} \right)\)     A2     N2

(ii) C on \({L_2}\) , so \(\left( {\begin{array}{*{20}{c}}
k\\
{ – k}\\
5
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
3
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
{ – 3}\\
4\\
2
\end{array}} \right)\)     (M1)

evidence of equating components (A1)

e.g. \(1 – 3t = k\) , \( – 2 + 4t = – k\) , \(5 = 3 + 2t\)

one correct value \(t = 1\) , \(k = – 2\) (seen anywhere)     (A1)

coordinates of C are \(( – 2{\text{, }}2{\text{, }}5)\)     A1     N3

[6 marks]

c(i) and (ii).

for setting up one (or more) correct equation using \(\left( {\begin{array}{*{20}{c}}
{ – 2}\\
2\\
5
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
3\\
{ – 8}\\
0
\end{array}} \right) + p\left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
{ – 1}
\end{array}} \right)\)     (M1)

e.g. \(3 + p = – 2\) , \( – 8 – 2p = 2\) , \( – p = 5\)

\(p = – 5\)     A1     N2

[2 marks]

d.

Question

Consider the points P(2, −1, 5) and Q(3, − 3, 8). Let \({L_1}\) be the line through P and Q.

Show that \(\overrightarrow {{\rm{PQ}}}  = \left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
3
\end{array}} \right)\) .

[1]
a.

The line \({L_1}\) may be represented by \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
  3 \\
  { – 3} \\
  8
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
  1 \\
  { – 2} \\
  3
\end{array}} \right)\) .

(i)     What information does the vector \(\left( {\begin{array}{*{20}{c}}
3\\
{ – 3}\\
8
\end{array}} \right)\) give about \({L_1}\) ?

(ii)    Write down another vector representation for \({L_1}\) using \(\left( {\begin{array}{*{20}{c}}
3\\
{ – 3}\\
8
\end{array}} \right)\) .

[3]
b.

The point \({\text{T}}( – 1{\text{, }}5{\text{, }}p)\) lies on \({L_1}\) .

Find the value of \(p\) .

[3]
c.

The point T also lies on \({L_2}\) with equation \(\left( {\begin{array}{*{20}{c}}
x\\
y\\
z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{ – 3}\\
9\\
2
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
q
\end{array}} \right)\) .

Show that \(q = – 3\) .

[3]
d.

Let \(\theta \) be the obtuse angle between \({L_1}\) and \({L_2}\) . Calculate the size of \(\theta \) .

[7]
e.
Answer/Explanation

Markscheme

evidence of correct approach     A1

e.g. \(\overrightarrow {{\rm{PQ}}}  = \overrightarrow {{\rm{OQ}}} – \overrightarrow {{\rm{OP}}} \) , \(\left( {\begin{array}{*{20}{c}}
3\\
{ – 3}\\
8
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
2\\
{ – 1}\\
5
\end{array}} \right)\)

\(\overrightarrow {{\rm{PQ}}}  = \left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
3
\end{array}} \right)\)     AG     N0

[1 mark]

a.

(i) correct description     R1     N1

e.g. reference to \(\left( {\begin{array}{*{20}{c}}
3\\
{ – 3}\\
8
\end{array}} \right)\) being the position vector of a point on the line, a vector to the line, a point on the line.

(ii) any correct expression in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\)    A2     N2

where \({\boldsymbol{a}}\) is \(\left( {\begin{array}{*{20}{c}}
  3 \\
  { – 3} \\
  8
\end{array}} \right)\) , and \({\boldsymbol{b}}\) is a scalar multiple of \(\left( {\begin{array}{*{20}{c}}
  1 \\
  { – 2} \\
  3
\end{array}} \right)\)

e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
  3 \\
  { – 3} \\
  8
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
  { – 1} \\
  2 \\
  { – 3}
\end{array}} \right)\) , \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
  {3 + 2s} \\
  { – 3 – 4s} \\
  {8 + 6s}
\end{array}} \right)\)

[3 marks]

b.

one correct equation     (A1)

e.g. \(3 + s = – 1\) , \( – 3 – 2s = 5\)

\(s = – 4\)     A1

\(p = – 4\)     A1     N2

[3 marks]

c.

one correct equation     A1

e.g. \( – 3 + t = – 1\) , \(9 – 2t = 5\)

\(t = 2\)     A1

substituting \(t = 2\)

e.g. \(2 + 2q = – 4\) , \(2q =  – 6\)     A1

\(q = – 3\)     AG     N0

[3 marks]

d.

choosing correct direction vectors \(\left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
3
\end{array}} \right)\) and \(\left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
{ – 3}
\end{array}} \right)\)     (A1)(A1)

finding correct scalar product and magnitudes     (A1)(A1)(A1)

scalar product \((1)(1) + ( – 2)( – 2) + ( – 3)(3)\)     \(( = – 4)\)

magnitudes \(\sqrt {{1^2} + {{( – 2)}^2} + {3^2}} \) \( = \sqrt {14} \) , \(\sqrt {{1^2} + {{( – 2)}^2} + {{( – 3)}^2}} \) \( = \sqrt {14} \)

evidence of substituting into scalar product     M1

e.g. \(\cos \theta  = \frac{{ – 4}}{{3.741 \ldots  \times 3.741 \ldots }}\)

\(\theta  = 1.86\) radians (or \(107^\circ \))    A1     N4

[7 marks]

e.

Question

Let \({\boldsymbol{v}} = \left( {\begin{array}{*{20}{c}}
2\\
{ – 3}\\
6
\end{array}} \right)\) and  \({\boldsymbol{w}} = \left( {\begin{array}{*{20}{c}}
k\\
{ – 2}\\
4
\end{array}} \right)\) , for \(k > 0\) . The angle between v and w is \(\frac{\pi }{3}\) .

Find the value of \(k\) .

Answer/Explanation

Markscheme

correct substitutions for \({\boldsymbol{v}} \bullet {\boldsymbol{w}}\) ; \(\left| {\boldsymbol{v}} \right|\) ; \(\left| {\boldsymbol{w}} \right|\)     (A1)(A1)(A1)

e.g. \(2k + ( – 3) \times ( – 2) + 6 \times 4\) , \(2k + 30\) ; \(\sqrt {{2^2} + {{( – 3)}^2} + {6^2}} \) , \(\sqrt {49} \) ; \(\sqrt {{k^2} + {{( – 2)}^2} + {4^2}} \) , \(\sqrt {{k^2} + 20} \)

evidence of substituting into the formula for scalar product     (M1)

e.g. \(\frac{{2k + 30}}{{7 \times \sqrt {{k^2} + 20} }}\)

correct substitution     A1

e.g. \(\cos \frac{\pi }{3} = \frac{{2k + 30}}{{7 \times \sqrt {{k^2} + 20} }}\)

\(k = 18.8\)     A2     N5

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

[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

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.

Question

Let \(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 4 \\ 1 \\ 2 \end{array}} \right)\).

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

[2]
a.

Let \(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 0 \\ 0 \end{array}} \right)\). Find \({\rm{B\hat AC}}\).

[4]
b.
Answer/Explanation

Markscheme

correct substitution     (A1)

eg\(\,\,\,\,\,\)\(\sqrt {{4^2} + {1^2} + {2^2}} \)

4.58257

\(\left| {\overrightarrow {{\text{AB}}} } \right| = \sqrt {21} \) (exact), 4.58     A1     N2

[2 marks]

a.

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

scalar product \( = (4 \times 3) + (1 \times 0) + (2 \times 0){\text{ }}( = 12)\)

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

substituting their values into cosine formula     (M1)

eg cos B\(\hat A\)C\({\text{ = }}\frac{{4 \times 3 + 0 + 0}}{{\sqrt {{3^2}}  \times \sqrt {21} }},{\text{ }}\frac{4}{{\sqrt {21} }},{\text{ }}\cos \theta  = 0.873\)

0.509739 (29.2059°)

\({\rm{B\hat AC}} = 0.510\) (29.2°)     A1     N2

[4 marks]

b.
Scroll to Top