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| Marks available | 2 |
| Reference code | 11M.1.sl.TZ1.2 |
Question
A line L passes through \({\text{A}}(1{\text{, }} – 1{\text{, }}2)\) and is parallel to the line \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
{ – 2}\\
1\\
5
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
1\\
3\\
{ – 2}
\end{array}} \right)\) .
The line L passes through point P when \(t = 2\) .
Write down a vector equation for L in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) .
Find
(i) \(\overrightarrow {{\rm{OP}}} \) ;
(ii) \(|\overrightarrow {{\rm{OP}}} |\) .
Answer/Explanation
Markscheme
correct equation in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) A2 N2
\({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
1\\
{ – 1}\\
2
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
1\\
3\\
{ – 2}
\end{array}} \right)\)
[2 marks]
(i) attempt to substitute \(t = 2\) into the equation (M1)
e.g. \(\left( {\begin{array}{*{20}{c}}
2\\
6\\
{ – 4}
\end{array}} \right)\) , \(\left( {\begin{array}{*{20}{c}}
1\\
{ – 1}\\
2
\end{array}} \right) + 2\left( {\begin{array}{*{20}{c}}
1\\
3\\
{ – 2}
\end{array}} \right)\)
\(\overrightarrow {{\rm{OP}}} = \left( {\begin{array}{*{20}{c}}
3\\
5\\
{ – 2}
\end{array}} \right)\) A1 N2
(ii) correct substitution into formula for magnitude A1
e.g. \(\sqrt {{3^2} + {5^2} + – {2^2}} \) , \(\sqrt {{3^2} + {5^2} + {2^2}} \)
\(|\overrightarrow {{\rm{OP}}}| = \sqrt {38} \) A1 N1
[4 marks]
| Marks available | 1 |
| Reference code | 12N.1.sl.TZ0.9 |
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]
Examiners report
Part (a) was answered correctly by nearly every candidate.
In part (b), the candidates who realized that the vectors must be perpendicular were successful using the scalar product to find p . Incorrect approaches included using magnitudes, or creating vector equations of lines for the sides and setting them equal to each other. In addition, there were a good number of candidates who worked backwards, using the given value of 3 for p to find the coordinates of point D. Candidates who work backwards on a “show that” question will earn no marks.
Part (c) was more difficult for candidates, and was left blank by some. Some candidates found \(\overrightarrow {{\rm{AC}}} \) rather than \(\overrightarrow {{\rm{OC}}} \) , as required. Many candidates recognized that the opposite sides of the rectangle must be equal, but did not consider the directions of the vectors for those sides. There were also a good number of candidates who mislabelled the vertices of their rectangles, which led to them working with a rectangle ABDC, rather than ABCD.
The majority of candidates who attempted part (d) were successful in multiplying the magnitudes of the sides. Unfortunately, there were some who set up their solutions correctly, then had arithmetic errors in their working.
| Marks available | 4 |
| Reference code | 15M.1.sl.TZ1.8 |
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]