Find

(i)     \(u \bullet v\);

(ii)     \(\left| {{u}} \right|\);

(iii)     \(\left| {{v}} \right|\).

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

(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]

Many candidates performed well in this question. Some candidates were unfamiliar with the basis vector notation and wrongly substituted the i-j-k into the formulas. Others occasionally assumed that the magnitude could be negative.

Many candidates performed well in this question. Some candidates were unfamiliar with the basis vector notation and wrongly substituted the i-j-k into the formulas. Others occasionally assumed that the magnitude could be negative.

Two lines with equations \({{\boldsymbol{r}}_1} = \left( {\begin{array}{*{20}{c}}
2\\
3\\
{ - 1}
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
5\\
{ - 3}\\
2
\end{array}} \right)\) and \({{\boldsymbol{r}}_2} = \left( {\begin{array}{*{20}{c}}
9\\
2\\
2
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
{ - 3}\\
5\\
{ - 1}
\end{array}} \right)\) intersect at the point P. Find the coordinates of P.

evidence of appropriate approach     (M1)

e.g. \(\left( {\begin{array}{*{20}{c}}
2\\
3\\
{ - 1}
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
5\\
{ - 3}\\
2
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
9\\
2\\
2
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
{ - 3}\\
5\\
{ - 1}
\end{array}} \right)\)

two correct equations     A1A1

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

attempting to solve the equations     (M1)

one correct parameter \(s = 2\) , \(t =  - 1\)     A1

P is \((12, - 3,3)\) (accept \(\left( {\begin{array}{*{20}{c}}
{12}\\
{ - 3}\\
3
\end{array}} \right)\))      A1     N3

[6 marks] 

If this topic had been taught well then the candidates scored highly. The question was either well answered or not at all. Many candidates did not understand what was needed and tried to find the length of vectors or mid-points of lines. The other most common mistake was to use the values of the parameters to write the coordinates as \({\text{P}}(2{\text{, }} - 1)\). 

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 \(k\) .

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]

For the most part, this question was well done and candidates had little difficulty finding the scalar product, the appropriate magnitudes and then correctly substituting into the formula for the angle between vectors. However, few candidates were able to solve the resulting equation using their GDCs to obtain the correct answer. Problems arose when candidates attempted to solve the resulting equation analytically.

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

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

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]

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]

(i)     Write down the coordinates of A.

(ii)    Find the speed of the airplane in \({\text{m}}{{\text{s}}^{ - 1}}\).

After seven seconds the airplane passes through a point B.

(i)     Find the coordinates of B.

(ii)    Find the distance the airplane has travelled during the seven seconds.

Airplane 2 passes through a point C. Its position q seconds after it passes through C is given by \(\left( {\begin{array}{*{20}{c}}
x\\
y\\
z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
2\\
{ - 5}\\
8
\end{array}} \right) + q\left( {\begin{array}{*{20}{c}}
{ - 1}\\
2\\
a
\end{array}} \right),a \in \mathbb{R}\) .

The angle between the flight paths of Airplane 1 and Airplane 2 is \({40^ \circ }\) . Find the two values of a.

(i) (3, \( - 4\), 0)     A1     N1

(ii) choosing velocity vector \(\left( {\begin{array}{*{20}{c}}
{ - 2}\\
3\\
1
\end{array}} \right)\)     (M1)

finding magnitude of velocity vector     (A1)

e.g. \(\sqrt {{{( - 2)}^2} + {3^2} + {1^2}} \) , \(\sqrt {4 + 9 + 1} \)

speed \(= 3.74\) \(\left( {\sqrt {14} } \right)\)     A1     N2

[4 marks]

(i) substituting \(p = 7\)     (M1)

\({\text{B}} = ( - 11{\text{, }}17{\text{, }}7)\)     A1     N2

(ii) METHOD 1

appropriate method to find \(\overrightarrow {{\rm{AB}}} \) or \(\overrightarrow {{\rm{BA}}} \)     (M1)

e.g. \(\overrightarrow {{\rm{AO}}}  + \overrightarrow {{\rm{OB}}} \) , \({\rm{A}} - {\rm{B}}\)

\(\overrightarrow {{\rm{AB}}}  = \left( {\begin{array}{*{20}{c}}
{ - 14}\\
{21}\\
7
\end{array}} \right)\) or \(\overrightarrow {{\rm{BA}}}  = \left( {\begin{array}{*{20}{c}}
{14}\\
{ - 21}\\
{ - 7}
\end{array}} \right)\)     (A1)

distance \(= 26.2\) \(\left( {7\sqrt {14} } \right)\)     A1     N3

METHOD 2

evidence of applying distance is speed × time     (M2)

e.g. \(3.74 \times 7\)

distance \(= 26.2\) \(\left( {7\sqrt {14} } \right)\)     A1     N3

METHOD 3

attempt to find AB² , AB     (M1)

e.g. \({(3 - ( - 11))^2} + {( - 4 - 17)^2} + (0 - 7){)^2}\) , \(\sqrt {{{(3 - ( - 11))}^2} + {{( - 4 - 17)}^2} + (0 - 7){)^2}} \)

AB² \(= 686\), AB \(= \sqrt {686} \)     (A1)

distance AB \(= 26.2\) \(\left( {7\sqrt {14} } \right)\)     A1     N3

[5 marks]

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

\(\left| {\begin{array}{*{20}{c}}
{ - 1}\\
2\\
a
\end{array}} \right| = \sqrt {{a^2} + 5} \) , \(\left( {\begin{array}{*{20}{c}}
{ - 2}\\
3\\
1
\end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}}
{ - 1}\\
2\\
a
\end{array}} \right) = a + 8\)      (A1)(A1)

substituting     M1

e.g. \(\cos {40^ \circ } = \frac{{a + 8}}{{\sqrt {14} \sqrt {{a^2} + 5} }}\)

\(a = 3.21\) , \(a = - 0.990\)     A1A1     N3

[7 marks]

Many candidates demonstrated a good understanding of the vector equation of a line and its application to a kinematics problem by correctly answering the first two parts of this question.

Many candidates demonstrated a good understanding of the vector equation of a line and its application to a kinematics problem by correctly answering the first two parts of this question.

Some knew that speed and distance were magnitudes of vectors but chose the wrong vectors to calculate magnitudes.

Very few candidates were able to get the two correct answers in (c) even if they set up the equation correctly. Much contorted algebra was seen and little evidence of using the GDC to solve the equation. Many made simple algebraic errors by combining unlike terms in working with the scalar product (often writing \(8a\) rather than \(8 + a\) ) or the magnitude (often writing \(5{a^2}\) rather than \(5 + {a^2}\) ).

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

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 \(\left( {\begin{array}{*{20}{c}}
3\\
{ - 3}\\
8
\end{array}} \right)\) .

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

Find the value of \(p\) .

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 \(q = - 3\) .

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

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]

(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]

one correct equation     (A1)

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

\(s = - 4\)     A1

\(p = - 4\)     A1     N2

[3 marks]

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]

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]

Most candidates answered part (a) easily.

For part (b), a number of candidates stated that the vector was a "starting point," which misses the idea that it is a position vector to some point on the line.

Parts (c) and (d) proved accessible to many.

Parts (c) and (d) proved accessible to many.

For part (e), a surprising number of candidates chose incorrect vectors. Few candidates seemed to have a good conceptual understanding of the vector equation of a line.

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

Find the coordinates of C.

Write down a vector equation for \(L\).

Given that \(\left| {\overrightarrow {{\text{AB}}} } \right| = k\left| {\overrightarrow {{\text{AC}}} } \right|\), find \(k\).

The point D lies on \(L\) such that \(\left| {\overrightarrow {{\text{AB}}} } \right| = \left| {\overrightarrow {{\text{BD}}} } \right|\). Find the possible coordinates of D.

valid approach     (M1)

eg\(\,\,\,\,\,\)\({\text{B}} - {\text{A}},{\text{ AO}} + {\text{OB}},{\text{ }}\left( {\begin{array}{*{20}{c}} { - 9} \\ 9 \\ { - 6} \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 1 \\ 5 \\ { - 7} \end{array}} \right)\)

\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} { - 10} \\ 4 \\ 1 \end{array}} \right)\)     A1     N2

[2 marks]

valid approach     (M1)

eg\(\,\,\,\,\,\)\({\text{OC}} = {\text{OA}} + {\text{AC}},{\text{ }}\left( {\begin{array}{*{20}{c}} {1 + 6} \\ {5 - 4} \\ { - 7 + 0} \end{array}} \right)\)

\({\text{C}}(7,{\text{ }}1,{\text{ }} - 7)\)    A1     N2

[2 marks]

any correct equation in the form r \( = \) a \( + t\)b (accept any parameter for \(t\)

where a is \(\left( {\begin{array}{*{20}{c}} { - 9} \\ 9 \\ { - 6} \end{array}} \right)\), and b is a scalar multiple of \(\left( {\begin{array}{*{20}{c}} 6 \\ { - 4} \\ 0 \end{array}} \right)\)     A2     N2

eg\(\,\,\,\,\,\)r \( = \left( {\begin{array}{*{20}{c}} { - 9} \\ 9 \\ { - 6} \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 6 \\ { - 4} \\ 0 \end{array}} \right)\), r \( =  - 9\)i \( + 9\)j \( - 6\)k \( + s(6\)i \( - 4\)j \( + 0\)k\()\)

[2 marks]

correct magnitudes     (A1)(A1)

eg\(\,\,\,\,\,\)\(\sqrt {{{( - 10)}^2} + {{( - 4)}^2} + {1^2}} ,{\text{ }}\sqrt {{6^2} + {{( - 4)}^2} + {{(0)}^2}} ,{\text{ }}\sqrt {{{10}^2} + {4^2} + 1} ,{\text{ }}\sqrt {{6^2} + {4^2}} \)

\(k = \frac{{\sqrt {117} }}{{\sqrt {52} }}{\text{ }}( = 1.5){\text{ }}({\text{exact}})\)    A1     N3

[3 marks]

correct interpretation of relationship between magnitudes     (A1)

eg\(\,\,\,\,\,\)\({\text{AB}} = 1.5{\text{AC}},{\text{ BD}} = 1.5{\text{AC}},{\text{ }}\sqrt {117}  = \sqrt {52{t^2}} \)

recognizing D can have two positions (may be seen in working)     R1

eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{BD}}}  = 1.5\overrightarrow {{\text{AC}}} \) and \(\overrightarrow {{\text{BD}}}  =  - 1.5\overrightarrow {{\text{AC}}} ,{\text{ }}t =  \pm 1.5\), diagram, two answers

valid approach (seen anywhere)     (M1)

eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{OD}}} = \overrightarrow {{\text{OB}}} + \overrightarrow {{\text{BD}}} {\text{, }}\left( {\begin{array}{*{20}{c}} { - 9} \\ 9 \\ { - 6} \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 6 \\ { - 4} \\ 0 \end{array}} \right),{\text{ }}\overrightarrow {{\text{BD}}} = k\left( {\begin{array}{*{20}{c}} 6 \\ { - 4} \\ 0 \end{array}} \right)\)

one correct expression for \(\overrightarrow {{\text{OD}}} \)     (A1)

eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{OD}}} = \left( {\begin{array}{*{20}{c}} { - 9} \\ 9 \\ { - 6} \end{array}} \right) + 1.5\left( {\begin{array}{*{20}{c}} 6 \\ { - 4} \\ 0 \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} { - 9} \\ 9 \\ { - 6} \end{array}} \right) - 1.5\left( {\begin{array}{*{20}{c}} 6 \\ { - 4} \\ 0 \end{array}} \right)\)

\({\text{D}} = (0,{\text{ }}3,{\text{ }} - 6),{\text{ D}} = ( - 18,{\text{ }}15,{\text{ }} - 6)\) (accept position vectors)     A1A1     N3

[6 marks]

Parts (a) and (b) were attempted by the great majority of the candidates and appropriate approaches were seen, earning at least the method marks.

Parts (a) and (b) were attempted by the great majority of the candidates and appropriate approaches were seen, earning at least the method marks.

Part (c) was generally well done, with many candidates writing the equation of the line as \(L = \) a \( + t\)b, losing one mark.

Part (d) was also well answered by a great majority of students. Even those candidates who had part (a) incorrect, could gain all the marks here.

Part (e) was the most challenging of the paper. For many a major problem was to set up the equation \(\sqrt {117}  = \sqrt {52{t^2}} \) and, hence, realize that D could have two positions.

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

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

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 \({L_1}\) .

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.

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.

(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]

two correct answers     A1A1

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

[2 marks]

(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]

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]

Part (ai) was done well by most students. Most knew how to approach finding the angle in part (aii). The problems occurred when the incorrect vectors were chosen. If the vectors being used were stated, then follow through marks could be given.

Part (b) was well done.

In part (ci), the error that occurred most often was the incorrect choice for the direction vector.

Those that were able to find the coordinates in part (cii) were also able to be successful in part (d).

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

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]

(i)     Find \(\overrightarrow {{\rm{OB}}} \) .

(ii)    Find \(\overrightarrow {{\rm{OF}}} \) .

(iii)   Show that \(\overrightarrow {{\rm{AG}}} = - 4{\boldsymbol{i}} + 3{\boldsymbol{j}} + 2{\boldsymbol{k}}\) .

Write down a vector equation for

(i)     the line OF;

(ii)    the line AG.

Find the obtuse angle between the lines OF and AG.

(i) valid approach     (M1)

e.g. \({\rm{OA + OB}}\)

\(\overrightarrow {{\rm{OB}}} = 4{\boldsymbol{i}} + 3{\boldsymbol{j}}\)     A1     N2

(ii) valid approach     (M1)

e.g. \(\overrightarrow {{\rm{OA}}} {\rm{ + }}\overrightarrow {{\rm{AB}}} {\rm{ + }}\overrightarrow {{\rm{BF}}} \) ; \(\overrightarrow {{\rm{OB}}} {\rm{ + }}\overrightarrow {{\rm{BF}}} \) ; \(\overrightarrow {{\rm{OC}}} {\rm{ + }}\overrightarrow {{\rm{CG}}} {\rm{ + }}\overrightarrow {{\rm{GF}}} \)

\(\overrightarrow {{\rm{OF}}} = 4{\boldsymbol{i}} + 3{\boldsymbol{j}} + 2{\boldsymbol{k}}\)     A1     N2

(iii) correct approach     A1

e.g. \(\overrightarrow {{\rm{AO}}} {\rm{ + }}\overrightarrow {{\rm{OC}}} {\rm{ + }}\overrightarrow {{\rm{CG}}} \) ; \(\overrightarrow {{\rm{AB}}} {\rm{ + }}\overrightarrow {{\rm{BF}}} {\rm{ + }}\overrightarrow {{\rm{FG}}} \) ; \(\overrightarrow {{\rm{AB}}} {\rm{ + }}\overrightarrow {{\rm{BC}}} {\rm{ + }}\overrightarrow {{\rm{CG}}} \)

\(\overrightarrow {{\rm{AG}}} = - 4{\boldsymbol{i}} + 3{\boldsymbol{j}} + 2{\boldsymbol{k}}\)     AG     N0

[5 marks]

(i) any correct equation for (OF) in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\)     A2     N2

where \({\boldsymbol{a}}\) is 0 or \(4{\boldsymbol{i}} + 3{\boldsymbol{j}} + 2{\boldsymbol{k}}\) , and \({\boldsymbol{b}}\) is a scalar multiple of \(4{\boldsymbol{i}} + 3{\boldsymbol{j}} + 2{\boldsymbol{k}}\)

e.g. \({\boldsymbol{r}} = t(4{\text{, }}3{\text{, }}2)\) , \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
{4t}\\
{3t}\\
{2t}
\end{array}} \right)\) , \({\boldsymbol{r}} = 4{\boldsymbol{i}} + 3{\boldsymbol{j}} + 2{\boldsymbol{k}} + t(4{\boldsymbol{i}} + 3{\boldsymbol{j}} + 2{\boldsymbol{k}})\)

(ii) any correct equation for (AG) in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + s{\boldsymbol{b}}\)     A2     N2

where \({\boldsymbol{a}}\) is \(4{\boldsymbol{i}}\) or \(3{\boldsymbol{j}} + 2{\boldsymbol{k}}\) and \({\boldsymbol{b}}\) is a scalar multiple of \( - 4{\boldsymbol{i}} + 3{\boldsymbol{j}} + 2{\boldsymbol{k}}\)

e.g. \({\boldsymbol{r}} = (4{\text{, }}0{\text{, }}0) + s( - 4{\text{, }}3{\text{, }}2)\) , \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
{4 - 4s}\\
{3s}\\
{2s}
\end{array}} \right)\) , \({\boldsymbol{r}} = 3{\boldsymbol{j}} + 2{\boldsymbol{k}} + s( - 4{\boldsymbol{i}} + 3{\boldsymbol{j}} + 2{\boldsymbol{k}})\)

[4 marks]

choosing correct direction vectors, \(\overrightarrow {{\rm{OF}}} \) and \(\overrightarrow {{\rm{AG}}} \)     (A1)(A1)

scalar product \( = - 16 + 9 + 4\) \(( = - 3)\)     (A1)

magnitudes \(\sqrt {{4^2} + {3^2} + {2^2}} \) , \(\sqrt {{{( - 4)}^2} + {3^2} + {2^2}} \) , \(\left( {\sqrt {29} ,\sqrt {29} } \right)\)     (A1)(A1)

substitution into formula     M1

e.g. \(\cos \theta  = \frac{{ - 16 + 9 + 4}}{{\left( {\sqrt {{4^2} + {3^2} + {2^2}} } \right) \times \sqrt {{{( - 4)}^2} + {3^2} + {2^2}} }} = \left( { - \frac{3}{{29}}} \right)\)

\(95.93777^\circ \) , \(1.67443{\text{ radians}}\)

\(\theta  = 95.9^\circ \) or \(1.67\)     A1     N4

[7 marks]

Although a large proportion of candidates managed to answer this question, their biggest challenge was the use of a proper notation to represent the vectors and vector equations of lines.

In part (a), finding \(\overrightarrow {{\rm{OB}}} \) and \(\overrightarrow {{\rm{OF}}} \) was generally well done, although many lost the mark for (iii) due to poor working or not clearly showing the result.

Part (b) was very poorly done. Not all the students recognized which correct position vectors they had to use to write the equations of the lines. It was seen that they frequently failed to present the equations in the required format, which prevented these candidates from achieving full marks. The notations generally seen were \({\text{AG}} = {\boldsymbol{a}} + {\boldsymbol{b}}t\) , \({\boldsymbol{r}} = 4 + t(4{\text{, }}3{\text{, }}2)\) or \(L = {\boldsymbol{a}} + {\boldsymbol{b}}t\) .

Most achieved the correct result in part (c) with many others gaining most of the marks as follow through from choosing incorrect vectors. Some students did not state which vectors had been used, another cause for losing marks. A few showed poor notation, including i, j and k in the working.

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 \({\rm{q}} = 1.2\) .

(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]

The majority of candidates successfully found the vectors between the given points in part (a).

In part (b), while most candidates correctly found the value of \(a\), many unnecessarily worked with the magnitudes of the vectors, sometimes leading to algebra errors.

Some candidates showed a minimum of working in part (c)(i); in a “show that” question, candidates need to ensure that their working clearly leads to the answer given. A common error was simplifying the magnitude of vector AC to \(\sqrt {20{a^2}} \) instead of \(\sqrt {20 + {a^2}} \) .

In part (c)(ii), a disappointing number of candidates embarked on a usually fruitless quest for an algebraic solution rather than simply solving the resulting equation with their GDC. Many of these candidates showed quite weak algebra manipulation skills, with errors involving the square root occurring in a myriad of ways.

In part (c)(ii), a disappointing number of candidates embarked on a usually fruitless quest for an algebraic solution rather than simply solving the resulting equation with their GDC. Many of these candidates showed quite weak algebra manipulation skills, with errors involving the square root occurring in a myriad of ways.

Let  \({\boldsymbol{v}} = 3{\boldsymbol{i}} + 4{\boldsymbol{j}} + {\boldsymbol{k}}\) and \({\boldsymbol{w}} = {\boldsymbol{i}} + 2{\boldsymbol{j}} - 3{\boldsymbol{k}}\) . The vector \({\boldsymbol{v}} + p{\boldsymbol{w}}\) is perpendicular to w. Find the value of p.

\(p{\boldsymbol{w}} = p{\boldsymbol{i}} + 2p{\boldsymbol{j}} - 3p{\boldsymbol{k}}\) (seen anywhere)     (A1)

attempt to find \({\boldsymbol{v}} + p{\boldsymbol{w}}\)     (M1)

e.g. \(3{\boldsymbol{i}} + 4{\boldsymbol{j}} + {\boldsymbol{k}} + p({\boldsymbol{i}} + 2{\boldsymbol{j}} - 3{\boldsymbol{k}})\)

collecting terms \((3 + p){\boldsymbol{i}} + (4 + 2p){\boldsymbol{j}} + (1 - 3p){\boldsymbol{k}}\)     A1

attempt to find the dot product     (M1)

e.g. \(1(3 + p) + 2(4 + 2p) - 3(1 - 3p)\)

setting their dot product equal to 0     (M1)

e.g. \(1(3 + p) + 2(4 + 2p) - 3(1 - 3p) = 0\)

simplifying     A1

e.g. \(3 + p + 8 + 4p - 3 + 9p = 0\) , \(14p + 8 = 0\)

\(p = - 0.571\) \(\left( { - \frac{8}{{14}}} \right)\)     A1     N3

[7 marks]

This question was very poorly done with many leaving it blank. Of those that did attempt it, most were able to find \({\boldsymbol{v}} + p{\boldsymbol{w}}\) but really did not know how to proceed from there. They tried many approaches, such as, finding magnitudes, using negative reciprocals, or calculating the angle between two vectors. A few had the idea that the scalar product should equal zero but had trouble trying to set it up.

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 \(k\).

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]

The majority of candidates had little difficulty with parts (a) and (c). The most common error in both these parts were unforced arithmetic errors and occasional misreads of the vectors.

In part (b), candidates who were successful used a variety of different approaches, and it was pleasing to see the vast majority of these being well reasoned, however, there were numerous unsuccessful responses including those who attempted to use the given vector to work backwards. A lack of appropriate vector notation often meant that ideas were not always clearly communicated.

The majority of candidates had little difficulty with parts (a) and (c). The most common error in both these parts were unforced arithmetic errors and occasional misreads of the vectors.

The majority of candidates struggled to make any progress in (d), with very few realizing that simple right-angled trigonometry could be used. Few were able to successfully express CD in terms of OC which was required to show the given result. Very few candidates attempted (d)(ii), with many unable to make the connection with results found in previous parts of the question.

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

Find an equation for \({L_1}\) in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) .

Find the angle between \({L_1}\) and \({L_2}\) .

The lines \({L_1}\) and \({L_2}\) intersect at point C. Find the coordinates of C.

appropriate approach     (M1)

e.g. \(\overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OB}}} \) , \(B - A\)

\(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
1\\
{ - 1}\\
1
\end{array}} \right)\)     A1     N2

[2 marks]

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

where \({\boldsymbol{b}}\) is a scalar multiple of \(\left( {\begin{array}{*{20}{c}}
1\\
{ - 1}\\
1
\end{array}} \right)\)

e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
1\\
{ - 1}\\
4
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
1\\
{ - 1}\\
1
\end{array}} \right)\) , \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
{2 + t}\\
{ - 2 - t}\\
{5 + t}
\end{array}} \right)\) , \({\boldsymbol{r}} = 2{\boldsymbol{i}} - 2{\boldsymbol{j}} + 5{\boldsymbol{k}} + t({\boldsymbol{i}} - {\boldsymbol{j}} + {\boldsymbol{k}})\)

 [2 marks]

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

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

scalar product \( = 1 \times 2 + - 1 \times 1 + 1 \times 3\) \(\left( { = 4} \right)\)

magnitudes \(\sqrt {{1^2} + {{( - 1)}^2} + {1^2}}{\text{  }}( = 1.73 \ldots )\) , \(\sqrt {4 + 1 + 9}{\text{  }}( = 3.74 \ldots )\)

substitution into \(\frac{{{\boldsymbol{u}} \bullet {\boldsymbol{v}}}}{{|{\boldsymbol{u}}||{\boldsymbol{v}}|}}\) (accept \(\theta = \frac{{{\boldsymbol{u}} \bullet {\boldsymbol{v}}}}{{|{\boldsymbol{u}}||{\boldsymbol{v}}|}}\) , but not \(\sin \theta = \frac{{{\boldsymbol{u}} \bullet {\boldsymbol{v}}}}{{|{\boldsymbol{u}}||{\boldsymbol{v}}|}}\) )     M1

e.g. \(\cos \theta = \frac{{1 \times 2 + - 1 \times 1 + 1 \times 3}}{{\sqrt {{1^2} + {{( - 1)}^2} + {1^2}} \sqrt {{2^2} + {1^2} + {3^2}} }}\) , \(\cos \theta = \frac{4}{{\sqrt {42} }}\)

\(\theta = 0.906\) \(({51.9^ \circ })\)     A1      N5

[7 marks]

METHOD 1 (from \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
  1 \\
  { - 1} \\
  4
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
  1 \\
  { - 1} \\
  1
\end{array}} \right)\) )

appropriate approach     (M1)

e.g. \({\boldsymbol{p}} = {\boldsymbol{r}}\) , \(\left( {\begin{array}{*{20}{c}}
1\\
{ - 1}\\
4
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
1\\
{ - 1}\\
1
\end{array}} \right)t = \left( {\begin{array}{*{20}{c}}
2\\
4\\
7
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
2\\
1\\
3
\end{array}} \right)s\)

two correct equations    A1A1

e.g. \(1 + t = 2 + 2s\) , \( - 1 - t = 4 + s\) , \(4 + t = 7 + 3s\)

attempt to solve     (M1)

one correct parameter     A1

e.g. \(t = - 3\) , \(s = - 2\)

C is \(( - 2{\text{, }}2{\text{, }}1)\)     A1     N3

METHOD 2 (from \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
  2 \\
  { - 2} \\
  5
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
  1 \\
  { - 1} \\
  1
\end{array}} \right)\) )

appropriate approach     (M1)

e.g. \({\boldsymbol{p}} = {\boldsymbol{r}}\) , \(\left( {\begin{array}{*{20}{c}}
2\\
{ - 2}\\
5
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
1\\
{ - 1}\\
1
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
2\\
4\\
7
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
2\\
1\\
3
\end{array}} \right)s\)

two correct equations     A1A1

e.g. \(2 + t = 2 + 2s\) , \( - 2 - t = 4 + s\) , \(5 + t = 7 + 3s\)

attempt to solve     (M1)

one correct parameter     A1

e.g. \(t = - 4\) , \(s = - 2\)

C is \(( - 2{\text{, }}2{\text{, }}1)\)     A1     N3

[6 marks]

Finding \(\overrightarrow {{\rm{AB}}} \) was generally well done, although some candidates reversed the subtraction.

In part (b) not all the candidates recognized that \(\overrightarrow {{\rm{AB}}} \) was the direction vector of the line, as some used the position vector of point B as the direction vector.

Many candidates successfully used scalar product and magnitudes in part (c), although a large number did choose vectors other than the direction vectors and many did not state clearly which vectors they were using.

Candidates who were comfortable on the first three parts often had little difficulty with the final part. While the resulting systems were easily solved algebraically, a surprising number of candidates did not check their solutions either manually or with technology. An occasionally seen error in the final part was using a midpoint to find C. Some candidates found the point of intersection in part (c) rather than in part (d), indicating a familiarity with the type of question but a lack of understanding of the concepts involved.

(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.

(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]

Candidates performed very well in this question, showing a strong ability to work with the algebra and geometry of vectors.

Candidates performed very well in this question, showing a strong ability to work with the algebra and geometry of vectors.

Some candidates were unable to find the scalar product in part (c), yet still managed to find the correct angle, able to use the formula in the information booklet without knowing that the scalar product is a part of that formula.

Few candidates considered that the area of the parallelogram is twice the area of a triangle, which is conveniently found using  \({\rm{B}}\widehat {\rm{A}}{\rm{D}}\) . In an effort to find base \(\times \) height , many candidates multiplied the magnitudes of  \(\overrightarrow {{\rm{AB}}} \) and  \(\overrightarrow {{\rm{AD}}} \) , missing that the height of a parallelogram is perpendicular to a base.

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 \(n\).

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]

Write down the line that is parallel to \({L_4}\) .

Write down the position vector of the point of intersection of \({L_1}\) and \({L_2}\) .

Given that \({L_1}\) is perpendicular to \({L_3}\) , find the value of a .

\({L_1}\)     A1     N1

[1 mark]

\(\left( \begin{array}{l}
1\\
2\\
3
\end{array} \right)\)     A1     N1

[1 mark]

choosing correct direction vectors     A1A1

e.g. \(\left( {\begin{array}{*{20}{c}}
3\\
{ - 2}\\
1
\end{array}} \right)\) , \(\left( {\begin{array}{*{20}{c}}
{ - 1}\\
2\\
{ - a}
\end{array}} \right)\)

recognizing that \({\boldsymbol{a}} \bullet {\boldsymbol{b}} = 0\)     M1

correct substitution     A1

e.g. \( - 3 - 4 - a = 0\)

\(a = - 7\)     A1     N3

[5 marks]

Find \(\mathop {{\text{PQ}}}\limits^ \to  \).

Find \(\left| {\mathop {{\text{PQ}}}\limits^ \to  } \right|\).

Find the angle between PQ and PR.

Find the area of triangle PQR.

Hence or otherwise find the shortest distance from R to the line through P and Q.

valid approach      (M1)

eg   (7, 4, 9) − (3, 2, 5)  A − B

\(\mathop {{\text{PQ}}}\limits^ \to   = \) 4i + 2j + 4k \(\left( { = \left( \begin{gathered}
4 \hfill \\
2 \hfill \\
4 \hfill \\
\end{gathered} \right)} \right)\)     A1 N2

[2 marks]

correct substitution into magnitude formula      (A1)
eg  \(\sqrt {{4^2} + {2^2} + {4^2}} \)

\(\left| {\mathop {{\text{PQ}}}\limits^ \to  } \right| = 6\)     A1 N2

[2 marks]

finding scalar product and magnitudes      (A1)(A1)

scalar product = (4 × 6) + (2 × (−1) + (4 × 3) (= 34)

magnitude of PR = \(\sqrt {36 + 1 + 9}  = \left( {6.782} \right)\)

correct substitution of their values to find cos \({\text{Q}}\mathop {\text{P}}\limits^ \wedge  {\text{R}}\)     M1

eg  cos \({\text{Q}}\mathop {\text{P}}\limits^ \wedge  {\text{R}}\,\,{\text{ = }}\frac{{24 - 2 + 12}}{{\left( 6 \right) \times \left( {\sqrt {46} } \right)}},\,\,0.8355\)

0.581746

\({\text{Q}}\mathop {\text{P}}\limits^ \wedge  {\text{R}}\) = 0.582 radians  or \({\text{Q}}\mathop {\text{P}}\limits^ \wedge  {\text{R}}\) = 33.3°     A1 N3

[4 marks]

correct substitution (A1)
eg    \(\frac{1}{2} \times \left| {\mathop {{\text{PQ}}}\limits^ \to  } \right| \times \left| {\mathop {{\text{PR}}}\limits^ \to  } \right| \times \,\,{\text{sin}}\,P,\,\,\frac{1}{2} \times 6 \times \sqrt {46}  \times \,\,{\text{sin}}\,0.582\)

area is 11.2 (sq. units)      A1 N2

[2 marks]

recognizing shortest distance is perpendicular distance from R to line through P and Q      (M1)

eg  sketch, height of triangle with base \(\left[ {{\text{PQ}}} \right],\,\,\frac{1}{2} \times 6 \times h,\,\,{\text{sin}}\,33.3^\circ  = \frac{h}{{\sqrt {46} }}\)

correct working      (A1)

eg  \(\frac{1}{2} \times 6 \times d = 11.2,\,\,\left| {\mathop {{\text{PR}}}\limits^ \to  } \right| \times \,\,{\text{sin}}\,P,\,\,\sqrt {46}  \times \,\,{\text{sin}}\,33.3^\circ \)

3.72677

distance = 3.73  (units)    A1 N2

[3 marks]

The line L₁ is represented by \({{\boldsymbol{r}}_1} = \left( {\begin{array}{*{20}{c}}
2\\
5\\
3
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
1\\
2\\
3
\end{array}} \right)\)  and the line L₂ by \({{\boldsymbol{r}}_2} = \left( {\begin{array}{*{20}{c}}
3\\
{ - 3}\\
8
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
{ - 1}\\
3\\
{ - 4}
\end{array}} \right)\) .

The lines L₁ and L₂ intersect at point T. Find the coordinates of T.

evidence of equating vectors     (M1)

e.g. \({L_1} = {L_2}\)

for any two correct equations     A1A1

e.g. \(2 + s = 3 - t\) , \(5 + 2s = - 3 + 3t\) , \(3 + 3s = 8 - 4t\)

attempting to solve the equations     (M1)

finding one correct parameter \((2 = - 1{\text{, }}t = 2)\)     A1

the coordinates of T are \((1{\text{, }}3{\text{, }}0)\)     A1     N3

[6 marks]

Those candidates prepared in this topic area answered the question particularly well, often only making some calculation error when solving the resulting system of equations. Curiously, a few candidates found correct values for s and t, but when substituting back into one of the vector equations, neglected to find the z-coordinate of T.

Line \({L_1}\) has equation \({\boldsymbol{r}_1} = \left( {\begin{array}{*{20}{c}}
{10}\\
6\\
{ - 1}
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
2\\
{ - 5}\\
{ - 2}
\end{array}} \right)\) and line \({L_2}\) has equation \({\boldsymbol{r}_2} = \left( {\begin{array}{*{20}{c}}
2\\
1\\
{ - 3}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
3\\
5\\
2
\end{array}} \right)\) .

Lines \({L_1}\) and \({L_2}\) intersect at point A. Find the coordinates of A.

appropriate approach     (M1)

eg   \(\left( {\begin{array}{*{20}{c}}
{10}\\
6\\
{ - 1}
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
2\\
{ - 5}\\
{ - 2}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
2\\
1\\
{ - 3}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
3\\
5\\
2
\end{array}} \right)\) , \({L_1} = {L_2}\)

any two correct equations     A1A1

eg   \(10 + 2s = 2 + 3t\) , \(6 - 5s = 1 + 5t\) , \( - 1 - 2s =  - 3 + 2t\)

attempt to solve     (M1)

eg substituting one equation into another

one correct parameter     A1

eg   \(s = - 1\) , \(t = 2\)

correct substitution     (A1)

eg   \(2 + 3(2)\) , \(1 + 5(2)\) , \( - 3 + 2(2)\)

A \( = \) (\(8\), \(11\), \(1\)) (accept column vector)     A1     N4

[7 marks]

Most students were able to set up one or more equations, but few chose to use their GDCs to solve the resulting system. Algebraic errors prevented many of these candidates from obtaining the final three marks. Some candidates stopped after finding the value of s and/or \(t\).