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Question
Let u \( = \left( {\begin{array}{*{20}{c}}
2\\
3\\
{ – 1}
\end{array}} \right)\) and w \( = \left( {\begin{array}{*{20}{c}}
3\\
{ – 1}\\
p
\end{array}} \right)\) . Given that u is perpendicular to w , find the value of p .
Let \({\boldsymbol{v}} = \left( {\begin{array}{*{20}{c}}
1 \\
q \\
5
\end{array}} \right)\) . Given that \(\left| {\boldsymbol{v}} \right| = \sqrt {42} \), find the possible values of \(q\) .
Answer/Explanation
Markscheme
evidence of equating scalar product to 0 (M1)
\(2 \times 3 + 3 \times ( – 1) + ( – 1) \times p = 0\) (\(6 – 3 – p = 0\), \(3 – p = 0\)) A1
\(p = 3\) A1 N2
[3 marks]
evidence of substituting into magnitude formula (M1)
e.g. \(\sqrt {1 + {q^2} + 25} \) , \(1 + {q^2} + 25\)
setting up a correct equation A1
e.g. \(\sqrt {1 + {q^2} + 25} = \sqrt {42} \) , \(1 + {q^2} + 25 = 42\) , \({q^2} = 16\)
\(q = \pm 4\) A1 N2
[3 marks]
Question
The line \({L_1}\) is represented by the vector equation \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
{ – 3}\\
{ – 1}\\
{ – 25}
\end{array}} \right) + p\left( {\begin{array}{*{20}{c}}
2\\
1\\
{ – 8}
\end{array}} \right)\) .
A second line \({L_2}\) is parallel to \({L_1}\) and passes through the point B(\( – 8\), \( – 5\), \(25\)) .
Write down a vector equation for \({L_2}\) in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) .
A third line \({L_3}\) is perpendicular to \({L_1}\) and is represented by \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
5\\
0\\
3
\end{array}} \right) + q\left( {\begin{array}{*{20}{c}}
{ – 7}\\
{ – 2}\\
k
\end{array}} \right)\) .
Show that \(k = – 2\) .
The lines \({L_1}\) and \({L_3}\) intersect at the point A.
Find the coordinates of A.
The lines \({L_2}\)and \({L_3}\)intersect at point C where \(\overrightarrow {{\rm{BC}}} = \left( {\begin{array}{*{20}{c}}
6\\
3\\
{ – 24}
\end{array}} \right)\) .
(i) Find \(\overrightarrow {{\rm{AB}}} \) .
(ii) Hence, find \(|\overrightarrow {{\rm{AC}}} |\) .
Answer/Explanation
Markscheme
any correct equation in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) (accept any parameter) A2 N2
e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
{ – 8}\\
{ – 5}\\
{25}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
2\\
1\\
{ – 8}
\end{array}} \right)\)
Note: Award A1 for \({\boldsymbol{a}} + t{\boldsymbol{b}}\) , A1 for \(L = {\boldsymbol{a}} + t{\boldsymbol{b}}\) , A0 for \({\boldsymbol{r}} = {\boldsymbol{b}} + t{\boldsymbol{a}}\) .
[2 marks]
recognizing scalar product must be zero (seen anywhere) R1
e.g. \({\boldsymbol{a}} \bullet {\boldsymbol{b}} = 0\)
evidence of choosing direction vectors \(\left( {\begin{array}{*{20}{c}}
2\\
1\\
{ – 8}
\end{array}} \right),\left( {\begin{array}{*{20}{c}}
{ – 7}\\
{ – 2}\\
k
\end{array}} \right)\) (A1)(A1)
correct calculation of scalar product (A1)
e.g. \(2( – 7) + 1( – 2) – 8k\)
simplification that clearly leads to solution A1
e.g. \( – 16 – 8k\) , \( – 16 – 8k = 0\)
\(k = – 2\) AG N0
[5 marks]
evidence of equating vectors (M1)
e.g. \({L_1} = {L_3}\) , \(\left( {\begin{array}{*{20}{c}}
{ – 3}\\
{ – 1}\\
{ – 25}
\end{array}} \right) + p\left( {\begin{array}{*{20}{c}}
2\\
1\\
{ – 8}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
5\\
0\\
3
\end{array}} \right) + q\left( {\begin{array}{*{20}{c}}
{ – 7}\\
{ – 2}\\
{ – 2}
\end{array}} \right)\)
any two correct equations A1A1
e.g. \( – 3 + 2p = 5 – 7q\) , \( – 1 + p = – 2q\) , \(- 25 – 8p = 3 – 2q\)
attempting to solve equations (M1)
finding one correct parameter (\(p = – 3\) , \(q = 2\) ) A1
the coordinates of A are \(( – 9, – 4, – 1)\) A1 N3
[6 marks]
(i) evidence of appropriate approach (M1)
e.g. \(\overrightarrow {{\rm{OA}}} + \overrightarrow {{\rm{AB}}} = \overrightarrow {{\rm{OB}}} \) , \(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
{ – 8}\\
{ – 5}\\
{25}
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
{ – 9}\\
{ – 4}\\
{ – 1}
\end{array}} \right)\)
\(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
1\\
{ – 1}\\
{26}
\end{array}} \right)\) A1 N2
(ii) finding \(\overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}}
7\\
2\\
2
\end{array}} \right)\) A1
evidence of finding magnitude (M1)
e.g. \(|\overrightarrow {{\rm{AC}}} | = \sqrt {{7^2} + {2^2} + {2^2}} \)
\(|\overrightarrow {{\rm{AC}}} | = \sqrt {57} \) A1 N3
[5 marks]
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]
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
Consider the vectors \(\boldsymbol{a} = \left( {\begin{array}{*{20}{c}}
2\\
{ – 3}
\end{array}} \right)\) and \(\boldsymbol{b} = \left( {\begin{array}{*{20}{c}}
1\\
4
\end{array}} \right)\) .
Let \(2\boldsymbol{a} + \boldsymbol{b} + \boldsymbol{c} = 0\) , where \(0\) is the zero vector.
(a) Find
(i) \(2\boldsymbol{a} + \boldsymbol{b}\) ;
(ii) \(\left| {2\boldsymbol{a} + \boldsymbol{b}} \right|\) .
Let \(2\boldsymbol{a} + \boldsymbol{b} + \boldsymbol{c} = 0\) , where \(0\) is the zero vector.
(b) Find \(\boldsymbol{c}\) .
Find
(i) \(2\boldsymbol{a} + \boldsymbol{b}\) ;
(ii) \(\left| {2\boldsymbol{a} + \boldsymbol{b}} \right|\) .
Find \(\boldsymbol{c}\) .
Answer/Explanation
Markscheme
(a) (i) \(2\boldsymbol{a} = \left( {\begin{array}{*{20}{c}}
4\\
{ – 6}
\end{array}} \right)\) (A1)
correct expression for \(2\boldsymbol{a} + \boldsymbol{b}\) A1 N2
eg \(\left( {\begin{array}{*{20}{c}}
5\\
{ – 2}
\end{array}} \right)\) , \((5, – 2)\) , \(5\boldsymbol{i} – 2\boldsymbol{j}\)
(ii) correct substitution into length formula (A1)
eg \(\sqrt {{5^2} + {2^2}} \) , \(\sqrt {{5^2} + – {2^2}} \)
\(\left| {2\boldsymbol{a} + \boldsymbol{b}} \right| = \sqrt {29} \) A1 N2
[4 marks]
(b) valid approach (M1)
eg \(\boldsymbol{c} = – (2\boldsymbol{a} + \boldsymbol{b})\) , \(5 + x = 0\) , \( – 2 + y = 0\)
\(\boldsymbol{c} = \left( {\begin{array}{*{20}{c}}
{ – 5}\\
2
\end{array}} \right)\) A1 N2
[2 marks]
(i) \(2\boldsymbol{a} = \left( {\begin{array}{*{20}{c}}
4\\
{ – 6}
\end{array}} \right)\) (A1)
correct expression for \(2\boldsymbol{a} + \boldsymbol{b}\) A1 N2
eg \(\left( {\begin{array}{*{20}{c}}
5\\
{ – 2}
\end{array}} \right)\) , \((5, – 2)\) , \(5\boldsymbol{i} – 2\boldsymbol{j}\)
(ii) correct substitution into length formula (A1)
eg \(\sqrt {{5^2} + {2^2}} \) , \(\sqrt {{5^2} + – {2^2}} \)
\(\left| {2\boldsymbol{a} + \boldsymbol{b}} \right| = \sqrt {29} \) A1 N2
[4 marks]
valid approach (M1)
eg \(\boldsymbol{c} = – (2\boldsymbol{a} + \boldsymbol{b})\) , \(5 + x = 0\) , \( – 2 + y = 0\)
\(\boldsymbol{c} = \left( {\begin{array}{*{20}{c}}
{ – 5}\\
2
\end{array}} \right)\) A1 N2
[2 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]