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Question
In this question, distance is in metres.
Toy airplanes fly in a straight line at a constant speed. Airplane 1 passes through a point A.
Its position, p seconds after it has passed through A, is given by \(\left( {\begin{array}{*{20}{c}}
x\\
y\\
z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
3\\
{ – 4}\\
0
\end{array}} \right) + p\left( {\begin{array}{*{20}{c}}
{ – 2}\\
3\\
1
\end{array}} \right)\) .
(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.
Answer/Explanation
Markscheme
(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]
Question
Line \({L_1}\) passes through points \({\text{A}}(1{\text{, }} – 1{\text{, }}4)\) and \({\text{B}}(2{\text{, }} – 2{\text{, }}5)\) .
Line \({L_2}\) has equation \({\boldsymbol{r}} = \left( \begin{array}{l}
2\\
4\\
7
\end{array} \right) + s\left( \begin{array}{l}
2\\
1\\
3
\end{array} \right)\) .
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.
Answer/Explanation
Markscheme
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]
Question
The following diagram shows the cuboid (rectangular solid) OABCDEFG, where O is the origin, and \(\overrightarrow {{\rm{OA}}} = 4\boldsymbol{i}\) , \(\overrightarrow {{\rm{OC}}} = 3\boldsymbol{j}\) , \(\overrightarrow {{\rm{OD}}} = 2\boldsymbol{k}\) .
(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.
Answer/Explanation
Markscheme
(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]
Question
The points A and B lie on a line \(L\), and have position vectors \(\left( {\begin{array}{*{20}{c}} { – 3} \\ { – 2} \\ 2 \end{array}} \right)\) and \(\left( {\begin{array}{*{20}{c}} 6 \\ 4 \\ { – 1} \end{array}} \right)\) respectively. Let O be the origin. This is shown on the following diagram.
The point C also lies on \(L\), such that \(\overrightarrow {{\text{AC}}} = 2\overrightarrow {{\text{CB}}} \).
Let \(\theta \) be the angle between \(\overrightarrow {{\text{AB}}} \) and \(\overrightarrow {{\text{OC}}} \).
Let D be a point such that \(\overrightarrow {{\text{OD}}} = k\overrightarrow {{\text{OC}}} \), where \(k > 1\). Let E be a point on \(L\) such that \({\rm{C\hat ED}}\) is a right angle. This is shown on the following diagram.
Find \(\overrightarrow {{\text{AB}}} \).
Show that \(\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 2 \\ 0 \end{array}} \right)\).
Find \(\theta \).
(i) Show that \(\left| {\overrightarrow {{\text{DE}}} } \right| = (k – 1)\left| {\overrightarrow {{\text{OC}}} } \right|\sin \theta \).
(ii) The distance from D to line \(L\) is less than 3 units. Find the possible values of \(k\).
Answer/Explanation
Markscheme
valid approach (addition or subtraction) (M1)
eg\(\,\,\,\,\,\)\({\text{AO}} + {\text{OB}},{\text{ B}} – {\text{A}}\)
\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 9 \\ 6 \\ { – 3} \end{array}} \right)\) A1 N2
[2 marks]
METHOD 1
valid approach using \(\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right)\) (M1)
eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}} {x + 3} \\ {y + 2} \\ {z – 2} \end{array}} \right),{\text{ }}\overrightarrow {{\text{CB}}} = \left( {\begin{array}{*{20}{c}} {6 – x} \\ {4 – y} \\ { – 1 – z} \end{array}} \right)\)
correct working A1
eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} {x + 3} \\ {y + 2} \\ {z – 2} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {12 – 2x} \\ {8 – 2y} \\ { – 2 – 2z} \end{array}} \right)\)
all three equations A1
eg\(\,\,\,\,\,\)\(x + 3 = 12 – 2x,{\text{ }}y + 2 = 8 – 2y,{\text{ }}z – 2 = – 2 – 2z\),
\(\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 2 \\ 0 \end{array}} \right)\) AG N0
METHOD 2
valid approach (M1)
eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{OC}}} – \overrightarrow {{\text{OA}}} = 2\left( {\overrightarrow {{\text{OB}}} – \overrightarrow {{\text{OC}}} } \right)\)
correct working A1
eg\(\,\,\,\,\,\)\(3\overrightarrow {{\text{OC}}} = 2\overrightarrow {{\text{OB}}} + \overrightarrow {{\text{OA}}} \)
correct substitution of \(\overrightarrow {{\text{OB}}} \) and \(\overrightarrow {{\text{OA}}} \) A1
eg\(\,\,\,\,\,\)\(3\overrightarrow {{\text{OC}}} = 2\left( {\begin{array}{*{20}{c}} 6 \\ 4 \\ { – 1} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} { – 3} \\ { – 2} \\ 2 \end{array}} \right),{\text{ }}3\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} 9 \\ 6 \\ 0 \end{array}} \right)\)
\(\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 2 \\ 0 \end{array}} \right)\) AG N0
METHOD 3
valid approach (M1)
eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{AC}}} = \frac{2}{3}\overrightarrow {{\text{AB}}} \), diagram, \(\overrightarrow {{\text{CB}}} = \frac{1}{3}\overrightarrow {{\text{AB}}} \)
correct working A1
eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}} 6 \\ 4 \\ { – 2} \end{array}} \right),{\text{ }}\overrightarrow {{\text{CB}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 2 \\ { – 1} \end{array}} \right)\)
correct working involving \(\overrightarrow {{\text{OC}}} \) A1
eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} { – 3} \\ { – 2} \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 6 \\ 4 \\ { – 2} \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} 6 \\ 4 \\ { – 1} \end{array}} \right) – \left( {\begin{array}{*{20}{c}} 3 \\ 2 \\ { – 1} \end{array}} \right)\)
\(\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 2 \\ 0 \end{array}} \right)\) AG N0
[3 marks]
finding scalar product and magnitudes (A1)(A1)(A1)
scalar product \( = (9 \times 3) + (6 \times 2) + ( – 3 \times 0){\text{ }}( = 39)\)
magnitudes \(\sqrt {81 + 36 + 9} {\text{ }}( = 11.22),{\text{ }}\sqrt {9 + 4} {\text{ }}( = 3.605)\)
substitution into formula M1
eg\(\,\,\,\,\,\)\(\cos \theta = \frac{{(9 \times 3) + 12}}{{\sqrt {126} \times \sqrt {13} }}\)
\(\theta = 0.270549{\text{ }}({\text{accept }}15.50135^\circ )\)
\(\theta = 0.271{\text{ }}({\text{accept }}15.5^\circ )\) A1 N4
[5 marks]
(i) attempt to use a trig ratio M1
eg\(\,\,\,\,\,\)\(\sin \theta = \frac{{{\text{DE}}}}{{{\text{CD}}}},{\text{ }}\left| {\overrightarrow {{\text{CE}}} } \right| = \left| {\overrightarrow {{\text{CD}}} } \right|\cos \theta \)
attempt to express \(\overrightarrow {{\text{CD}}} \) in terms of \(\overrightarrow {{\text{OC}}} \) M1
eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{OC}}} + \overrightarrow {{\text{CD}}} = \overrightarrow {{\text{OD}}} ,{\text{ OC}} + {\text{CD}} = {\text{OD}}\)
correct working A1
eg\(\,\,\,\,\,\)\(\left| {k\overrightarrow {{\text{OC}}} – \overrightarrow {{\text{OC}}} } \right|\sin \theta \)
\(\left| {\overrightarrow {{\text{DE}}} } \right| = (k – 1)\left| {\overrightarrow {{\text{OC}}} } \right|\sin \theta \) AG N0
(ii) valid approach involving the segment DE (M1)
eg\(\,\,\,\,\,\)recognizing \(\left| {\overrightarrow {{\text{DE}}} } \right| < 3,{\text{ DE}} = 3\)
correct working (accept equation) (A1)
eg\(\,\,\,\,\,\)\((k – 1)(\sqrt {13} )\sin 0.271 < 3,{\text{ }}k – 1 = 3.11324\)
\(1 < k < 4.11{\text{ }}({\text{accept }}k < 4.11{\text{ but not }}k = 4.11)\) A1 N2
[6 marks]