Question
Consider the points A (1 , 5 , 4) , B (3 , 1 , 2) and D (3 , k , 2) , with (AD) perpendicular to (AB) .
The point O has coordinates (0 , 0 , 0) , point A has coordinates (1 , – 2 , 3) and point B has coordinates (– 3 , 4 , 2) .
Find
(i) \(\overrightarrow {{\rm{AB}}} \) ;
(ii) \(\overrightarrow {{\rm{AD}}} \) giving your answer in terms of k .
[3 marks]
Show that \(k = 7\) .
The point C is such that \(\overrightarrow {{\rm{BC}}} = \frac{1}{2}\overrightarrow {{\rm{AD}}} \) .
Find the position vector of C.
Find \(\cos {\rm{A}}\widehat {\rm{B}}{\rm{C}}\) .
Answer/Explanation
Markscheme
(i) evidence of combining vectors (M1)
e.g. \(\overrightarrow {{\rm{AB}}} = \overrightarrow {{\rm{OB}}} – \overrightarrow {{\rm{OA}}} \) (or \(\overrightarrow {{\rm{AD}}} = \overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OD}}} \) in part (ii))
\(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
2\\
{ – 4}\\
{ – 2}
\end{array}} \right)\) A1 N2
(ii) \(\overrightarrow {{\rm{AD}}} = \left( {\begin{array}{*{20}{c}}
2\\
{k – 5}\\
{ – 2}
\end{array}} \right)\) A1 N1
[3 marks]
evidence of using perpendicularity \( \Rightarrow \) scalar product = 0 (M1)
e.g. \(\left( {\begin{array}{*{20}{c}}
2\\
{ – 4}\\
{ – 2}
\end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}}
2\\
{k – 5}\\
{ – 2}
\end{array}} \right) = 0\)
\(4 – 4(k – 5) + 4 = 0\) A1
\( – 4k + 28 = 0\) (accept any correct equation clearly leading to \(k = 7\) ) A1
\(k = 7\) AG N0
[3 marks]
\(\overrightarrow {{\rm{AD}}} = \left( {\begin{array}{*{20}{c}}
2\\
2\\
{ – 2}
\end{array}} \right)\) (A1)
\(\overrightarrow {{\rm{BC}}} = \left( {\begin{array}{*{20}{c}}
1\\
1\\
{ – 1}
\end{array}} \right)\) A1
evidence of correct approach (M1)
e.g. \(\overrightarrow {{\rm{OC}}} = \overrightarrow {{\rm{OB}}} + \overrightarrow {{\rm{BC}}} \) , \(\left( {\begin{array}{*{20}{c}}
3\\
1\\
2
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
1\\
1\\
{ – 1}
\end{array}} \right)\) , \(\left( {\begin{array}{*{20}{c}}
{x – 3}\\
{y – 1}\\
{z – 2}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
1\\
1\\
{ – 1}
\end{array}} \right)\)
\(\overrightarrow {{\rm{OC}}} = \left( {\begin{array}{*{20}{c}}
4\\
2\\
1
\end{array}} \right)\) A1 N3
[4 marks]
METHOD 1
choosing appropriate vectors, \(\overrightarrow {{\rm{BA}}} \) , \(\overrightarrow {{\rm{BC}}} \) (A1)
finding the scalar product M1
e.g. \( – 2(1) + 4(1) + 2( – 1)\) , \(2(1) + ( – 4)(1) + ( – 2)( – 1)\)
\(\cos {\rm{A}}\widehat {\rm{B}}{\rm{C = 0}}\) A1 N1
METHOD 2
\(\overrightarrow {{\rm{BC}}} \) parallel to \(\overrightarrow {{\rm{AD}}} \) (may show this on a diagram with points labelled) R1
\(\overrightarrow {{\rm{BC}}} \bot \overrightarrow {{\rm{AB}}} \) (may show this on a diagram with points labelled) R1
\({\rm{A}}\widehat {\rm{B}}{\rm{C}} = 90^\circ \)
\(\cos {\rm{A}}\widehat {\rm{B}}{\rm{C = 0}}\) A1 N1
[3 marks]
Examiners report
This question was well done by many candidates. Most found \(\overrightarrow {{\rm{AB}}} \) and \(\overrightarrow {{\rm{AD}}} \) correctly.
The majority of candidates correctly used the scalar product to show \(k = 7\) .
Some confusion arose in substituting \(k = 7\) into \(\overrightarrow {{\rm{AD}}} \) , but otherwise part (c) was well done, though finding the position vector of C presented greater difficulty.
Owing to \(\overrightarrow {{\rm{AB}}} \) and \(\overrightarrow {{\rm{BC}}} \) being perpendicular, no problems were created by using these two vectors to find \(\cos {\rm{A}}\widehat {\rm{B}}{\rm{C}} = 0\) , and the majority of candidates answering part (d) did exactly that.
Question
Let u \( = \left( {\begin{array}{*{20}{c}}
2\\
3\\
{ – 1}
\end{array}} \right)\) and w \( = \left( {\begin{array}{*{20}{c}}
3\\
{ – 1}\\
p
\end{array}} \right)\) . Given that u is perpendicular to w , find the value of p .
Let \({\boldsymbol{v}} = \left( {\begin{array}{*{20}{c}}
1 \\
q \\
5
\end{array}} \right)\) . Given that \(\left| {\boldsymbol{v}} \right| = \sqrt {42} \), find the possible values of \(q\) .
Answer/Explanation
Markscheme
evidence of equating scalar product to 0 (M1)
\(2 \times 3 + 3 \times ( – 1) + ( – 1) \times p = 0\) (\(6 – 3 – p = 0\), \(3 – p = 0\)) A1
\(p = 3\) A1 N2
[3 marks]
evidence of substituting into magnitude formula (M1)
e.g. \(\sqrt {1 + {q^2} + 25} \) , \(1 + {q^2} + 25\)
setting up a correct equation A1
e.g. \(\sqrt {1 + {q^2} + 25} = \sqrt {42} \) , \(1 + {q^2} + 25 = 42\) , \({q^2} = 16\)
\(q = \pm 4\) A1 N2
[3 marks]
Question
The line \({L_1}\) is represented by the vector equation \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
{ – 3}\\
{ – 1}\\
{ – 25}
\end{array}} \right) + p\left( {\begin{array}{*{20}{c}}
2\\
1\\
{ – 8}
\end{array}} \right)\) .
A second line \({L_2}\) is parallel to \({L_1}\) and passes through the point B(\( – 8\), \( – 5\), \(25\)) .
Write down a vector equation for \({L_2}\) in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) .
A third line \({L_3}\) is perpendicular to \({L_1}\) and is represented by \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
5\\
0\\
3
\end{array}} \right) + q\left( {\begin{array}{*{20}{c}}
{ – 7}\\
{ – 2}\\
k
\end{array}} \right)\) .
Show that \(k = – 2\) .
The lines \({L_1}\) and \({L_3}\) intersect at the point A.
Find the coordinates of A.
The lines \({L_2}\)and \({L_3}\)intersect at point C where \(\overrightarrow {{\rm{BC}}} = \left( {\begin{array}{*{20}{c}}
6\\
3\\
{ – 24}
\end{array}} \right)\) .
(i) Find \(\overrightarrow {{\rm{AB}}} \) .
(ii) Hence, find \(|\overrightarrow {{\rm{AC}}} |\) .
Answer/Explanation
Markscheme
any correct equation in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) (accept any parameter) A2 N2
e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
{ – 8}\\
{ – 5}\\
{25}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
2\\
1\\
{ – 8}
\end{array}} \right)\)
Note: Award A1 for \({\boldsymbol{a}} + t{\boldsymbol{b}}\) , A1 for \(L = {\boldsymbol{a}} + t{\boldsymbol{b}}\) , A0 for \({\boldsymbol{r}} = {\boldsymbol{b}} + t{\boldsymbol{a}}\) .
[2 marks]
recognizing scalar product must be zero (seen anywhere) R1
e.g. \({\boldsymbol{a}} \bullet {\boldsymbol{b}} = 0\)
evidence of choosing direction vectors \(\left( {\begin{array}{*{20}{c}}
2\\
1\\
{ – 8}
\end{array}} \right),\left( {\begin{array}{*{20}{c}}
{ – 7}\\
{ – 2}\\
k
\end{array}} \right)\) (A1)(A1)
correct calculation of scalar product (A1)
e.g. \(2( – 7) + 1( – 2) – 8k\)
simplification that clearly leads to solution A1
e.g. \( – 16 – 8k\) , \( – 16 – 8k = 0\)
\(k = – 2\) AG N0
[5 marks]
evidence of equating vectors (M1)
e.g. \({L_1} = {L_3}\) , \(\left( {\begin{array}{*{20}{c}}
{ – 3}\\
{ – 1}\\
{ – 25}
\end{array}} \right) + p\left( {\begin{array}{*{20}{c}}
2\\
1\\
{ – 8}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
5\\
0\\
3
\end{array}} \right) + q\left( {\begin{array}{*{20}{c}}
{ – 7}\\
{ – 2}\\
{ – 2}
\end{array}} \right)\)
any two correct equations A1A1
e.g. \( – 3 + 2p = 5 – 7q\) , \( – 1 + p = – 2q\) , \(- 25 – 8p = 3 – 2q\)
attempting to solve equations (M1)
finding one correct parameter (\(p = – 3\) , \(q = 2\) ) A1
the coordinates of A are \(( – 9, – 4, – 1)\) A1 N3
[6 marks]
(i) evidence of appropriate approach (M1)
e.g. \(\overrightarrow {{\rm{OA}}} + \overrightarrow {{\rm{AB}}} = \overrightarrow {{\rm{OB}}} \) , \(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
{ – 8}\\
{ – 5}\\
{25}
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
{ – 9}\\
{ – 4}\\
{ – 1}
\end{array}} \right)\)
\(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
1\\
{ – 1}\\
{26}
\end{array}} \right)\) A1 N2
(ii) finding \(\overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}}
7\\
2\\
2
\end{array}} \right)\) A1
evidence of finding magnitude (M1)
e.g. \(|\overrightarrow {{\rm{AC}}} | = \sqrt {{7^2} + {2^2} + {2^2}} \)
\(|\overrightarrow {{\rm{AC}}} | = \sqrt {57} \) A1 N3
[5 marks]
Question
Let \(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
6\\
{ – 2}\\
3
\end{array}} \right)\) and \(\overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}}
{ – 2}\\
{ – 3}\\
2
\end{array}} \right)\) .
Find \(\overrightarrow {{\rm{BC}}} \) .
Find a unit vector in the direction of \(\overrightarrow {{\rm{AB}}} \) .
Show that \(\overrightarrow {{\rm{AB}}} \) is perpendicular to \(\overrightarrow {{\rm{AC}}} \) .
Answer/Explanation
Markscheme
evidence of appropriate approach (M1)
e.g. \(\overrightarrow {{\rm{BC}}} {\rm{ = }}\overrightarrow {{\rm{BA}}} + \overrightarrow {{\rm{AC}}} \) , \(\left( {\begin{array}{*{20}{c}}
{ – 2}\\
{ – 3}\\
2
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
6\\
{ – 2}\\
3
\end{array}} \right)\)
\(\overrightarrow {{\rm{BC}}} = \left( {\begin{array}{*{20}{c}}
{ – 8}\\
{ – 1}\\
{ – 1}
\end{array}} \right)\) A1 N2
[2 marks]
attempt to find the length of \(\overrightarrow {{\rm{AB}}} \) (M1)
\(\overrightarrow {|{\rm{AB}}} | = \sqrt {{6^2} + {{( – 2)}^2} + {3^2}} \) \(( = \sqrt {36 + 4 + 9} = \sqrt {49} = 7)\) (A1)
unit vector is \(\frac{1}{7}\left( {\begin{array}{*{20}{c}}
6\\
{ – 2}\\
3
\end{array}} \right)\) \(\left( { = \left( {\begin{array}{*{20}{c}}
{\frac{6}{7}}\\
{ – \frac{2}{7}}\\
{\frac{3}{7}}
\end{array}} \right)} \right)\) A1 N2
[3 marks]
recognizing that the dot product or \(\cos \theta \) being 0 implies perpendicular (M1)
correct substitution in a scalar product formula A1
e.g. \((6) \times ( – 2) + ( – 2) \times ( – 3) + (3) \times (2)\) , \(\cos \theta = \frac{{ – 12 + 6 + 6}}{{7 \times \sqrt {17} }}\)
correct calculation A1
e.g. \(\overrightarrow {{\rm{AB}}} \bullet \overrightarrow {{\rm{AC}}} = 0\) , \(\cos \theta = 0\)
therefore, they are perpendicular AG N0
[3 marks]
Question
The diagram shows quadrilateral ABCD with vertices A(1, 0), B(1, 5), C(5, 2) and D(4, −1) .
(i) Show that \(\overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}}
4\\
2
\end{array}} \right)\) .
(ii) Find \(\overrightarrow {{\rm{BD}}} \) .
(iii) Show that \(\overrightarrow {{\rm{AC}}} \) is perpendicular to \(\overrightarrow {{\rm{BD}}} \) .
The line (AC) has equation \({\boldsymbol{r}} = {\boldsymbol{u}} + s{\boldsymbol{v}}\) .
(i) Write down vector u and vector v .
(ii) Find a vector equation for the line (BD).
The lines (AC) and (BD) intersect at the point \({\text{P}}(3{\text{, }}k)\) .
Show that \(k = 1\) .
The lines (AC) and (BD) intersect at the point \({\text{P}}(3{\text{, }}k)\) .
Hence find the area of triangle ACD.
Answer/Explanation
Markscheme
(i) correct approach A1
e.g. \(\overrightarrow {{\rm{OC}}} – \overrightarrow {{\rm{OA}}} \) , \(\left( {\begin{array}{*{20}{c}}
5\\
2
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
1\\
0
\end{array}} \right)\)
\(\overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}}
4\\
2
\end{array}} \right)\) AG N0
(ii) appropriate approach (M1)
e.g. \({\mathop{\rm D}\nolimits} – {\rm{B}}\) , \(\left( {\begin{array}{*{20}{c}}
4\\
{ – 1}
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
1\\
5
\end{array}} \right)\) , move 3 to the right and 6 down
\(\overrightarrow {{\rm{BD}}} = \left( {\begin{array}{*{20}{c}}
3\\
{ – 6}
\end{array}} \right)\) A1 N2
(iii) finding the scalar product A1
e.g. \(4(3) + 2( – 6)\) , \(12 – 12\)
valid reasoning R1
e.g. \(4(3) + 2( – 6) = 0\) , scalar product is zero
\(\overrightarrow {{\rm{AC}}} \) is perpendicular to \(\overrightarrow {{\rm{BD}}} \) AG N0
[5 marks]
(i) correct “position” vector for u; “direction” vector for v A1A1 N2
e.g. \({\boldsymbol{u}} = \left( {\begin{array}{*{20}{c}}
5\\
2
\end{array}} \right)\) , \({\boldsymbol{u}} = \left( {\begin{array}{*{20}{c}}
1\\
0
\end{array}} \right)\) ; \({\boldsymbol{v}} = \left( {\begin{array}{*{20}{c}}
4\\
2
\end{array}} \right)\) , \({\boldsymbol{v}} = \left( {\begin{array}{*{20}{c}}
{ – 2}\\
{ – 1}
\end{array}} \right)\)
accept in equation e.g. \(\left( {\begin{array}{*{20}{c}}
5\\
2
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
{ – 4}\\
{ – 2}
\end{array}} \right)\)
(ii) any correct equation in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) , where \({\boldsymbol{b}} = \overrightarrow {{\rm{BD}}} \)
\({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
1\\
5
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
3\\
{ – 6}
\end{array}} \right)\) , \(\left( {\begin{array}{*{20}{c}}
x\\
y
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
4\\
{ – 1}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
{ – 1}\\
2
\end{array}} \right)\) A2 N2
[4 marks]
METHOD 1
substitute (3, k) into equation for (AC) or (BD) (M1)
e.g. \(3 = 1 + 4s\) , \(3 = 1 + 3t\)
value of t or s A1
e.g. \(s = \frac{1}{2}\) , \( – \frac{1}{2}\) , \(t = \frac{2}{3}\) , \( – \frac{1}{3}\)
substituting A1
e.g. \(k = 0 + \frac{1}{2}(2)\)
\(k = 1\) AG N0
METHOD 2
setting up two equations (M1)
e.g. \(1 + 4s = 4 + 3t\) , \(2s = – 1 – 6t\) ; setting vector equations of lines equal
value of t or s A1
e.g. \(s = \frac{1}{2}\) , \( – \frac{1}{2}\) , \(t = \frac{2}{3}\) , \( – \frac{1}{3}\)
substituting A1
e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
4\\
{ – 1}
\end{array}} \right) – \frac{1}{3}\left( {\begin{array}{*{20}{c}}
3\\
{ – 6}
\end{array}} \right)\)
\(k = 1\) AG N0
[3 marks]
\(\overrightarrow {{\rm{PD}}} = \left( {\begin{array}{*{20}{c}}
1\\
{ – 2}
\end{array}} \right)\) (A1)
\(|\overrightarrow {{\rm{PD}}} | = \sqrt {{2^2} + {1^2}} \) \(( = \sqrt 5 )\) (A1)
\(|\overrightarrow {{\rm{AC}}} | = \sqrt {{4^2} + {2^2}} \) \(( = \sqrt {20} )\) (A1)
area \( = \frac{1}{2} \times |\overrightarrow {{\rm{AC}}} | \times |\overrightarrow {{\rm{PD}}} |\) (\( = \frac{1}{2} \times \sqrt {20} \times \sqrt 5 \)) M1
\( = 5\) A1 N4
[5 marks]
Question
The following diagram shows the obtuse-angled triangle ABC such that \(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
{ – 3}\\
0\\
{ – 4}
\end{array}} \right)\) and \(\overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}}
{ – 2}\\
2\\
{ – 6}
\end{array}} \right)\) .
(i) Write down \(\overrightarrow {{\rm{BA}}} \) .
(ii) Find \(\overrightarrow {{\rm{BC}}} \) .
(i) Find \(\cos {\rm{A}}\widehat {\rm{B}}{\rm{C}}\) .
(ii) Hence, find \({\rm{sinA}}\widehat {\rm{B}}{\rm{C}}\) .
The point D is such that \(\overrightarrow {{\rm{CD}}} = \left( {\begin{array}{*{20}{c}}
{ – 4}\\
5\\
p
\end{array}} \right)\) , where \(p > 0\) .
(i) Given that \(\overrightarrow {|{\rm{CD}}|} = \sqrt {50} \) , show that \(p = 3\) .
(ii) Hence, show that \(\overrightarrow {{\rm{CD}}} \) is perpendicular to \(\overrightarrow {{\rm{BC}}} \) .
Answer/Explanation
Markscheme
(i) \(\overrightarrow {{\rm{BA}}} = \left( {\begin{array}{*{20}{c}}
3\\
0\\
4
\end{array}} \right)\) A1 N1
(ii) evidence of combining vectors (M1)
e.g. \(\overrightarrow {{\rm{AB}}} + \overrightarrow {{\rm{BC}}} = \overrightarrow {{\rm{AC}}} \) , \(\overrightarrow {{\rm{BA}}} + \overrightarrow {{\rm{AC}}} \) , \(\left( {\begin{array}{*{20}{c}}
{ – 2}\\
2\\
{ – 6}
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
{ – 3}\\
0\\
{ – 4}
\end{array}} \right)\)
\(\overrightarrow {{\rm{BC}}} = \left( {\begin{array}{*{20}{c}}
1\\
2\\
{ – 2}
\end{array}} \right)\) A1 N2
[3 marks]
(i) METHOD 1
finding \(\overrightarrow {{\rm{BA}}} \bullet \overrightarrow {{\rm{BC}}} \) , \(|\overrightarrow {{\rm{BA}}} |\) , \(|\overrightarrow {{\rm{BC}}} |\)
e.g. \(\overrightarrow {{\rm{BA}}} \bullet \overrightarrow {{\rm{BC}}} = 3 \times 1 + 0 + 4 \times – 2\) , \(\overrightarrow {|{\rm{BA}}} | = \sqrt {{3^2} + {4^2}} \) , \(\overrightarrow {|{\rm{BC}}} | = 3\)
substituting into formula for \(\cos \theta \) M1
e.g. \(\frac{{3 \times 1 + 0 + 4 \times – 2}}{{3\sqrt {{3^2} + 0 + {4^2}} }}\) , \(\frac{{ – 5}}{{5 \times 3}}\)
\(cos {\rm{A}}\widehat {\rm{B}}{\rm{C}} = – \frac{5}{{15}}\) \(\left( { = – \frac{1}{3}} \right)\) A1 N3
METHOD 2
finding \(|\overrightarrow {{\rm{AC}}} |\) , \(|\overrightarrow {{\rm{BA}}} |\) , \(|\overrightarrow {{\rm{BC}}} |\) (A1)(A1)(A1)
e.g. \(|\overrightarrow {{\rm{AC}}} | = \sqrt {{2^2} + {2^2} + {6^2}} \) , \(|\overrightarrow {{\rm{BA}}} | = \sqrt {{3^2} + {4^2}} \) , \(|\overrightarrow {{\rm{BC}}} | = 3\)
substituting into cosine rule M1
e.g. \(\frac{{{5^2} + {3^2} – {{\left( {\sqrt {44} } \right)}^2}}}{{2 \times 5 \times 3}}\) , \(\frac{{25 + 9 – 44}}{{30}}\)
\(\cos {\rm{A}}\widehat {\rm{B}}{\rm{C}} = – \frac{{10}}{{30}}\) \(\left( { = – \frac{1}{3}} \right)\) A1 N3
(ii) evidence of using Pythagoras (M1)
e.g. right-angled triangle with values, \({\sin ^2}x + {\cos ^2}x = 1\)
\(\sin {\rm{A}}\widehat {\rm{B}}{\rm{C}} = \frac{{\sqrt 8 }}{3}\) \(\left( { = \frac{{2\sqrt 2 }}{3}} \right)\) A1 N2
[7 marks]
(i) attempt to find an expression for \(|\overrightarrow {{\rm{CD}}} |\) (M1)
e.g. \(\sqrt {{{( – 4)}^2} + {5^2} + {p^2}} \) , \(|\overrightarrow {{\rm{CD}}} {|^2} = {4^2} + {5^2} + {p^2}\)
correct equation A1
e.g. \(\sqrt {{{( – 4)}^2} + {5^2} + {p^2}} = \sqrt {50} \) , \({4^2} + {5^2} + {p^2} = 50\)
\({p^2} = 9\) A1
\(p = 3\) AG N0
(ii) evidence of scalar product (M1)
e.g. \(\left( {\begin{array}{*{20}{c}}
{ – 4}\\
5\\
3
\end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}}
1\\
2\\
{ – 2}
\end{array}} \right)\) , \(\overrightarrow {{\rm{CD}}} \bullet \overrightarrow {{\rm{BC}}} \)
correct substitution
e.g. \( – 4 \times 1 + 5 \times 2 + 3 \times – 2\) , \( – 4 + 10 – 6\) A1
\(\overrightarrow {{\rm{CD}}} \bullet \overrightarrow {{\rm{BC}}} = 0\) A1
\(\overrightarrow {{\rm{CD}}} \) is perpendicular to \(\overrightarrow {{\rm{BC}}} \) AG N0
[6 marks]
Question
The following diagram shows quadrilateral ABCD, with \(\overrightarrow {{\rm{AD}}} = \overrightarrow {{\rm{BC}}} \) , \(\overrightarrow {{\rm{AB}}} = \left( \begin{array}{l}
3\\
1
\end{array} \right)\) , and \(\overrightarrow {{\rm{AC}}} = \left( \begin{array}{l}
4\\
4
\end{array} \right)\) .
Find \(\overrightarrow {{\rm{BC}}} \) .
Show that \(\overrightarrow {{\rm{BD}}} = \left( {\begin{array}{*{20}{c}}
{ – 2}\\
2
\end{array}} \right)\) .
Show that vectors \(\overrightarrow {{\rm{BD}}} \) and \(\overrightarrow {{\rm{AC}}} \) are perpendicular.
Answer/Explanation
Markscheme
evidence of appropriate approach (M1)
e.g. \(\overrightarrow {{\rm{AC}}} – \overrightarrow {{\rm{AB}}} \) , \(\left( \begin{array}{l}
4 – 3\\
4 – 1
\end{array} \right)\)
\(\overrightarrow {{\rm{BC}}} = \left( \begin{array}{l}
1\\
3
\end{array} \right)\) A1 N2
[2 marks]
METHOD 1
\(\overrightarrow {{\rm{AD}}} = \left( {\begin{array}{*{20}{c}}
1\\
3
\end{array}} \right)\) (A1)
correct approach A1
e.g. \(\overrightarrow {{\rm{AD}}} – \overrightarrow {{\rm{AB}}} \) , \(\left( \begin{array}{l}
1 – 3\\
3 – 1
\end{array} \right)\)
\(\overrightarrow {{\rm{BD}}} = \left( {\begin{array}{*{20}{c}}
{ – 2}\\
2
\end{array}} \right)\) AG N0
METHOD 2
recognizing \(\overrightarrow {{\rm{CD}}} = \overrightarrow {{\rm{BA}}} \) (A1)
correct approach A1
e.g. \(\overrightarrow {{\rm{BC}}} + \overrightarrow {{\rm{CD}}} \) , \(\left( \begin{array}{l}
1 – 3\\
3 – 1
\end{array} \right)\)
\(\overrightarrow {{\rm{BD}}} = \left( {\begin{array}{*{20}{c}}
{ – 2}\\
2
\end{array}} \right)\) AG N0
[2 marks]
METHOD 1
evidence of scalar product (M1)
e.g. \(\overrightarrow {{\rm{BD}}} \bullet \overrightarrow {{\rm{AC}}} \) , \(\left( {\begin{array}{*{20}{c}}
{ – 2}\\
2
\end{array}} \right) \bullet \left( \begin{array}{l}
4\\
4
\end{array} \right)\)
correct substitution A1
e.g. \(( – 2)(4) + (2)(4)\) , \( – 8 + 8\)
\(\overrightarrow {{\rm{BD}}} \bullet \overrightarrow {{\rm{AC}}} = 0\) A1
therefore vectors \(\overrightarrow {{\rm{BD}}} \) and \(\overrightarrow {{\rm{AC}}} \) are perpendicular AG N0
METHOD 2
attempt to find angle between two vectors (M1)
e.g. \(\frac{{{\boldsymbol{a}} \bullet {\boldsymbol{b}}}}{{{\boldsymbol{ab}}}}\)
correct substitution A1
e.g. \(\frac{{( – 2)(4) + (2)(4)}}{{\sqrt 8 \sqrt {32} }}\) , \(\cos \theta = 0\)
\(\theta = {90^ \circ }\) A1
therefore vectors \(\overrightarrow {{\rm{BD}}} \) and \(\overrightarrow {{\rm{AC}}} \) are perpendicular AG N0
[3 marks]
Question
The line \({L_1}\) passes through the points P(2, 4, 8) and Q(4, 5, 4) .
The line \({L_2}\) is perpendicular to \({L_1}\) , and parallel to \(\left( {\begin{array}{*{20}{c}}
{3p}\\
{2p}\\
4
\end{array}} \right)\) , where \(p \in \mathbb{Z}\) .
(i) Find \(\overrightarrow {{\rm{PQ}}} \) .
(ii) Hence write down a vector equation for \({L_1}\) in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + s{\boldsymbol{b}}\) .
(i) Find the value of p .
(ii) Given that \({L_2}\) passes through \({\text{R}}(10{\text{, }}6{\text{, }}- 40)\) , write down a vector equation for \({L_2}\) .
The lines \({L_1}\) and \({L_2}\) intersect at the point A. Find the x-coordinate of A.
Answer/Explanation
Markscheme
(i) evidence of approach (M1)
e.g. \(\overrightarrow {{\rm{PO}}} + \overrightarrow {{\rm{OQ}}} \) , \({\rm{P}} – {\rm{Q}}\)
\(\overrightarrow {{\rm{PQ}}} = \left( {\begin{array}{*{20}{c}}
2\\
1\\
{ – 4}
\end{array}} \right)\) A1 N2
(ii) any correct equation in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + s{\boldsymbol{b}}\) (accept any parameter for s)
where a is \(\left( {\begin{array}{*{20}{c}}
2\\
4\\
8
\end{array}} \right)\) or \(\left( {\begin{array}{*{20}{c}}
4\\
5\\
4
\end{array}} \right)\) , and b is a scalar multiple of \(\left( {\begin{array}{*{20}{c}}
2\\
1\\
{ – 4}
\end{array}} \right)\) A2 N2
e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
2\\
4\\
8
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
2\\
1\\
{ – 4}
\end{array}} \right)\) , \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
{4 + 2s}\\
{5 + 1s}\\
{4 – 4s}
\end{array}} \right)\) , \({\boldsymbol{r}} = 2{\boldsymbol{i}} + 4{\boldsymbol{j}} + 8{\boldsymbol{k}} + s(2{\boldsymbol{i}} + 1{\boldsymbol{j}} – 4{\boldsymbol{k}})\)
Note: Award A1 for the form \({\boldsymbol{a}} + s{\boldsymbol{b}}\) , A1 for \({\boldsymbol{L}} = {\boldsymbol{a}} + s{\boldsymbol{b}}\) , A0 for \({\boldsymbol{r}} = {\boldsymbol{b}} + s{\boldsymbol{a}}\) .
[4 marks]
(i) choosing correct direction vectors for \({L_1}\) and \({L_2}\) (A1) (A1)
e.g. \(\left( {\begin{array}{*{20}{c}}
2\\
1\\
{ – 4}
\end{array}} \right)\) , \(\left( {\begin{array}{*{20}{c}}
{3p}\\
{2p}\\
4
\end{array}} \right)\)
evidence of equating scalar product to 0 (M1)
correct calculation of scalar product A1
e.g. \(2 \times 3p + 1 \times 2p + ( – 4) \times 4\) , \(8p – 16 = 0\)
\(p = 2\) A1 N3
(ii) any correct expression in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) (accept any parameter for t)
where a is \(\left( {\begin{array}{*{20}{c}}
{10}\\
6\\
{ – 40}
\end{array}} \right)\) , and b is a scalar multiple of \(\left( {\begin{array}{*{20}{c}}
6\\
4\\
4
\end{array}} \right)\) A2 N2
e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
{10}\\
6\\
{ – 40}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
6\\
4\\
4
\end{array}} \right)\) , \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
{10 + 6s}\\
{6 + 4s}\\
{ – 40 + 4s}
\end{array}} \right)\) , \({\boldsymbol{r}} = 10{\boldsymbol{i}} + 6{\boldsymbol{j}} – 40{\boldsymbol{k}} + s(6{\boldsymbol{i}} + 4{\boldsymbol{j}} + 4{\boldsymbol{k}})\)
Note: Award A1 for the form \({\boldsymbol{a}} + t{\boldsymbol{b}}\) , A1 for \({\boldsymbol{L}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) (unless they have been penalised for \({\boldsymbol{L}} = {\boldsymbol{a}} + s{\boldsymbol{b}}\) in part (a)), A0 for \({\boldsymbol{r}} = {\boldsymbol{b}} + t{\boldsymbol{a}}\) .
[7 marks]
appropriate approach (M1)
e.g. \(\left( {\begin{array}{*{20}{c}}
2\\
4\\
8
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
2\\
1\\
{ – 4}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{10}\\
6\\
{ – 40}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
6\\
4\\
4
\end{array}} \right)\)
any two correct equations with different parameters A1A1
e.g. \(2 + 2s = 10 + 6t\) , \(4 + s = 6 + 4t\) , \(8 – 4s = – 40 + 4t\)
attempt to solve simultaneous equations (M1)
correct working (A1)
e.g. \( – 6 = – 2 – 2t\) , \(4 = 2t\) , \( – 4 + 5s = 46\) , \(5s = 50\)
one correct parameter \(s = 10\) , \(t = 2\) A1
\(x = 22\) (accept (22, 14, −32)) A1 N4
[7 marks]
Question
Let A and B be points such that \(\overrightarrow {{\rm{OA}}} = \left( {\begin{array}{*{20}{c}}
5\\
2\\
1
\end{array}} \right)\) and \(\overrightarrow {{\rm{OB}}} = \left( {\begin{array}{*{20}{c}}
6\\
0\\
3
\end{array}} \right)\) .
Show that \(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
2
\end{array}} \right)\) .
Let C and D be points such that ABCD is a rectangle.
Given that \(\overrightarrow {{\rm{AD}}} = \left( {\begin{array}{*{20}{c}}
4\\
p\\
1
\end{array}} \right)\) , show that \(p = 3\) .
Let C and D be points such that ABCD is a rectangle.
Find the coordinates of point C.
Let C and D be points such that ABCD is a rectangle.
Find the area of rectangle ABCD.
Answer/Explanation
Markscheme
correct approach A1
e.g. \(\overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OB}}} ,\left( {\begin{array}{*{20}{c}}
6\\
0\\
3
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
5\\
2\\
1
\end{array}} \right)\)
\(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
2
\end{array}} \right)\) AG N0
[1 mark]
recognizing \(\overrightarrow {{\rm{AD}}} \) is perpendicular to \(\overrightarrow {{\rm{AB}}} \) (may be seen in sketch) (R1)
e.g. adjacent sides of rectangle are perpendicular
recognizing dot product must be zero (R1)
e.g. \(\overrightarrow {{\rm{AD}}} \bullet \overrightarrow {AB} = 0\)
correct substitution (A1)
e.g. \((1 \times 4) + ( – 2 \times p) + (2 \times 1)\) , \(4 – 2p + 2 = 0\)
equation which clearly leads to \(p = 3\) A1
e.g. \(6 – 2p = 0\) , \(2p = 6\)
\(p = 3\) AG N0
[4 marks]
correct approach (seen anywhere including sketch) (A1)
e.g. \(\overrightarrow {{\rm{OC}}} = \overrightarrow {{\rm{OB}}} + \overrightarrow {{\rm{BC}}} \) , \(\overrightarrow {{\rm{OD}}} + \overrightarrow {{\rm{DC}}} \)
recognizing opposite sides are equal vectors (may be seen in sketch) (R1)
e.g. \(\overrightarrow {{\rm{BC}}} = \overrightarrow {{\rm{AD}}} \) , \(\overrightarrow {{\rm{DC}}} = \overrightarrow {{\rm{AB}}} \) , \(\left( {\begin{array}{*{20}{c}}
6\\
0\\
3
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
4\\
3\\
1
\end{array}} \right)\) , \(\left( {\begin{array}{*{20}{c}}
9\\
5\\
2
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
2
\end{array}} \right)\)
coordinates of point C are (10, 3, 4) (accept \(\left( {\begin{array}{*{20}{c}}
{10}\\
3\\
4
\end{array}} \right)\) ) A2 N4
Note: Award A1 for two correct values.
[4 marks]
attempt to find one side of the rectangle (M1)
e.g. substituting into magnitude formula
two correct magnitudes A1A1
e.g. \(\sqrt {{{(1)}^2} + {{( – 2)}^2} + {2^2}} \) , 3 ; \(\sqrt {16 + 9 + 1} \) , \(\sqrt {26} \)
multiplying magnitudes (M1)
e.g. \(\sqrt {26} \times \sqrt 9 \)
\({\rm{area}} = \sqrt {234} ( = 3\sqrt {26} )\) (accept \(3 \times \sqrt {26} \) ) A1 N3
[5 marks]
Question
Consider points A(\(1\), \( – 2\), \( -1\)) , B(\(7\), \( – 4\), \(3\)) and C(\(1\), \( -2\), \(3\)) . The line \({L_1}\) passes through C and is parallel to \(\overrightarrow {{\rm{AB}}} \) .
A second line, \({L_2}\) , is given by \(\boldsymbol{r} = \left( {\begin{array}{*{20}{c}}
{ – 1}\\
2\\
{15}
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
3\\
{ – 3}\\
p
\end{array}} \right)\) .
Find \(\overrightarrow {{\rm{AB}}} \) .
Hence, write down a vector equation for \({L_1}\) .
Given that \({L_1}\) is perpendicular to \({L_2}\) , show that \(p = – 6\) .
The line \({L_1}\) intersects the line \({L_2}\) at point Q. Find the \(x\)-coordinate of Q.
Answer/Explanation
Markscheme
valid approach (M1)
eg \(\left( {\begin{array}{*{20}{c}}
7\\
{ – 4}\\
3
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
{ – 1}
\end{array}} \right)\) , \({\rm{A}} – {\rm{B}}\) , \(\overrightarrow {{\rm{AB}}} = \overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OB}}} \)
\(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
6\\
{ – 2}\\
4
\end{array}} \right)\) A1 N2
[2 marks]
any correct equation in the form \(\boldsymbol{r} = \boldsymbol{a} + t\boldsymbol{b}\) (accept any parameter for \(t\))
where \(\boldsymbol{a} = \left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
3
\end{array}} \right)\) and \(\boldsymbol{b}\) is a scalar multiple of \(\overrightarrow {{\rm{AB}}} \) A2 N2
eg \(\boldsymbol{r} = \left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
3
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
6\\
{ – 2}\\
4
\end{array}} \right)\) , \((x,y,z) = (1, – 2,3) + t(3, – 1,2)\) , \(\boldsymbol{r} = \left( {\begin{array}{*{20}{c}}
{1 + 6t}\\
{ – 2 – 2t}\\
{3 + 4t}
\end{array}} \right)\)
Note: Award A1 for \(\boldsymbol{a} + t\boldsymbol{b}\) , A1 for \({L_1} = \boldsymbol{a} + t\boldsymbol{b}\) , A0 for \(\boldsymbol{r} = \boldsymbol{b} + t\boldsymbol{a}\) .
[2 marks]
recognizing that scalar product \( = 0\) (seen anywhere) R1
correct calculation of scalar product (A1)
eg \(6(3) – 2( – 3) + 4p\) , \(18 + 6 + 4p\)
correct working A1
eg \(24 + 4p = 0\) , \(4p = – 24\)
\(p = – 6\) AG N0
[3 marks]
setting lines equal (M1)
eg \({L_1} = {L_2}\) , \(\left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
3
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
6\\
{ – 2}\\
4
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{ – 1}\\
2\\
{15}
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
3\\
{ – 3}\\
{ – 6}
\end{array}} \right)\)
any two correct equations with different parameters A1A1
eg \(1 + 6t = 1 + 3s\) , \( – 2 – 2t = 2 – 3s\) , \(3 + 4t = 15 – 6s\)
attempt to solve their simultaneous equations (M1)
one correct parameter A1
eg \(t = \frac{1}{2}\) , \(s = \frac{5}{3}\)
attempt to substitute parameter into vector equation (M1)
eg \(\left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
3
\end{array}} \right) + \frac{1}{2}\left( {\begin{array}{*{20}{c}}
6\\
{ – 2}\\
4
\end{array}} \right)\) , \(1 + \frac{1}{2} \times 6\)
\(x = 4\) (accept (4, -3, 5), ignore incorrect values for \(y\) and \(z\)) A1 N3
[7 marks]
Question
The line \(L\) is parallel to the vector \(\left( \begin{array}{l}3\\2\end{array} \right)\).
The line \(L\) passes through the point \((9, 4)\).
Find the gradient of the line \(L\).
Find the equation of the line \(L\) in the form \(y = ax + b\).
Write down a vector equation for the line \(L\).
Answer/Explanation
Markscheme
attempt to find gradient (M1)
eg reference to change in \(x\) is \(3\) and/or \(y\) is \(2\), \(\frac{3}{2}\)
gradient \( = \frac{2}{3}\) A1 N2
[2 marks]
attempt to substitute coordinates and/or gradient into Cartesian equation
for a line (M1)
eg \(y – 4 = m(x – 9),{\text{ }}y = \frac{2}{3}x + b,{\text{ }}9 = a(4) + c\)
correct substitution (A1)
eg \(4 = \frac{2}{3}(9) + c,{\text{ }}y – 4 = \frac{2}{3}(x – 9)\)
\(y = \frac{2}{3}x – 2{\text{ }}\left( {{\text{accept }}a = \frac{2}{3},{\text{ }}b = – 2} \right)\) A1 N2
[3 marks]
any correct equation in the form r = a + tb (any parameter for t), where a indicates position eg \(\left( \begin{array}{l}9\\4\end{array} \right)\) or \(\left( \begin{array}{c}0\\ – 2\end{array} \right)\), and b is a scalar multiple of \(\left( \begin{array}{l}3\\2\end{array} \right)\)
eg r = \(\left( \begin{array}{c}9\\4\end{array} \right) + t\left( \begin{array}{c}3\\2\end{array} \right),\left( \begin{array}{c}x\\y\end{array} \right) = \left( \begin{array}{c}3t + 9\\2t + 4\end{array} \right)\), r = 0i − 2 j + s(3i + 2 j) A2 N2
Note: Award A1 for a + tb, A1 for L = a + tb, A0 for r = b + ta.
[2 marks]
Question
Distances in this question are in metres.
Ryan and Jack have model airplanes, which take off from level ground. Jack’s airplane takes off after Ryan’s.
The position of Ryan’s airplane \(t\) seconds after it takes off is given by \(\boldsymbol{r}=\left( \begin{array}{c}5\\6\\0\end{array} \right) + t\left( \begin{array}{c} – 4\\2\\4\end{array} \right)\).
Find the speed of Ryan’s airplane.
Find the height of Ryan’s airplane after two seconds.
The position of Jack’s airplane \(s\) seconds after it takes off is given by r = \(\left( \begin{array}{c} – 39\\44\\0\end{array} \right) + s\left( \begin{array}{c}4\\ – 6\\7\end{array} \right)\).
Show that the paths of the airplanes are perpendicular.
The two airplanes collide at the point \((-23, 20, 28)\).
How long after Ryan’s airplane takes off does Jack’s airplane take off?
Answer/Explanation
Markscheme
valid approach (M1)
eg magnitude of direction vector
correct working (A1)
eg \(\sqrt {{{( – 4)}^2} + {2^2} + {4^2}} ,{\text{ }}\sqrt { – {4^2} + {2^2} + {4^2}} \)
\(6{\text{ (m}}{{\text{s}}^{ – 1}})\) A1 N2
[3 marks]
substituting \(2\) for \(t\) (A1)
eg \(0 + 2(4)\), r = \(\left( \begin{array}{c}5\\6\\0\end{array} \right) + 2\left( \begin{array}{c} – 4\\2\\4\end{array} \right),\left( \begin{array}{c} – 3\\10\\8\end{array} \right)\), \(y = 10\)
\(8\) (metres) A1 N2
[2 marks]
METHOD 1
choosing correct direction vectors \(\left( \begin{array}{c} – 4\\2\\4\end{array} \right)\) and \(\left( \begin{array}{c}4\\ – 6\\7\end{array} \right)\) (A1)(A1)
evidence of scalar product M1
eg a \( \cdot \) b
correct substitution into scalar product (A1)
eg \(( – 4 \times 4) + (2 \times – 6) + (4 \times 7)\)
evidence of correct calculation of the scalar product as \(0\) A1
eg \( – 16 – 12 + 28 = 0\)
directions are perpendicular AG N0
METHOD 2
choosing correct direction vectors \(\left( \begin{array}{c} – 4\\2\\4\end{array} \right)\) and \(\left( \begin{array}{c}4\\ – 6\\7\end{array} \right)\) (A1)(A1)
attempt to find angle between vectors M1
correct substitution into numerator A1
eg \(\cos \theta = \frac{{ – 16 – 12 + 28}}{{\left| a \right|\left| b \right|}},{\text{ }}\cos \theta = 0\)
\(\theta = 90^\circ \) A1
directions are perpendicular AG N0
[5 marks]
METHOD 1
one correct equation for Ryan’s airplane (A1)
eg \(5 – 4t = – 23,{\text{ }}6 + 2t = 20,{\text{ }}0 + 4t = 28\)
\(t = 7\) A1
one correct equation for Jack’s airplane (A1)
eg \( – 39 + 4s = – 23,{\text{ }}44 – 6s = 20,{\text{ }}0 + 7s = 28\)
\(s = 4\) A1
\(3\) (seconds later) A1 N2
METHOD 2
valid approach (M1)
eg \(\left( \begin{array}{c}5\\6\\0\end{array} \right) + t\left( \begin{array}{c} – 4\\2\\4\end{array} \right) = \left( \begin{array}{c} – 39\\44\\0\end{array} \right) + s\left( \begin{array}{c}4\\ – 6\\7\end{array} \right)\), one correct equation
two correct equations (A1)
eg \(5 – 4t = – 39 + 4s,{\text{ }}6 + 2t = 44 – 6s,{\text{ }}4t = 7s\)
\(t = 7\) A1
\(s = 4\) A1
\(3\) (seconds later) A1 N2
[5 marks]