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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
Consider the points A(\(5\), \(2\), \(1\)) , B(\(6\), \(5\), \(3\)) , and C(\(7\), \(6\), \(a + 1\)) , \(a \in{\mathbb{R}}\) .
Let \({\rm{q}}\) be the angle between \(\overrightarrow {{\rm{AB}}} \) and \(\overrightarrow {{\rm{AC}}} \) .
Find
(i) \(\overrightarrow {{\rm{AB}}} \) ;
(ii) \(\overrightarrow {{\rm{AC}}} \) .
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 \({\rm{q}} = 1.2\) .
Answer/Explanation
Markscheme
(i) appropriate approach (M1)
eg \(\overrightarrow {{\rm{AO}}} {\rm{ + }}\overrightarrow {{\rm{OB}}} \) , \({\rm{B}} – {\rm{A}}\)
\(\overrightarrow {{\rm{AB}}} = \left( \begin{array}{l}
1\\
3\\
2
\end{array} \right)\) A1 N2
(ii) \(\overrightarrow {{\rm{AC}}} = \left( \begin{array}{l}
2\\
4\\
a
\end{array} \right)\) A1 N1
[3 marks]
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]