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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]
Question
The vertices of the triangle PQR are defined by the position vectors
\(\overrightarrow {{\rm{OP}}} = \left( {\begin{array}{*{20}{c}}
4\\
{ – 3}\\
1
\end{array}} \right)\) , \(\overrightarrow {{\rm{OQ}}} = \left( {\begin{array}{*{20}{c}}
3\\
{ – 1}\\
2
\end{array}} \right)\) and \(\overrightarrow {{\rm{OR}}} = \left( {\begin{array}{*{20}{c}}
6\\
{ – 1}\\
5
\end{array}} \right)\) .
Find
(i) \(\overrightarrow {{\rm{PQ}}} \) ;
(ii) \(\overrightarrow {{\rm{PR}}} \) .
Show that \(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{1}{2}\) .
(i) Find \({\rm{sinR}}\widehat {\rm{P}}{\rm{Q}}\) .
(ii) Hence, find the area of triangle PQR, giving your answer in the form \(a\sqrt 3 \) .
Answer/Explanation
Markscheme
(i) evidence of approach (M1)
e.g. \(\overrightarrow {{\rm{PQ}}} = \overrightarrow {{\rm{PO}}} + \overrightarrow {{\rm{OQ}}} \) , \({\rm{Q}} – {\rm{P}}\)
\(\overrightarrow {{\rm{PQ}}} = \left( {\begin{array}{*{20}{c}}
{ – 1}\\
2\\
1
\end{array}} \right)\) A1 N2
(ii) \(\overrightarrow {{\rm{PR}}} = \left( {\begin{array}{*{20}{c}}
2\\
2\\
4
\end{array}} \right)\) A1 N1
[3 marks]
METHOD 1
choosing correct vectors \(\overrightarrow {{\rm{PQ}}} \) and \(\overrightarrow {{\rm{PR}}} \) (A1)(A1)
finding \(\overrightarrow {{\rm{PQ}}} \bullet \overrightarrow {{\rm{PR}}} \) , \(\left| {\overrightarrow {{\rm{PQ}}} } \right|\) , \(\left| {\overrightarrow {{\rm{PR}}} } \right|\) (A1) (A1)(A1)
\(\overrightarrow {{\rm{PQ}}} \bullet \overrightarrow {{\rm{PR}}} = – 2 + 4 + 4( = 6)\)
\(\left| {\overrightarrow {{\rm{PQ}}} } \right| = \sqrt {{{( – 1)}^2} + {2^2} + {1^2}} \) \(\left( { = \sqrt 6 } \right)\) , \(\left| {\overrightarrow {{\rm{PR}}} } \right| = \sqrt {{2^2} + {2^2} + {4^2}} \) \(\left( { = \sqrt {24} } \right)\)
substituting into formula for angle between two vectors M1
e.g. \(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{6}{{\sqrt 6 \times \sqrt {24} }}\)
simplifying to expression clearly leading to \(\frac{1}{2}\) A1
e.g. \(\frac{6}{{\sqrt 6 \times 2\sqrt 6 }}\) , \(\frac{6}{{\sqrt {144} }}\) , \(\frac{6}{{12}}\)
\(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{1}{2}\) AG N0
METHOD 2
evidence of choosing cosine rule (seen anywhere) (M1)
\(\overrightarrow {{\rm{QR}}} = \left( {\begin{array}{*{20}{c}}
3\\
0\\
3
\end{array}} \right)\) A1
\(\left| {\overrightarrow {{\rm{QR}}} } \right| = \sqrt {18} \) , \(\left| {\overrightarrow {{\rm{PQ}}} } \right| = \sqrt 6 \) and \(\left| {\overrightarrow {{\rm{PR}}} } \right| = \sqrt {24} \) (A1)(A1)(A1)
\(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{{{{\left( {\sqrt 6 } \right)}^2} + {{\left( {\sqrt {24} } \right)}^2} – {{\left( {\sqrt {18} } \right)}^2}}}{{2\sqrt 6 \times \sqrt {24} }}\) A1
\(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{{6 + 24 – 18}}{{24}}\) \(\left( { = \frac{{12}}{{24}}} \right)\) A1
\(\cos {\rm{R}}\widehat {\rm{P}}{\rm{Q}} = \frac{1}{2}\) AG N0
[7 marks]
(i) METHOD 1
evidence of appropriate approach (M1)
e.g. using \({\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{R}}\widehat {\rm{P}}{\rm{Q + co}}{{\rm{s}}^{\rm{2}}}{\rm{R}}\widehat {\rm{P}}{\rm{Q}} = 1\) , diagram
substituting correctly (A1)
e.g. \({\rm{sinR}}\widehat {\rm{P}}{\rm{Q}} = \sqrt {1 – {{\left( {\frac{1}{2}} \right)}^2}} \)
\({\rm{sinR}}\widehat {\rm{P}}{\rm{Q}} = \sqrt {\frac{3}{4}} \) \(\left( { = \frac{{\sqrt 3 }}{2}} \right)\) A1 N3
METHOD 2
since \(\cos \widehat {\rm{P}} = \frac{1}{2}\) , \(\widehat {\rm{P}} = 60^\circ \) (A1)
evidence of approach
e.g. drawing a right triangle, finding the missing side (A1)
\(\sin \widehat {\rm{P}} = \frac{{\sqrt 3 }}{2}\) A1 N3
(ii) evidence of appropriate approach (M1)
e.g. attempt to substitute into \(\frac{1}{2}ab\sin C\)
correct substitution
e.g. area \( = \frac{1}{2}\sqrt 6 \times \sqrt {24} \times \frac{{\sqrt 3 }}{2}\) A1
area \( = 3\sqrt 3 \) A1 N2
[6 marks]
Question
Find the cosine of the angle between the two vectors \(3{\boldsymbol{i}} + 4{\boldsymbol{j}} + 5{\boldsymbol{k}}\) and \(4{\boldsymbol{i}} – 5{\boldsymbol{j}} – 3{\boldsymbol{k}}\) .
Answer/Explanation
Markscheme
finding scalar product and magnitudes (A1)(A1)(A1)
scalar product \( = 12 – 20 – 15\) (\( = – 23\))
magnitudes \( = \sqrt {{3^2} + {4^2} + {5^2}} \) , \( = \sqrt {{4^2} + {{( – 5)}^2} + {{( – 3)}^2}} \) , \(\left( {\sqrt {50} ,\sqrt {50} } \right)\)
substitution into formula M1
e.g. \(\cos \theta = \frac{{12 – 20 – 15}}{{\left( {\sqrt {{3^2} + {4^2} + {5^2}} } \right) \times \left( {\sqrt {{4^2} + {{( – 5)}^2} + {{( – 3)}^2}} } \right)}}\)
\(\cos \theta = – \frac{{23}}{{50}}\) \(( = – 0.46)\) A2 N4
[6 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 triangle \(ABC\).
Let \(\overrightarrow {{\text{AB}}} \bullet \overrightarrow {{\text{AC}}} = – 5\sqrt 3 \) and \(\left| {\overrightarrow {{\text{AB}}} } \right|\left| {\overrightarrow {{\text{AC}}} } \right| = 10\). Find the area of triangle \(ABC\).
Answer/Explanation
Markscheme
attempt to find \(\cos {\rm{C\hat AB}}\) (seen anywhere) (M1)
eg\(\;\;\;\cos \theta = \frac{{\overrightarrow {{\text{AB}}} \bullet \overrightarrow {{\text{AC}}} }}{{\left| {\overrightarrow {{\text{AB}}} } \right|\left| {\overrightarrow {{\text{AC}}} } \right|}}\)
\(\cos {\rm{C\hat AB}} = \frac{{ – 5\sqrt 3 }}{{10}}\;\;\;\left( { = – \frac{{\sqrt 3 }}{2}} \right)\) A1
valid attempt to find \(\sin {\rm{C\hat AB}}\) (M1)
eg\(\;\;\;\)triangle, Pythagorean identity, \({\rm{C\hat AB}} = \frac{{5\pi }}{6},{\text{ }}150^\circ \)
\(\sin {\rm{C\hat AB}} = \frac{1}{2}\) (A1)
correct substitution into formula for area (A1)
eg\(\;\;\;\frac{1}{2} \times 10 \times \frac{1}{2},{\text{ }}\frac{1}{2} \times 10 \times \sin \frac{\pi }{6}\)
\({\text{area}} = \frac{{10}}{4}\;\;\;\left( { = \frac{5}{2}} \right)\) A1 N3
[6 marks]
Question
A line \(L\) passes through points \({\text{A}}( – 2,{\text{ }}4,{\text{ }}3)\) and \({\text{B}}( – 1,{\text{ }}3,{\text{ }}1)\).
(i) Show that \(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 1 \\ { – 1} \\ { – 2} \end{array}} \right)\).
(ii) Find \(\left| {\overrightarrow {{\text{AB}}} } \right|\).
Find a vector equation for \(L\).
The following diagram shows the line \(L\) and the origin \(O\). The point \(C\) also lies on \(L\).
Point \(C\) has position vector \(\left( {\begin{array}{*{20}{c}} 0 \\ y \\ { – 1} \end{array}} \right)\).
Show that \(y = 2\).
(i) Find \(\overrightarrow {{\text{OC}}} \bullet \overrightarrow {{\text{AB}}} \).
(ii) Hence, write down the size of the angle between \(C\) and \(L\).
Hence or otherwise, find the area of triangle \(OAB\).
Answer/Explanation
Markscheme
(i) correct approach A1
eg\(\;\;\;{\text{B}} – {\text{A, AO}} + {\text{OB}}\)
\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 1 \\ { – 1} \\ { – 2} \end{array}} \right)\) AG N0
(ii) correct substitution (A1)
eg\(\;\;\;\sqrt {{{(1)}^2} + {{( – 1)}^2} + {{( – 2)}^2}} ,{\text{ }}\sqrt {1 + 1 + 4} \)
\(\left| {\overrightarrow {{\text{AB}}} } \right| = \sqrt 6 \) A1 N2
[3 marks]
any correct equation in the form \(r = a + tb\) (any parameter for \(t\))
where \(a\) is \(\left( {\begin{array}{*{20}{c}} { – 2} \\ 4 \\ 3 \end{array}} \right)\) or \(\left( {\begin{array}{*{20}{c}} { – 1} \\ 3 \\ 1 \end{array}} \right)\) and \(b\) is a scalar multiple of \(\left( {\begin{array}{*{20}{c}} 1 \\ { – 1} \\ { – 2} \end{array}} \right)\) A2 N2
eg\(\;\;\;\(r\) = \left( {\begin{array}{*{20}{c}} { – 2} \\ 4 \\ 3 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 1 \\ { – 1} \\ { – 2} \end{array}} \right),{\text{ }}(x,{\text{ }}y,{\text{ }}z) = ( – 1,{\text{ }}3,{\text{ }}1) + t(1,{\text{ }} – 1,{\text{ }} – 2),{\text{ }}{\mathbf{r}} = \left( {\begin{array}{*{20}{c}} { – 1 + t} \\ {3 – t} \\ {1 – 2t} \end{array}} \right)\)
Note: Award A1 for the form \({\mathbf{a}} + t{\mathbf{b}}\), A1 for the form \(L = {\mathbf{a}} + t{\mathbf{b}}\), A0 for the form \({\mathbf{r}} = {\mathbf{b}} + t{\mathbf{a}}\).
METHOD 1
valid approach (M1)
eg\(\;\;\;\left( {\begin{array}{*{20}{c}} { – 1} \\ 3 \\ 1 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} 1 \\ { – 1} \\ { – 2} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0 \\ y \\ { – 1} \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} 0 \\ y \\ { – 1} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { – 2} \\ 4 \\ 3 \end{array}} \right) + s\left( {\begin{array}{*{20}{c}} 1 \\ { – 1} \\ { – 2} \end{array}} \right)\)
one correct equation from their approach A1
eg\(\;\;\; – 1 + t = 0,{\text{ }}1 – 2t = – 1,{\text{ }} – 2 + s = 0,{\text{ }}3 – 2s = – 1\)
one correct value for their parameter and equation A1
eg\(\;\;\;t = 1,{\text{ }}s = 2\)
correct substitution A1
eg\(\;\;\;3 + 1( – 1),{\text{ }}4 + 2( – 1)\)
\(y = 2\) AG N0
METHOD 2
valid approach (M1)
eg\(\;\;\;\overrightarrow {{\text{AC}}} = k\overrightarrow {{\text{AB}}} \)
correct working A1
eg\(\;\;\;\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}} 2 \\ {y – 4} \\ { – 4} \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} 2 \\ {y – 4} \\ { – 4} \end{array}} \right) = k\left( {\begin{array}{*{20}{c}} 1 \\ { – 1} \\ { – 2} \end{array}} \right)\)
\(k = 2\) A1
correct substitution A1
eg\(\;\;\;y – 4 = – 2\)
\(y = 2\) AG N0
[4 marks]
(i) correct substitution A1
eg\(\;\;\;0(1) + 2( – 1) – 1( – 2),{\text{ }}0 – 2 + 2\)
\(\overrightarrow {{\text{OC}}} \bullet \overrightarrow {{\text{AB}}} = 0\) A1 N1
(ii) \(9{0^ \circ }\) or \(\frac{\pi }{2}\) A1 N1
[3 marks]
METHOD 1 \({\text{(area}} = 0.5 \times {\text{height}} \times {\text{base)}}\)
\(\left| {\overrightarrow {{\text{OC}}} } \right| = \sqrt {0 + {2^2} + {{( – 1)}^2}} \;\;\;\left( { = \sqrt 5 } \right)\;\;\;\)(seen anywhere) A1
valid approach (M1)
eg\(\;\;\;\frac{1}{2} \times \left| {\overrightarrow {{\text{AB}}} } \right| \times \left| {\overrightarrow {{\text{OC}}} } \right|,{\text{ }}\left| {\overrightarrow {{\text{OC}}} } \right|\) is height of triangle
correct substitution A1
eg\(\;\;\;\frac{1}{2} \times \sqrt 6 \times \sqrt {0 + {{(2)}^2} + {{( – 1)}^2}} ,{\text{ }}\frac{1}{2} \times \sqrt 6 \times \sqrt 5 \)
area is \(\frac{{\sqrt {30} }}{2}\) A1 N2
METHOD 2 (difference of two areas)
one correct magnitude (seen anywhere) A1
eg\(\;\;\;\left| {\overrightarrow {{\text{OC}}} } \right| = \sqrt {{2^2} + {{( – 1)}^2}} \;\;\;\left( { = \sqrt 5 } \right),\;\;\;\left| {\overrightarrow {{\text{AC}}} } \right| = \sqrt {4 + 4 + 16} \;\;\;\left( { = \sqrt {24} } \right),\;\;\;\left| {\overrightarrow {{\text{BC}}} } \right| = \sqrt 6 \)
valid approach (M1)
eg\(\;\;\;\Delta {\text{OAC}} – \Delta {\text{OBC}}\)
correct substitution A1
eg\(\;\;\;\frac{1}{2} \times \sqrt {24} \times \sqrt 5 – \frac{1}{2} \times \sqrt 5 \times \sqrt 6 \)
area is \(\frac{{\sqrt {30} }}{2}\) A1 N2
METHOD 3 \({\text{(area}} = \frac{1}{2}ab\sin C{\text{ for }}\Delta {\text{OAB)}}\)
one correct magnitude of \(\overrightarrow {{\text{OA}}} \) or \(\overrightarrow {{\text{OB}}} \) (seen anywhere) A1
eg\(\;\;\;\left| {\overrightarrow {{\text{OA}}} } \right| = \sqrt {{{( – 2)}^2} + {4^2} + {3^2}} \;\;\;\left( { = \sqrt {29} } \right),\;\;\;\left| {\overrightarrow {{\text{OB}}} } \right| = \sqrt {1 + 9 + 1} \;\;\;\left( { = \sqrt {11} } \right)\)
valid attempt to find \(\cos \theta \) or \(\sin \theta \) (M1)
eg\(\;\;\;\cos {\text{C}} = \frac{{ – 1 – 3 – 2}}{{\sqrt 6 \times \sqrt {11} }}\;\;\;\left( { = \frac{{ – 6}}{{\sqrt {66} }}} \right),\;\;\;29 = 6 + 11 – 2\sqrt 6 \sqrt {11} \cos \theta ,{\text{ }}\frac{{\sin \theta }}{{\sqrt 5 }} = \frac{{\sin 90}}{{\sqrt {29} }}\)
correct substitution into \(\frac{1}{2}ab\sin {\text{C}}\) A1
eg\(\;\;\;\frac{1}{2} \times \sqrt 6 \times \sqrt {11} \times \sqrt {1 – \frac{{36}}{{66}}} ,{\text{ }}0.5 \times \sqrt 6 \times \sqrt {29} \times \frac{{\sqrt 5 }}{{\sqrt {29} }}\)
area is \(\frac{{\sqrt {30} }}{2}\) A1 N2
[4 marks]
Total [16 marks]
Question
A line \({L_1}\) passes through the points \({\text{A}}(0,{\text{ }} – 3,{\text{ }}1)\) and \({\text{B}}( – 2,{\text{ }}5,{\text{ }}3)\).
(i) Show that \(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} { – 2} \\ 8 \\ 2 \end{array}} \right)\).
(ii) Write down a vector equation for \({L_1}\).
A line \({L_2}\) has equation \({\mathbf{r}} = \left( {\begin{array}{*{20}{c}} { – 1} \\ 7 \\ { – 4} \end{array}} \right) + s\left( {\begin{array}{*{20}{c}} 0 \\ 1 \\ { – 1} \end{array}} \right)\). The lines \({L_1}\) and \({L_2}\) intersect at a point \(C\).
Show that the coordinates of \(C\) are \(( – 1,{\text{ }}1,{\text{ }}2)\).
A point \(D\) lies on line \({L_2}\) so that \(\left| {\overrightarrow {{\text{CD}}} } \right| = \sqrt {18} \) and \(\overrightarrow {{\text{CA}}} \bullet \overrightarrow {{\text{CD}}} = – 9\). Find \({\rm{A\hat CD}}\).
Answer/Explanation
Markscheme
(i) correct approach A1
eg\(\;\;\;{\text{OB}} – {\text{OA, }}\left( {\begin{array}{*{20}{c}} { – 2} \\ 5 \\ 3 \end{array}} \right) – \left( {\begin{array}{*{20}{c}} 0 \\ { – 3} \\ 1 \end{array}} \right),{\text{ B}} – {\text{A}}\)
\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} { – 2} \\ 8 \\ 2 \end{array}} \right)\) AG N0
(ii) any correct equation in the form \(r = a + \) t\(b\) (accept any parameter for \(t\))
where \(a\) is \(\left( {\begin{array}{*{20}{c}} 0 \\ { – 3} \\ 1 \end{array}} \right)\) or \(\left( {\begin{array}{*{20}{c}} { – 2} \\ 5 \\ 3 \end{array}} \right)\), and \(b\) is a scalar multiple of \(\left( {\begin{array}{*{20}{c}} { – 2} \\ 8 \\ 2 \end{array}} \right)\) A2 N2
eg\(r\) = \left( {\begin{array}{*{20}{c}} 0 \\ { – 3} \\ 1 \end{array}} \right) + t\left( {\begin{array}{*{20}{c}} { – 2} \\ 8 \\ 2 \end{array}} \right),\(r\) = \left( {\begin{array}{*{20}{c}} { – 2 – 2s} \\ {5 + 8s} \\ {3 + 2s} \end{array}} \right),\(r = 2i + 5j + 3k + \) t\(( – 2i + 8j + 2k)\)
Note: Award A1 for the form \(a\) + t\(b\), A1 for the form \(L = \(a\) + t\(b\),
A0 for the form \(r\) = \(b\) + t\(a\).
[3 marks]
valid approach (M1)
eg\(\;\;\;\)equating lines, \({L_1} = {L_2}\)
one correct equation in one variable A1
eg\(\;\;\; – 2t = – 1,{\text{ }} – 2 – 2t = – 1\)
valid attempt to solve (M1)
eg\(\;\;\;2t = 1,{\text{ }} – 2t = 1\)
one correct parameter A1
eg\(\;\;\;t = \frac{1}{2},{\text{ }}t = – \frac{1}{2},{\text{ }}s = – 6\)
correct substitution of either parameter A1
eg\(\;\;\;r = \left( {\begin{array}{*{20}{c}} 0 \\ { – 3} \\ 1 \end{array}} \right) + \frac{1}{2}\left( {\begin{array}{*{20}{c}} { – 2} \\ 8 \\ 2 \end{array}} \right),{\text{ }}r = \left( {\begin{array}{*{20}{c}} { – 2} \\ 5 \\ 3 \end{array}} \right) – \frac{1}{2}\left( {\begin{array}{*{20}{c}} { – 2} \\ 8 \\ 2 \end{array}} \right),{\text{ }}r = \left( {\begin{array}{*{20}{c}} { – 1} \\ 7 \\ { – 4} \end{array}} \right) – 6\left( {\begin{array}{*{20}{c}} 0 \\ 1 \\ { – 1} \end{array}} \right)\)
the coordinates of \(C\) are \(( – 1,{\text{ }}1,{\text{ }}2)\), or position vector of \(C\) is \(\left( {\begin{array}{*{20}{c}} { – 1} \\ 1 \\ 2 \end{array}} \right)\) AG N0
Note: If candidate uses the same parameter in both vector equations and working shown, award M1A1M1A0A0.
[5 marks]
valid approach (M1)
eg\(\;\;\;\)attempt to find \(\overrightarrow {{\text{CA}}} ,{\text{ }}\cos {\rm{A\hat CD}} = \frac{{\overrightarrow {{\text{CA}}} \bullet \overrightarrow {{\text{CD}}} }}{{\left| {\overrightarrow {{\text{CA}}} } \right|\left| {\overrightarrow {{\text{CD}}} } \right|}},{\rm{ A\hat CD}}\) formed by \(\overrightarrow {{\text{CA}}} \) and \(\overrightarrow {{\text{CD}}} \)
\(\overrightarrow {{\text{CA}}} = \left( {\begin{array}{*{20}{c}} 1 \\ { – 4} \\ { – 1} \end{array}} \right)\) (A1)
Notes: Exceptions to FT:
1 if candidate indicates that they are finding \(\overrightarrow {{\text{CA}}} \), but makes an error, award M1A0;
2 if candidate finds an incorrect vector (including \(\overrightarrow {{\text{AC}}} \)), award M0A0.
In both cases, if working shown, full FT may be awarded for subsequent correct FT work.
Award the final (A1) for simplification of their value for \({\rm{A\hat CD}}\).
Award the final A2 for finding their arc cos. If their value of cos does not allow them to find an angle, they cannot be awarded this A2.
finding \(\left| {\overrightarrow {{\text{CA}}} } \right|\) (may be seen in cosine formula) A1
eg\(\;\;\;\sqrt {{1^2} + {{( – 4)}^2} + {{( – 1)}^2}} ,{\text{ }}\sqrt {18} \)
correct substitution into cosine formula (A1)
eg\(\;\;\;\frac{{ – 9}}{{\sqrt {18} \sqrt {18} }}\)
finding \(\cos {\rm{A\hat CD}} – \frac{1}{2}\) (A1)
\({\rm{A\hat CD}} = \frac{{2\pi }}{3}\;\;\;(120^\circ )\) A2 N2
Notes: Award A1 if additional answers are given.
Award A1 for answer \(\frac{\pi }{3}{\rm{ (60^\circ )}}\).
[7 marks]
Total [15 marks]