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Question
The point O has coordinates (0 , 0 , 0) , point A has coordinates (1 , – 2 , 3) and point B has coordinates (– 3 , 4 , 2) .
(i) Show that \(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
{ – 4}\\
6\\
{ – 1}
\end{array}} \right)\) .
(ii) Find \({\rm{B}}\widehat {\rm{A}}{\rm{O}}\) .
The line \({L_1}\) has equation \(\left( {\begin{array}{*{20}{c}}
x\\
y\\
z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{ – 3}\\
4\\
2
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
{ – 4}\\
6\\
{ – 1}
\end{array}} \right)\) .
Write down the coordinates of two points on \({L_1}\) .
The line \({L_2}\) passes through A and is parallel to \(\overrightarrow {{\rm{OB}}} \) .
(i) Find a vector equation for \({L_2}\) , giving your answer in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) .
(ii) Point \(C(k, – k,5)\) is on \({L_2}\) . Find the coordinates of C.
The line \({L_3}\) has equation \(\left( {\begin{array}{*{20}{c}}
x\\
y\\
z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
3\\
{ – 8}\\
0
\end{array}} \right) + p\left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
{ – 1}
\end{array}} \right)\) and passes through the point C.
Find the value of p at C.
Answer/Explanation
Markscheme
(i) evidence of approach M1
e.g. \(\overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OB}}} = \overrightarrow {{\rm{AB}}} \)
\(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
{ – 4}\\
6\\
{ – 1}
\end{array}} \right)\) AG N0
(ii) for choosing correct vectors, (\(\overrightarrow {{\rm{AO}}} \) with \(\overrightarrow {{\rm{AB}}} \) or \(\overrightarrow {{\rm{OA}}} \) with \(\overrightarrow {{\rm{BA}}} \) ) (A1)(A1)
Note: Using \(\overrightarrow {{\rm{AO}}} \) with \(\overrightarrow {{\rm{BA}}} \) will lead to \(\pi – 0.799\) . If they then say \({\rm{B}}\widehat {\rm{A}}{\rm{O}} = 0.799\) , this is a correct solution.
calculating \(\overrightarrow {{\rm{AO}}} \bullet \overrightarrow {{\rm{AB}}} \) , \(\left| {\overrightarrow {{\rm{AO}}} } \right|\) , \(\left| {\overrightarrow {{\rm{AB}}} } \right|\) (A1)(A1)(A1)
e.g. \({d_1} \bullet {d_2} = ( – 1)( – 4) + (2)(6) + ( – 3)( – 1)( = 19)\)
\(\left| {{d_1}} \right| = \sqrt {{{( – 1)}^2} + {2^2} + {{( – 3)}^2}} ( = \sqrt {14} )\) , \(\left| {{d_2}} \right| = \sqrt {{{( – 4)}^2} + {6^2} + {{( – 1)}^2}} ( = \sqrt {53} )\)
evidence of using the formula to find the angle M1
e.g. \(\cos \theta = \frac{{( – 1)( – 4) + (2)(6) + ( – 3)( – 1)}}{{\sqrt {{{( – 1)}^2} + {2^2} + {{( – 3)}^2}} \sqrt {{{( – 4)}^2} + {6^2} + {{( – 1)}^2}} }}\) , \(\frac{{19}}{{\sqrt {14} \sqrt {53} }}\) , \(0.69751 \ldots \)
\({\rm{B}}\widehat {\rm{A}}{\rm{O}} = 0.799\) radians (accept \(45.8^\circ \)) A1 N3
[8 marks]
two correct answers A1A1
e.g. (1, \( – 2\), 3) , (\( – 3\), 4, 2) , (\( – 7\), 10, 1), (\( – 11\), 16, 0) N2
[2 marks]
(i) \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
3
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
{ – 3}\\
4\\
2
\end{array}} \right)\) A2 N2
(ii) C on \({L_2}\) , so \(\left( {\begin{array}{*{20}{c}}
k\\
{ – k}\\
5
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
3
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
{ – 3}\\
4\\
2
\end{array}} \right)\) (M1)
evidence of equating components (A1)
e.g. \(1 – 3t = k\) , \( – 2 + 4t = – k\) , \(5 = 3 + 2t\)
one correct value \(t = 1\) , \(k = – 2\) (seen anywhere) (A1)
coordinates of C are \(( – 2{\text{, }}2{\text{, }}5)\) A1 N3
[6 marks]
for setting up one (or more) correct equation using \(\left( {\begin{array}{*{20}{c}}
{ – 2}\\
2\\
5
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
3\\
{ – 8}\\
0
\end{array}} \right) + p\left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
{ – 1}
\end{array}} \right)\) (M1)
e.g. \(3 + p = – 2\) , \( – 8 – 2p = 2\) , \( – p = 5\)
\(p = – 5\) A1 N2
[2 marks]
Question
Consider the points P(2, −1, 5) and Q(3, − 3, 8). Let \({L_1}\) be the line through P and Q.
Show that \(\overrightarrow {{\rm{PQ}}} = \left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
3
\end{array}} \right)\) .
The line \({L_1}\) may be represented by \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
3 \\
{ – 3} \\
8
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
1 \\
{ – 2} \\
3
\end{array}} \right)\) .
(i) What information does the vector \(\left( {\begin{array}{*{20}{c}}
3\\
{ – 3}\\
8
\end{array}} \right)\) give about \({L_1}\) ?
(ii) Write down another vector representation for \({L_1}\) using \(\left( {\begin{array}{*{20}{c}}
3\\
{ – 3}\\
8
\end{array}} \right)\) .
The point \({\text{T}}( – 1{\text{, }}5{\text{, }}p)\) lies on \({L_1}\) .
Find the value of \(p\) .
The point T also lies on \({L_2}\) with equation \(\left( {\begin{array}{*{20}{c}}
x\\
y\\
z
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{ – 3}\\
9\\
2
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
q
\end{array}} \right)\) .
Show that \(q = – 3\) .
Let \(\theta \) be the obtuse angle between \({L_1}\) and \({L_2}\) . Calculate the size of \(\theta \) .
Answer/Explanation
Markscheme
evidence of correct approach A1
e.g. \(\overrightarrow {{\rm{PQ}}} = \overrightarrow {{\rm{OQ}}} – \overrightarrow {{\rm{OP}}} \) , \(\left( {\begin{array}{*{20}{c}}
3\\
{ – 3}\\
8
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
2\\
{ – 1}\\
5
\end{array}} \right)\)
\(\overrightarrow {{\rm{PQ}}} = \left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
3
\end{array}} \right)\) AG N0
[1 mark]
(i) correct description R1 N1
e.g. reference to \(\left( {\begin{array}{*{20}{c}}
3\\
{ – 3}\\
8
\end{array}} \right)\) being the position vector of a point on the line, a vector to the line, a point on the line.
(ii) any correct expression in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) A2 N2
where \({\boldsymbol{a}}\) is \(\left( {\begin{array}{*{20}{c}}
3 \\
{ – 3} \\
8
\end{array}} \right)\) , and \({\boldsymbol{b}}\) is a scalar multiple of \(\left( {\begin{array}{*{20}{c}}
1 \\
{ – 2} \\
3
\end{array}} \right)\)
e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
3 \\
{ – 3} \\
8
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
{ – 1} \\
2 \\
{ – 3}
\end{array}} \right)\) , \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
{3 + 2s} \\
{ – 3 – 4s} \\
{8 + 6s}
\end{array}} \right)\)
[3 marks]
one correct equation (A1)
e.g. \(3 + s = – 1\) , \( – 3 – 2s = 5\)
\(s = – 4\) A1
\(p = – 4\) A1 N2
[3 marks]
one correct equation A1
e.g. \( – 3 + t = – 1\) , \(9 – 2t = 5\)
\(t = 2\) A1
substituting \(t = 2\)
e.g. \(2 + 2q = – 4\) , \(2q = – 6\) A1
\(q = – 3\) AG N0
[3 marks]
choosing correct direction vectors \(\left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
3
\end{array}} \right)\) and \(\left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
{ – 3}
\end{array}} \right)\) (A1)(A1)
finding correct scalar product and magnitudes (A1)(A1)(A1)
scalar product \((1)(1) + ( – 2)( – 2) + ( – 3)(3)\) \(( = – 4)\)
magnitudes \(\sqrt {{1^2} + {{( – 2)}^2} + {3^2}} \) \( = \sqrt {14} \) , \(\sqrt {{1^2} + {{( – 2)}^2} + {{( – 3)}^2}} \) \( = \sqrt {14} \)
evidence of substituting into scalar product M1
e.g. \(\cos \theta = \frac{{ – 4}}{{3.741 \ldots \times 3.741 \ldots }}\)
\(\theta = 1.86\) radians (or \(107^\circ \)) A1 N4
[7 marks]
Question
Let \({\boldsymbol{v}} = \left( {\begin{array}{*{20}{c}}
2\\
{ – 3}\\
6
\end{array}} \right)\) and \({\boldsymbol{w}} = \left( {\begin{array}{*{20}{c}}
k\\
{ – 2}\\
4
\end{array}} \right)\) , for \(k > 0\) . The angle between v and w is \(\frac{\pi }{3}\) .
Find the value of \(k\) .
Answer/Explanation
Markscheme
correct substitutions for \({\boldsymbol{v}} \bullet {\boldsymbol{w}}\) ; \(\left| {\boldsymbol{v}} \right|\) ; \(\left| {\boldsymbol{w}} \right|\) (A1)(A1)(A1)
e.g. \(2k + ( – 3) \times ( – 2) + 6 \times 4\) , \(2k + 30\) ; \(\sqrt {{2^2} + {{( – 3)}^2} + {6^2}} \) , \(\sqrt {49} \) ; \(\sqrt {{k^2} + {{( – 2)}^2} + {4^2}} \) , \(\sqrt {{k^2} + 20} \)
evidence of substituting into the formula for scalar product (M1)
e.g. \(\frac{{2k + 30}}{{7 \times \sqrt {{k^2} + 20} }}\)
correct substitution A1
e.g. \(\cos \frac{\pi }{3} = \frac{{2k + 30}}{{7 \times \sqrt {{k^2} + 20} }}\)
\(k = 18.8\) A2 N5
[7 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]
Question
Let \(u = 6i + 3j + 6k\) and \(v = 2i + 2j + k\).
Find
(i) \(u \bullet v\);
(ii) \(\left| {{u}} \right|\);
(iii) \(\left| {{v}} \right|\).
Find the angle between \({{u}}\) and \({{v}}\).
Answer/Explanation
Markscheme
(i) correct substitution (A1)
eg\(\;\;\;6 \times 2 + 3 \times 2 + 6 \times 1\)
\(u \bullet v = 24\) A1 N2
(ii) correct substitution into magnitude formula for \({{u}}\) or \({{v}}\) (A1)
eg\(\;\;\;\sqrt {{6^2} + {3^2} + {6^2}} ,{\text{ }}\sqrt {{2^2} + {2^2} + {1^2}} \), correct value for \(\left| {{v}} \right|\)
\(\left| {{u}} \right| = 9\) A1 N2
(iii) \(\left| {{v}} \right| = 3\) A1 N1
[5 marks]
correct substitution into angle formula (A1)
eg\(\;\;\;\frac{{24}}{{9 \times 3}},{\text{ }}0.\bar 8\)
\(0.475882,{\text{ }}27.26604^\circ \) A1 N2
\(0.476,{\text{ }}27.3^\circ \)
[2 marks]
Total [7 marks]
Question
The points A and B lie on a line \(L\), and have position vectors \(\left( {\begin{array}{*{20}{c}} { – 3} \\ { – 2} \\ 2 \end{array}} \right)\) and \(\left( {\begin{array}{*{20}{c}} 6 \\ 4 \\ { – 1} \end{array}} \right)\) respectively. Let O be the origin. This is shown on the following diagram.
The point C also lies on \(L\), such that \(\overrightarrow {{\text{AC}}} = 2\overrightarrow {{\text{CB}}} \).
Let \(\theta \) be the angle between \(\overrightarrow {{\text{AB}}} \) and \(\overrightarrow {{\text{OC}}} \).
Let D be a point such that \(\overrightarrow {{\text{OD}}} = k\overrightarrow {{\text{OC}}} \), where \(k > 1\). Let E be a point on \(L\) such that \({\rm{C\hat ED}}\) is a right angle. This is shown on the following diagram.
Find \(\overrightarrow {{\text{AB}}} \).
Show that \(\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 2 \\ 0 \end{array}} \right)\).
Find \(\theta \).
(i) Show that \(\left| {\overrightarrow {{\text{DE}}} } \right| = (k – 1)\left| {\overrightarrow {{\text{OC}}} } \right|\sin \theta \).
(ii) The distance from D to line \(L\) is less than 3 units. Find the possible values of \(k\).
Answer/Explanation
Markscheme
valid approach (addition or subtraction) (M1)
eg\(\,\,\,\,\,\)\({\text{AO}} + {\text{OB}},{\text{ B}} – {\text{A}}\)
\(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 9 \\ 6 \\ { – 3} \end{array}} \right)\) A1 N2
[2 marks]
METHOD 1
valid approach using \(\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right)\) (M1)
eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}} {x + 3} \\ {y + 2} \\ {z – 2} \end{array}} \right),{\text{ }}\overrightarrow {{\text{CB}}} = \left( {\begin{array}{*{20}{c}} {6 – x} \\ {4 – y} \\ { – 1 – z} \end{array}} \right)\)
correct working A1
eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} {x + 3} \\ {y + 2} \\ {z – 2} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {12 – 2x} \\ {8 – 2y} \\ { – 2 – 2z} \end{array}} \right)\)
all three equations A1
eg\(\,\,\,\,\,\)\(x + 3 = 12 – 2x,{\text{ }}y + 2 = 8 – 2y,{\text{ }}z – 2 = – 2 – 2z\),
\(\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 2 \\ 0 \end{array}} \right)\) AG N0
METHOD 2
valid approach (M1)
eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{OC}}} – \overrightarrow {{\text{OA}}} = 2\left( {\overrightarrow {{\text{OB}}} – \overrightarrow {{\text{OC}}} } \right)\)
correct working A1
eg\(\,\,\,\,\,\)\(3\overrightarrow {{\text{OC}}} = 2\overrightarrow {{\text{OB}}} + \overrightarrow {{\text{OA}}} \)
correct substitution of \(\overrightarrow {{\text{OB}}} \) and \(\overrightarrow {{\text{OA}}} \) A1
eg\(\,\,\,\,\,\)\(3\overrightarrow {{\text{OC}}} = 2\left( {\begin{array}{*{20}{c}} 6 \\ 4 \\ { – 1} \end{array}} \right) + \left( {\begin{array}{*{20}{c}} { – 3} \\ { – 2} \\ 2 \end{array}} \right),{\text{ }}3\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} 9 \\ 6 \\ 0 \end{array}} \right)\)
\(\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 2 \\ 0 \end{array}} \right)\) AG N0
METHOD 3
valid approach (M1)
eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{AC}}} = \frac{2}{3}\overrightarrow {{\text{AB}}} \), diagram, \(\overrightarrow {{\text{CB}}} = \frac{1}{3}\overrightarrow {{\text{AB}}} \)
correct working A1
eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}} 6 \\ 4 \\ { – 2} \end{array}} \right),{\text{ }}\overrightarrow {{\text{CB}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 2 \\ { – 1} \end{array}} \right)\)
correct working involving \(\overrightarrow {{\text{OC}}} \) A1
eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} { – 3} \\ { – 2} \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 6 \\ 4 \\ { – 2} \end{array}} \right),{\text{ }}\left( {\begin{array}{*{20}{c}} 6 \\ 4 \\ { – 1} \end{array}} \right) – \left( {\begin{array}{*{20}{c}} 3 \\ 2 \\ { – 1} \end{array}} \right)\)
\(\overrightarrow {{\text{OC}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 2 \\ 0 \end{array}} \right)\) AG N0
[3 marks]
finding scalar product and magnitudes (A1)(A1)(A1)
scalar product \( = (9 \times 3) + (6 \times 2) + ( – 3 \times 0){\text{ }}( = 39)\)
magnitudes \(\sqrt {81 + 36 + 9} {\text{ }}( = 11.22),{\text{ }}\sqrt {9 + 4} {\text{ }}( = 3.605)\)
substitution into formula M1
eg\(\,\,\,\,\,\)\(\cos \theta = \frac{{(9 \times 3) + 12}}{{\sqrt {126} \times \sqrt {13} }}\)
\(\theta = 0.270549{\text{ }}({\text{accept }}15.50135^\circ )\)
\(\theta = 0.271{\text{ }}({\text{accept }}15.5^\circ )\) A1 N4
[5 marks]
(i) attempt to use a trig ratio M1
eg\(\,\,\,\,\,\)\(\sin \theta = \frac{{{\text{DE}}}}{{{\text{CD}}}},{\text{ }}\left| {\overrightarrow {{\text{CE}}} } \right| = \left| {\overrightarrow {{\text{CD}}} } \right|\cos \theta \)
attempt to express \(\overrightarrow {{\text{CD}}} \) in terms of \(\overrightarrow {{\text{OC}}} \) M1
eg\(\,\,\,\,\,\)\(\overrightarrow {{\text{OC}}} + \overrightarrow {{\text{CD}}} = \overrightarrow {{\text{OD}}} ,{\text{ OC}} + {\text{CD}} = {\text{OD}}\)
correct working A1
eg\(\,\,\,\,\,\)\(\left| {k\overrightarrow {{\text{OC}}} – \overrightarrow {{\text{OC}}} } \right|\sin \theta \)
\(\left| {\overrightarrow {{\text{DE}}} } \right| = (k – 1)\left| {\overrightarrow {{\text{OC}}} } \right|\sin \theta \) AG N0
(ii) valid approach involving the segment DE (M1)
eg\(\,\,\,\,\,\)recognizing \(\left| {\overrightarrow {{\text{DE}}} } \right| < 3,{\text{ DE}} = 3\)
correct working (accept equation) (A1)
eg\(\,\,\,\,\,\)\((k – 1)(\sqrt {13} )\sin 0.271 < 3,{\text{ }}k – 1 = 3.11324\)
\(1 < k < 4.11{\text{ }}({\text{accept }}k < 4.11{\text{ but not }}k = 4.11)\) A1 N2
[6 marks]
Question
Let \(\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 4 \\ 1 \\ 2 \end{array}} \right)\).
Find \(\left| {\overrightarrow {{\text{AB}}} } \right|\).
Let \(\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 0 \\ 0 \end{array}} \right)\). Find \({\rm{B\hat AC}}\).
Answer/Explanation
Markscheme
correct substitution (A1)
eg\(\,\,\,\,\,\)\(\sqrt {{4^2} + {1^2} + {2^2}} \)
4.58257
\(\left| {\overrightarrow {{\text{AB}}} } \right| = \sqrt {21} \) (exact), 4.58 A1 N2
[2 marks]
finding scalar product and \(\left| {\overrightarrow {{\text{AC}}} } \right|\) (A1)(A1)
scalar product \( = (4 \times 3) + (1 \times 0) + (2 \times 0){\text{ }}( = 12)\)
\(\left| {\overrightarrow {{\text{AC}}} } \right| = \sqrt {{3^2} + 0 + 0} {\text{ }}( = 3)\)
substituting their values into cosine formula (M1)
eg cos B\(\hat A\)C\({\text{ = }}\frac{{4 \times 3 + 0 + 0}}{{\sqrt {{3^2}} \times \sqrt {21} }},{\text{ }}\frac{4}{{\sqrt {21} }},{\text{ }}\cos \theta = 0.873\)
0.509739 (29.2059°)
\({\rm{B\hat AC}} = 0.510\) (29.2°) A1 N2
[4 marks]