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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
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
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]
Question
Let u \( = – 3\)i \( + \) j \( + \) k and v \( = m\)j \( + {\text{ }}n\)k , where \(m,{\text{ }}n \in \mathbb{R}\). Given that v is a unit vector perpendicular to u, find the possible values of \(m\) and of \(n\).
Answer/Explanation
Markscheme
correct scalar product (A1)
eg\(\,\,\,\,\,\)\(m + n\)
setting up their scalar product equal to 0 (seen anywhere) (M1)
eg\(\,\,\,\,\,\)u \( \bullet \) v \( = 0,{\text{ }} – 3(0) + 1(m) + 1(n) = 0,{\text{ }}m = – n\)
correct interpretation of unit vector (A1)
eg\(\,\,\,\,\,\)\(\sqrt {{0^2} + {m^2} + {n^2}} = 1,{\text{ }}{m^2} + {n^2} = 1\)
valid attempt to solve their equations (must be in one variable) M1
eg\(\,\,\,\,\,\)\({( – n)^2} + {n^2} = 1,{\text{ }}\sqrt {1 – {n^2}} + n = 0,{\text{ }}{m^2} + {( – m)^2} = 1,{\text{ }}m – \sqrt {1 – {m^2}} = 0\)
correct working A1
eg\(\,\,\,\,\,\)\(2{n^2} = 1,{\text{ }}2{m^2} = 1,{\text{ }}\sqrt 2 = \frac{1}{n},{\text{ }}m = \pm \frac{1}{{\sqrt 2 }}\)
both correct pairs A2 N3
eg\(\,\,\,\,\,\)\(m = \frac{1}{{\sqrt 2 }}\) and \(n = – \frac{1}{{\sqrt 2 }},{\text{ }}m = – \frac{1}{{\sqrt 2 }}\) and \(n = \frac{1}{{\sqrt 2 }}\),
\(m = {(0.5)^{\frac{1}{2}}}\) and \(n = – {(0.5)^{\frac{1}{2}}},{\text{ }}m = – \sqrt {\frac{1}{2}} \) and \(n = \sqrt {\frac{1}{2}} \)
Note: Award A0 for \(m = \pm \frac{1}{{\sqrt 2 }},{\text{ }}n = \pm \frac{1}{{\sqrt 2 }}\), or any other answer that does not clearly indicate the correct pairs.
[7 marks]
Examiners report
Most of the candidates recognized that the scalar product of the vectors must be zero. However, some did not find the correct scalar product because they did not multiply the correct corresponding vector components of u and v. In addition, the majority of candidates did not attempt to use the fact that the unit vector v has a magnitude of 1. For the small number of candidates who were successful in solving for \(m\) and/or \(n\), some did not clearly present the correct pairs of answers.
Question
The position vectors of points P and Q are i \( + \) 2 j \( – \) k and 7i \( + \) 3j \( – \) 4k respectively.
Find a vector equation of the line that passes through P and Q.
The line through P and Q is perpendicular to the vector 2i \( + \) nk. Find the value of \(n\).
Answer/Explanation
Markscheme
valid attempt to find direction vector (M1)
eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{PQ}}} ,{\text{ }}\overrightarrow {{\text{QP}}} \)
correct direction vector (or multiple of) (A1)
eg\(\,\,\,\,\,\)6i \( + \) j \( – \) 3k
any correct equation in the form r \( = \) a \( + \) tb (any parameter for \(t\)) A2 N3
where a is i \( + \) 2j \( – \) k or 7i \( + \) 3j \( – \) 4k , and b is a scalar multiple of 6i \( + \) j \( – \) 3k
eg\(\,\,\,\,\,\)r \( = \) 7i \( + \) 3j \( – \) 4k \( + \) t(6i \( + \) j \( – \) 3k), r \( = \left( {\begin{array}{*{20}{c}} {1 + 6s} \\ {2 + 1s} \\ { – 1 – 3s} \end{array}} \right),{\text{ }}r = \left( {\begin{array}{*{20}{c}} 1 \\ 2 \\ { – 1} \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} { – 6} \\ { – 1} \\ 3 \end{array}} \right)\)
Notes: Award A1 for the form a \( + \) tb, A1 for the form L \( = \) a \( + \) tb, A0 for the form r \( = \) b \( + \) ta.
[4 marks]
correct expression for scalar product (A1)
eg\(\,\,\,\,\,\)\(6 \times 2 + 1 \times 0 + ( – 3) \times n,{\text{ }} – 3n + 12\)
setting scalar product equal to zero (seen anywhere) (M1)
eg\(\,\,\,\,\,\)u \( \bullet \) v \( = 0,{\text{ }} – 3n + 12 = 0\)
\(n = 4\) A1 N2
[3 marks]
Question
The vectors a = \(\left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right)\) and b = \(\left( {\begin{array}{*{20}{c}} {k + 3} \\ k \end{array}} \right)\) are perpendicular to each other.
Find the value of \(k\).
Given that c = a + 2b, find c.
Answer/Explanation
Markscheme
evidence of scalar product M1
eg\(\,\,\,\,\,\)a \( \bullet \) b, \(4(k + 3) + 2k\)
recognizing scalar product must be zero (M1)
eg\(\,\,\,\,\,\)a \( \bullet \) b \( = 0,{\text{ }}4k + 12 + 2k = 0\)
correct working (must involve combining terms) (A1)
eg \(6k + 12,\,\,\,6k = – 12\)
\(\,\,\,\,\,\)\(k = – 2\) A1 N2
[4 marks]
attempt to substitute their value of \(k\) (seen anywhere) (M1)
eg\(\,\,\,\,\,\)b = \(\left( {\begin{array}{*{20}{c}} { – 2 + 3} \\ { – 2} \end{array}} \right)\), 2b = \(\left( {\begin{array}{*{20}{c}} 2 \\ { – 4} \end{array}} \right)\)
correct working (A1)
eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 2 \\ { – 4} \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} {4 + 2k + 6} \\ {2 + 2k} \end{array}} \right)\)
c = \(\left( {\begin{array}{*{20}{c}} 6 \\ { – 2} \end{array}} \right)\) A1 N2
[3 marks]
Question
A line \(L\) passes through points \({\text{A}}( – 3,{\text{ }}4,{\text{ }}2)\) and \({\text{B}}( – 1,{\text{ }}3,{\text{ }}3)\).
The line \(L\) also passes through the point \({\text{C}}(3,{\text{ }}1,{\text{ }}p)\).
Show that \(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 2 \\ { – 1} \\ 1 \end{array}} \right)\).
Find a vector equation for \(L\).
Find the value of \(p\).
The point D has coordinates \(({q^2},{\text{ }}0,{\text{ }}q)\). Given that \(\overrightarrow {{\text{DC}}} \) is perpendicular to \(L\), find the possible values of \(q\).
Answer/Explanation
Markscheme
correct approach A1
eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} { – 1} \\ 3 \\ 3 \end{array}} \right) – \left( {\begin{array}{*{20}{c}} { – 3} \\ 4 \\ 2 \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} 3 \\ { – 4} \\ { – 2} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} { – 1} \\ 3 \\ 3 \end{array}} \right)\)
\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 2 \\ { – 1} \\ 1 \end{array}} \right)\) AG N0
[1 mark]
any correct equation in the form \(r = a + tb\) (any parameter for \(t\))
where \(a\) is \(\left( {\begin{array}{*{20}{c}} { – 3} \\ 4 \\ 2 \end{array}} \right)\) or \(\left( {\begin{array}{*{20}{c}} { – 1} \\ 3 \\ 3 \end{array}} \right)\) and \(b\) is a scalar multiple of \(\left( {\begin{array}{*{20}{c}} 2 \\ { – 1} \\ 1 \end{array}} \right)\) A2 N2
eg\(\,\,\,\,\,\)\(r = \left( {\begin{array}{*{20}{c}} { – 3} \\ 4 \\ 2 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 2 \\ { – 1} \\ 1 \end{array}} \right),{\text{ }}(x,{\text{ }}y,{\text{ }}z) = ( – 1,{\text{ }}3,{\text{ }}3) + s( – 2,{\text{ }}1,{\text{ }} – 1),{\text{ }}r = \left( {\begin{array}{*{20}{c}} { – 3 + 2t} \\ {4 – t} \\ {2 + t} \end{array}} \right)\)
Note: Award A1 for the form \(a + tb\), A1 for the form \(L = a + tb\), A0 for the form \(r = b + ta\).
[2 marks]
METHOD 1 – finding value of parameter
valid approach (M1)
eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} { – 3} \\ 4 \\ 2 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 2 \\ { – 1} \\ 1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ p \end{array}} \right),{\text{ }}( – 1,{\text{ }}3,{\text{ }}3) + s( – 2,{\text{ }}1,{\text{ }} – 1) = (3,{\text{ }}1,{\text{ }}p)\)
one correct equation (not involving \(p\)) (A1)
eg\(\,\,\,\,\,\)\( – 3 + 2t = 3,{\text{ }} – 1 – 2s = 3,{\text{ }}4 – t = 1,{\text{ }}3 + s = 1\)
correct parameter from their equation (may be seen in substitution) A1
eg\(\,\,\,\,\,\)\(t = 3,{\text{ }}s = – 2\)
correct substitution (A1)
eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} { – 3} \\ 4 \\ 2 \end{array}} \right) + 3\left( {\begin{array}{*{20}{c}} 2 \\ { – 1} \\ 1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ p \end{array}} \right),{\text{ }}3 – ( – 2)\)
\(p = 5\,\,\,\,\,\left( {{\text{accept }}\left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ 5 \end{array}} \right)} \right)\) A1 N2
METHOD 2 – eliminating parameter
valid approach (M1)
eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} { – 3} \\ 4 \\ 2 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 2 \\ { – 1} \\ 1 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ p \end{array}} \right),{\text{ }}( – 1,{\text{ }}3,{\text{ }}3) + s( – 2,{\text{ }}1,{\text{ }} – 1) = (3,{\text{ }}1,{\text{ }}p)\)
one correct equation (not involving \(p\)) (A1)
eg\(\,\,\,\,\,\)\( – 3 + 2t = 3,{\text{ }} – 1 – 2s = 3,{\text{ }}4 – t = 1,{\text{ }}3 + s = 1\)
correct equation (with \(p\)) A1
eg\(\,\,\,\,\,\)\(2 + t = p,{\text{ }}3 – s = p\)
correct working to solve for \(p\) (A1)
eg\(\,\,\,\,\,\)\(7 = 2p – 3,{\text{ }}6 = 1 + p\)
\(p = 5\,\,\,\,\,\left( {{\text{accept }}\left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ 5 \end{array}} \right)} \right)\) A1 N2
[5 marks]
valid approach to find \(\overrightarrow {{\text{DC}}} \) or \(\overrightarrow {{\text{CD}}} \) (M1)
eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ 5 \end{array}} \right) – \left( {\begin{array}{*{20}{c}} {{q^2}} \\ 0 \\ q \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} {{q^2}} \\ 0 \\ q \end{array}} \right) – \left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ 5 \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} {{q^2}} \\ 0 \\ q \end{array}} \right) – \left( {\begin{array}{*{20}{c}} 3 \\ 1 \\ p \end{array}} \right)\)
correct vector for \(\overrightarrow {{\text{DC}}} \) or \(\overrightarrow {{\text{CD}}} \) (may be seen in scalar product) A1
eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} {3 – {q^2}} \\ 1 \\ {5 – q} \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} {{q^2} – 3} \\ { – 1} \\ {q – 5} \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} {3 – {q^2}} \\ 1 \\ {p – q} \end{array}} \right)\)
recognizing scalar product of \(\overrightarrow {{\text{DC}}} \) or \(\overrightarrow {{\text{CD}}} \) with direction vector of \(L\) is zero (seen anywhere) (M1)
eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} {3 – {q^2}} \\ 1 \\ {p – q} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 2 \\ { – 1} \\ 1 \end{array}} \right) = 0,{\text{ }}\overrightarrow {{\text{DC}}} \bullet \overrightarrow {{\text{AC}}} = 0,{\text{ }}\left( {\begin{array}{*{20}{c}} {3 – {q^2}} \\ 1 \\ {5 – q} \end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}} 2 \\ { – 1} \\ 1 \end{array}} \right) = 0\)
correct scalar product in terms of only \(q\) A1
eg\(\,\,\,\,\,\)\(6 – 2{q^2} – 1 + 5 – q,{\text{ }}2{q^2} + q – 10 = 0,{\text{ }}2(3 – {q^2}) – 1 + 5 – q\)
correct working to solve quadratic (A1)
eg\(\,\,\,\,\,\)\((2q + 5)(q – 2),{\text{ }}\frac{{ – 1 \pm \sqrt {1 – 4(2)( – 10)} }}{{2(2)}}\)
\(q = – \frac{5}{2},{\text{ }}2\) A1A1 N3
[7 marks]