| Marks available | 2 |
| Reference code | 10M.1.sl.TZ1.10 |
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 \(|\overrightarrow {{\rm{AC}}} |\) .
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]
Examiners report
Many candidates gave a correct vector equation for the line.
A common error was to misplace the initial position and direction vectors. Those who set the scalar product of the direction vectors to zero typically solved for k successfully. Those who substituted \(k = – 2\) earned fewer marks for working backwards in a “show that” question.
Many went on to find the coordinates of point A, however some used the same letter, say p, for each parameter and thus could not solve the system.
Part (d) proved challenging as many candidates did not consider that \(\overrightarrow {{\rm{AB}}} + \overrightarrow {{\rm{BC}}} = \overrightarrow {{\rm{AC}}} \) . Rather, many attempted to find the coordinates of point C, which became a more arduous and error-prone task.
| Marks available | 2 |
| Reference code | 10M.1.sl.TZ2.2 |
Question
Let \(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
6\\
{ – 2}\\
3
\end{array}} \right)\) and \(\overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}}
{ – 2}\\
{ – 3}\\
2
\end{array}} \right)\) .
Find \(\overrightarrow {{\rm{BC}}} \) .
Find a unit vector in the direction of \(\overrightarrow {{\rm{AB}}} \) .
Show that \(\overrightarrow {{\rm{AB}}} \) is perpendicular to \(\overrightarrow {{\rm{AC}}} \) .
Markscheme
evidence of appropriate approach (M1)
e.g. \(\overrightarrow {{\rm{BC}}} {\rm{ = }}\overrightarrow {{\rm{BA}}} + \overrightarrow {{\rm{AC}}} \) , \(\left( {\begin{array}{*{20}{c}}
{ – 2}\\
{ – 3}\\
2
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
6\\
{ – 2}\\
3
\end{array}} \right)\)
\(\overrightarrow {{\rm{BC}}} = \left( {\begin{array}{*{20}{c}}
{ – 8}\\
{ – 1}\\
{ – 1}
\end{array}} \right)\) A1 N2
[2 marks]
attempt to find the length of \(\overrightarrow {{\rm{AB}}} \) (M1)
\(\overrightarrow {|{\rm{AB}}} | = \sqrt {{6^2} + {{( – 2)}^2} + {3^2}} \) \(( = \sqrt {36 + 4 + 9} = \sqrt {49} = 7)\) (A1)
unit vector is \(\frac{1}{7}\left( {\begin{array}{*{20}{c}}
6\\
{ – 2}\\
3
\end{array}} \right)\) \(\left( { = \left( {\begin{array}{*{20}{c}}
{\frac{6}{7}}\\
{ – \frac{2}{7}}\\
{\frac{3}{7}}
\end{array}} \right)} \right)\) A1 N2
[3 marks]
recognizing that the dot product or \(\cos \theta \) being 0 implies perpendicular (M1)
correct substitution in a scalar product formula A1
e.g. \((6) \times ( – 2) + ( – 2) \times ( – 3) + (3) \times (2)\) , \(\cos \theta = \frac{{ – 12 + 6 + 6}}{{7 \times \sqrt {17} }}\)
correct calculation A1
e.g. \(\overrightarrow {{\rm{AB}}} \bullet \overrightarrow {{\rm{AC}}} = 0\) , \(\cos \theta = 0\)
therefore, they are perpendicular AG N0
[3 marks]
Examiners report
Part (a) was generally done well with candidates employing different correct methods to find the vector \(\overrightarrow {{\rm{BC}}} \) . Some candidates subtracted the given vectors in the wrong order and others simply added them. Calculation errors were seen with some frequency.
Many candidates did not appear to know how to find a unit vector in part (b). Some tried to write down the vector equation of a line, indicating no familiarity with the concept of unit vectors while others gave the vector (1, 1, 1) or wrote the same vector \(\overrightarrow {{\rm{AB}}} \) as a linear combination of i, j and k. A number of candidates correctly found the magnitude but did not continue on to write the unit vector.
Candidates were generally successful in showing that the vectors in part (c) were perpendicular. Many used the efficient approach of showing that the scalar product equalled zero, while others worked a little harder than necessary and used the cosine rule to find the angle between the two vectors.
| Marks available | 5 |
| Reference code | 10N.1.sl.TZ0.8 |
Question
The diagram shows quadrilateral ABCD with vertices A(1, 0), B(1, 5), C(5, 2) and D(4, −1) .
(i) Show that \(\overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}}
4\\
2
\end{array}} \right)\) .
(ii) Find \(\overrightarrow {{\rm{BD}}} \) .
(iii) Show that \(\overrightarrow {{\rm{AC}}} \) is perpendicular to \(\overrightarrow {{\rm{BD}}} \) .
The line (AC) has equation \({\boldsymbol{r}} = {\boldsymbol{u}} + s{\boldsymbol{v}}\) .
(i) Write down vector u and vector v .
(ii) Find a vector equation for the line (BD).
The lines (AC) and (BD) intersect at the point \({\text{P}}(3{\text{, }}k)\) .
Show that \(k = 1\) .
The lines (AC) and (BD) intersect at the point \({\text{P}}(3{\text{, }}k)\) .
Hence find the area of triangle ACD.
Markscheme
(i) correct approach A1
e.g. \(\overrightarrow {{\rm{OC}}} – \overrightarrow {{\rm{OA}}} \) , \(\left( {\begin{array}{*{20}{c}}
5\\
2
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
1\\
0
\end{array}} \right)\)
\(\overrightarrow {{\rm{AC}}} = \left( {\begin{array}{*{20}{c}}
4\\
2
\end{array}} \right)\) AG N0
(ii) appropriate approach (M1)
e.g. \({\mathop{\rm D}\nolimits} – {\rm{B}}\) , \(\left( {\begin{array}{*{20}{c}}
4\\
{ – 1}
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
1\\
5
\end{array}} \right)\) , move 3 to the right and 6 down
\(\overrightarrow {{\rm{BD}}} = \left( {\begin{array}{*{20}{c}}
3\\
{ – 6}
\end{array}} \right)\) A1 N2
(iii) finding the scalar product A1
e.g. \(4(3) + 2( – 6)\) , \(12 – 12\)
valid reasoning R1
e.g. \(4(3) + 2( – 6) = 0\) , scalar product is zero
\(\overrightarrow {{\rm{AC}}} \) is perpendicular to \(\overrightarrow {{\rm{BD}}} \) AG N0
[5 marks]
(i) correct “position” vector for u; “direction” vector for v A1A1 N2
e.g. \({\boldsymbol{u}} = \left( {\begin{array}{*{20}{c}}
5\\
2
\end{array}} \right)\) , \({\boldsymbol{u}} = \left( {\begin{array}{*{20}{c}}
1\\
0
\end{array}} \right)\) ; \({\boldsymbol{v}} = \left( {\begin{array}{*{20}{c}}
4\\
2
\end{array}} \right)\) , \({\boldsymbol{v}} = \left( {\begin{array}{*{20}{c}}
{ – 2}\\
{ – 1}
\end{array}} \right)\)
accept in equation e.g. \(\left( {\begin{array}{*{20}{c}}
5\\
2
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
{ – 4}\\
{ – 2}
\end{array}} \right)\)
(ii) any correct equation in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) , where \({\boldsymbol{b}} = \overrightarrow {{\rm{BD}}} \)
\({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
1\\
5
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
3\\
{ – 6}
\end{array}} \right)\) , \(\left( {\begin{array}{*{20}{c}}
x\\
y
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
4\\
{ – 1}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
{ – 1}\\
2
\end{array}} \right)\) A2 N2
[4 marks]
METHOD 1
substitute (3, k) into equation for (AC) or (BD) (M1)
e.g. \(3 = 1 + 4s\) , \(3 = 1 + 3t\)
value of t or s A1
e.g. \(s = \frac{1}{2}\) , \( – \frac{1}{2}\) , \(t = \frac{2}{3}\) , \( – \frac{1}{3}\)
substituting A1
e.g. \(k = 0 + \frac{1}{2}(2)\)
\(k = 1\) AG N0
METHOD 2
setting up two equations (M1)
e.g. \(1 + 4s = 4 + 3t\) , \(2s = – 1 – 6t\) ; setting vector equations of lines equal
value of t or s A1
e.g. \(s = \frac{1}{2}\) , \( – \frac{1}{2}\) , \(t = \frac{2}{3}\) , \( – \frac{1}{3}\)
substituting A1
e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
4\\
{ – 1}
\end{array}} \right) – \frac{1}{3}\left( {\begin{array}{*{20}{c}}
3\\
{ – 6}
\end{array}} \right)\)
\(k = 1\) AG N0
[3 marks]
\(\overrightarrow {{\rm{PD}}} = \left( {\begin{array}{*{20}{c}}
1\\
{ – 2}
\end{array}} \right)\) (A1)
\(|\overrightarrow {{\rm{PD}}} | = \sqrt {{2^2} + {1^2}} \) \(( = \sqrt 5 )\) (A1)
\(|\overrightarrow {{\rm{AC}}} | = \sqrt {{4^2} + {2^2}} \) \(( = \sqrt {20} )\) (A1)
area \( = \frac{1}{2} \times |\overrightarrow {{\rm{AC}}} | \times |\overrightarrow {{\rm{PD}}} |\) (\( = \frac{1}{2} \times \sqrt {20} \times \sqrt 5 \)) M1
\( = 5\) A1 N4
[5 marks]
Examiners report
The majority of candidates were successful on part (a), finding vectors between two points and using the scalar product to show two vectors to be perpendicular.
Although a large number of candidates answered part (b) correctly, there were many who had trouble with the vector equation of a line. Most notably, there were those who confused the position vector with the direction vector, and those who wrote their equation in an incorrect form.
In part (c), most candidates seemed to know what was required, though there were many who made algebraic errors when solving for the parameters. A few candidates worked backward, using \(k = 1\) , which is not allowed on a “show that” question.
In part (d), candidates attempted many different geometric and vector methods to find the area of the triangle. As the question said “hence”, it was required that candidates should use answers from their previous working – i.e. \({\rm{AC}} \bot {\rm{BD}}\) and \({\text{P}}(3{\text{, }}1)\) . Some geometric approaches, while leading to the correct answer, did not use “hence” or lacked the required justification.
| Marks available | 3 |
| Reference code | 11M.1.sl.TZ1.9 |
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 \(\overrightarrow {{\rm{BC}}} \) .
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]
Examiners report
Many candidates answered (a) correctly, although some reversed the vectors when finding \(\overrightarrow {{\rm{BC}}} \) , while others miscopied the vectors from the question paper.
Students had no difficulty finding the scalar product and magnitudes of the vectors used in finding the cosine. However, few recognized that \(\overrightarrow {{\rm{BA}}} \) is the vector to apply in the formula to find the cosine value. Most used \(\overrightarrow {{\rm{AB}}} \) to obtain a positive cosine, which neglects that the angle is obtuse and thus has a negative cosine. Surprisingly few students could then take a value for cosine and use it to find a value for sine. Most left (bii) blank entirely.
Part (c) proved accessible for many candidates. Some created an expression for \(|\overrightarrow {{\rm{CD}}} |\) and then substituted the given \(p = 3\) to obtain \(\sqrt {50} \) , which does not satisfy the “show that” instruction. Many students recognized that the scalar product must be zero for vectors to be perpendicular, and most provided the supporting calculations.
| Marks available | 2 |
| Reference code | 11M.1.sl.TZ2.3 |
Question
The following diagram shows quadrilateral ABCD, with \(\overrightarrow {{\rm{AD}}} = \overrightarrow {{\rm{BC}}} \) , \(\overrightarrow {{\rm{AB}}} = \left( \begin{array}{l}
3\\
1
\end{array} \right)\) , and \(\overrightarrow {{\rm{AC}}} = \left( \begin{array}{l}
4\\
4
\end{array} \right)\) .
Find \(\overrightarrow {{\rm{BC}}} \) .
Show that \(\overrightarrow {{\rm{BD}}} = \left( {\begin{array}{*{20}{c}}
{ – 2}\\
2
\end{array}} \right)\) .
Show that vectors \(\overrightarrow {{\rm{BD}}} \) and \(\overrightarrow {{\rm{AC}}} \) are perpendicular.
Markscheme
evidence of appropriate approach (M1)
e.g. \(\overrightarrow {{\rm{AC}}} – \overrightarrow {{\rm{AB}}} \) , \(\left( \begin{array}{l}
4 – 3\\
4 – 1
\end{array} \right)\)
\(\overrightarrow {{\rm{BC}}} = \left( \begin{array}{l}
1\\
3
\end{array} \right)\) A1 N2
[2 marks]
METHOD 1
\(\overrightarrow {{\rm{AD}}} = \left( {\begin{array}{*{20}{c}}
1\\
3
\end{array}} \right)\) (A1)
correct approach A1
e.g. \(\overrightarrow {{\rm{AD}}} – \overrightarrow {{\rm{AB}}} \) , \(\left( \begin{array}{l}
1 – 3\\
3 – 1
\end{array} \right)\)
\(\overrightarrow {{\rm{BD}}} = \left( {\begin{array}{*{20}{c}}
{ – 2}\\
2
\end{array}} \right)\) AG N0
METHOD 2
recognizing \(\overrightarrow {{\rm{CD}}} = \overrightarrow {{\rm{BA}}} \) (A1)
correct approach A1
e.g. \(\overrightarrow {{\rm{BC}}} + \overrightarrow {{\rm{CD}}} \) , \(\left( \begin{array}{l}
1 – 3\\
3 – 1
\end{array} \right)\)
\(\overrightarrow {{\rm{BD}}} = \left( {\begin{array}{*{20}{c}}
{ – 2}\\
2
\end{array}} \right)\) AG N0
[2 marks]
METHOD 1
evidence of scalar product (M1)
e.g. \(\overrightarrow {{\rm{BD}}} \bullet \overrightarrow {{\rm{AC}}} \) , \(\left( {\begin{array}{*{20}{c}}
{ – 2}\\
2
\end{array}} \right) \bullet \left( \begin{array}{l}
4\\
4
\end{array} \right)\)
correct substitution A1
e.g. \(( – 2)(4) + (2)(4)\) , \( – 8 + 8\)
\(\overrightarrow {{\rm{BD}}} \bullet \overrightarrow {{\rm{AC}}} = 0\) A1
therefore vectors \(\overrightarrow {{\rm{BD}}} \) and \(\overrightarrow {{\rm{AC}}} \) are perpendicular AG N0
METHOD 2
attempt to find angle between two vectors (M1)
e.g. \(\frac{{{\boldsymbol{a}} \bullet {\boldsymbol{b}}}}{{{\boldsymbol{ab}}}}\)
correct substitution A1
e.g. \(\frac{{( – 2)(4) + (2)(4)}}{{\sqrt 8 \sqrt {32} }}\) , \(\cos \theta = 0\)
\(\theta = {90^ \circ }\) A1
therefore vectors \(\overrightarrow {{\rm{BD}}} \) and \(\overrightarrow {{\rm{AC}}} \) are perpendicular AG N0
[3 marks]
Examiners report
This question on two-dimensional vectors was generally very well done.
This question on two-dimensional vectors was generally very well done. A very small number of candidates had trouble with the “show that” in part (b) of the question.
Nearly all candidates knew to use the scalar product in part (c) to show that the vectors are perpendicular.
| Marks available | 4 |
| Reference code | 11N.1.sl.TZ0.8 |
Question
The line \({L_1}\) passes through the points P(2, 4, 8) and Q(4, 5, 4) .
The line \({L_2}\) is perpendicular to \({L_1}\) , and parallel to \(\left( {\begin{array}{*{20}{c}}
{3p}\\
{2p}\\
4
\end{array}} \right)\) , where \(p \in \mathbb{Z}\) .
(i) Find \(\overrightarrow {{\rm{PQ}}} \) .
(ii) Hence write down a vector equation for \({L_1}\) in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + s{\boldsymbol{b}}\) .
(i) Find the value of p .
(ii) Given that \({L_2}\) passes through \({\text{R}}(10{\text{, }}6{\text{, }}- 40)\) , write down a vector equation for \({L_2}\) .
The lines \({L_1}\) and \({L_2}\) intersect at the point A. Find the x-coordinate of A.
Markscheme
(i) evidence of approach (M1)
e.g. \(\overrightarrow {{\rm{PO}}} + \overrightarrow {{\rm{OQ}}} \) , \({\rm{P}} – {\rm{Q}}\)
\(\overrightarrow {{\rm{PQ}}} = \left( {\begin{array}{*{20}{c}}
2\\
1\\
{ – 4}
\end{array}} \right)\) A1 N2
(ii) any correct equation in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + s{\boldsymbol{b}}\) (accept any parameter for s)
where a is \(\left( {\begin{array}{*{20}{c}}
2\\
4\\
8
\end{array}} \right)\) or \(\left( {\begin{array}{*{20}{c}}
4\\
5\\
4
\end{array}} \right)\) , and b is a scalar multiple of \(\left( {\begin{array}{*{20}{c}}
2\\
1\\
{ – 4}
\end{array}} \right)\) A2 N2
e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
2\\
4\\
8
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
2\\
1\\
{ – 4}
\end{array}} \right)\) , \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
{4 + 2s}\\
{5 + 1s}\\
{4 – 4s}
\end{array}} \right)\) , \({\boldsymbol{r}} = 2{\boldsymbol{i}} + 4{\boldsymbol{j}} + 8{\boldsymbol{k}} + s(2{\boldsymbol{i}} + 1{\boldsymbol{j}} – 4{\boldsymbol{k}})\)
Note: Award A1 for the form \({\boldsymbol{a}} + s{\boldsymbol{b}}\) , A1 for \({\boldsymbol{L}} = {\boldsymbol{a}} + s{\boldsymbol{b}}\) , A0 for \({\boldsymbol{r}} = {\boldsymbol{b}} + s{\boldsymbol{a}}\) .
[4 marks]
(i) choosing correct direction vectors for \({L_1}\) and \({L_2}\) (A1) (A1)
e.g. \(\left( {\begin{array}{*{20}{c}}
2\\
1\\
{ – 4}
\end{array}} \right)\) , \(\left( {\begin{array}{*{20}{c}}
{3p}\\
{2p}\\
4
\end{array}} \right)\)
evidence of equating scalar product to 0 (M1)
correct calculation of scalar product A1
e.g. \(2 \times 3p + 1 \times 2p + ( – 4) \times 4\) , \(8p – 16 = 0\)
\(p = 2\) A1 N3
(ii) any correct expression in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) (accept any parameter for t)
where a is \(\left( {\begin{array}{*{20}{c}}
{10}\\
6\\
{ – 40}
\end{array}} \right)\) , and b is a scalar multiple of \(\left( {\begin{array}{*{20}{c}}
6\\
4\\
4
\end{array}} \right)\) A2 N2
e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
{10}\\
6\\
{ – 40}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
6\\
4\\
4
\end{array}} \right)\) , \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
{10 + 6s}\\
{6 + 4s}\\
{ – 40 + 4s}
\end{array}} \right)\) , \({\boldsymbol{r}} = 10{\boldsymbol{i}} + 6{\boldsymbol{j}} – 40{\boldsymbol{k}} + s(6{\boldsymbol{i}} + 4{\boldsymbol{j}} + 4{\boldsymbol{k}})\)
Note: Award A1 for the form \({\boldsymbol{a}} + t{\boldsymbol{b}}\) , A1 for \({\boldsymbol{L}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\) (unless they have been penalised for \({\boldsymbol{L}} = {\boldsymbol{a}} + s{\boldsymbol{b}}\) in part (a)), A0 for \({\boldsymbol{r}} = {\boldsymbol{b}} + t{\boldsymbol{a}}\) .
[7 marks]
appropriate approach (M1)
e.g. \(\left( {\begin{array}{*{20}{c}}
2\\
4\\
8
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
2\\
1\\
{ – 4}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{10}\\
6\\
{ – 40}
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
6\\
4\\
4
\end{array}} \right)\)
any two correct equations with different parameters A1A1
e.g. \(2 + 2s = 10 + 6t\) , \(4 + s = 6 + 4t\) , \(8 – 4s = – 40 + 4t\)
attempt to solve simultaneous equations (M1)
correct working (A1)
e.g. \( – 6 = – 2 – 2t\) , \(4 = 2t\) , \( – 4 + 5s = 46\) , \(5s = 50\)
one correct parameter \(s = 10\) , \(t = 2\) A1
\(x = 22\) (accept (22, 14, −32)) A1 N4
[7 marks]
Examiners report
In part (a), nearly all the candidates correctly found the vector PQ, and the majority went onto find the correct vector equation of the line. There are still many candidates who do not write this equation in the correct form, using “r = “, and these candidates were penalized one mark.
In part (b), the majority of candidates knew to set the scalar product equal to zero for the perpendicular vectors, and were able to find the correct value of p.
A good number of candidates used the correct method to find the intersection of the two lines, though some algebraic and arithmetic errors kept some from finding the correct final answer.
| Marks available | 2 |
| Reference code | 13M.1.sl.TZ1.8 |
Question
Consider points A(\(1\), \( – 2\), \( -1\)) , B(\(7\), \( – 4\), \(3\)) and C(\(1\), \( -2\), \(3\)) . The line \({L_1}\) passes through C and is parallel to \(\overrightarrow {{\rm{AB}}} \) .
A second line, \({L_2}\) , is given by \(\boldsymbol{r} = \left( {\begin{array}{*{20}{c}}
{ – 1}\\
2\\
{15}
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
3\\
{ – 3}\\
p
\end{array}} \right)\) .
Find \(\overrightarrow {{\rm{AB}}} \) .
Hence, write down a vector equation for \({L_1}\) .
Given that \({L_1}\) is perpendicular to \({L_2}\) , show that \(p = – 6\) .
The line \({L_1}\) intersects the line \({L_2}\) at point Q. Find the \(x\)-coordinate of Q.
Markscheme
valid approach (M1)
eg \(\left( {\begin{array}{*{20}{c}}
7\\
{ – 4}\\
3
\end{array}} \right) – \left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
{ – 1}
\end{array}} \right)\) , \({\rm{A}} – {\rm{B}}\) , \(\overrightarrow {{\rm{AB}}} = \overrightarrow {{\rm{AO}}} + \overrightarrow {{\rm{OB}}} \)
\(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
6\\
{ – 2}\\
4
\end{array}} \right)\) A1 N2
[2 marks]
any correct equation in the form \(\boldsymbol{r} = \boldsymbol{a} + t\boldsymbol{b}\) (accept any parameter for \(t\))
where \(\boldsymbol{a} = \left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
3
\end{array}} \right)\) and \(\boldsymbol{b}\) is a scalar multiple of \(\overrightarrow {{\rm{AB}}} \) A2 N2
eg \(\boldsymbol{r} = \left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
3
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
6\\
{ – 2}\\
4
\end{array}} \right)\) , \((x,y,z) = (1, – 2,3) + t(3, – 1,2)\) , \(\boldsymbol{r} = \left( {\begin{array}{*{20}{c}}
{1 + 6t}\\
{ – 2 – 2t}\\
{3 + 4t}
\end{array}} \right)\)
Note: Award A1 for \(\boldsymbol{a} + t\boldsymbol{b}\) , A1 for \({L_1} = \boldsymbol{a} + t\boldsymbol{b}\) , A0 for \(\boldsymbol{r} = \boldsymbol{b} + t\boldsymbol{a}\) .
[2 marks]
recognizing that scalar product \( = 0\) (seen anywhere) R1
correct calculation of scalar product (A1)
eg \(6(3) – 2( – 3) + 4p\) , \(18 + 6 + 4p\)
correct working A1
eg \(24 + 4p = 0\) , \(4p = – 24\)
\(p = – 6\) AG N0
[3 marks]
setting lines equal (M1)
eg \({L_1} = {L_2}\) , \(\left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
3
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
6\\
{ – 2}\\
4
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{ – 1}\\
2\\
{15}
\end{array}} \right) + s\left( {\begin{array}{*{20}{c}}
3\\
{ – 3}\\
{ – 6}
\end{array}} \right)\)
any two correct equations with different parameters A1A1
eg \(1 + 6t = 1 + 3s\) , \( – 2 – 2t = 2 – 3s\) , \(3 + 4t = 15 – 6s\)
attempt to solve their simultaneous equations (M1)
one correct parameter A1
eg \(t = \frac{1}{2}\) , \(s = \frac{5}{3}\)
attempt to substitute parameter into vector equation (M1)
eg \(\left( {\begin{array}{*{20}{c}}
1\\
{ – 2}\\
3
\end{array}} \right) + \frac{1}{2}\left( {\begin{array}{*{20}{c}}
6\\
{ – 2}\\
4
\end{array}} \right)\) , \(1 + \frac{1}{2} \times 6\)
\(x = 4\) (accept (4, -3, 5), ignore incorrect values for \(y\) and \(z\)) A1 N3
[7 marks]
Examiners report
While many candidates can find a vector given two points, few could write down a fully correct vector equation of a line.
While many candidates can find a vector given two points, few could write down a fully correct vector equation of a line. Most candidates wrote their equation as “ \({L_1} = \) ”, which misrepresents that the resulting equation must still be a vector.
Those who recognized that vector perpendicularity means the scalar product is zero found little difficulty answering part (b). Occasionally a candidate would use the given \(p = 6\) to show the scalar product is zero. However, working backward from the given answer earns no marks in a question that requires candidates to show that this value is achieved.
While many candidates knew to set the lines equal to find an intersection point, a surprising number could not carry the process to correct completion. Some could not solve a simultaneous pair of equations, and for those who did, some did not know what to do with the parameter value. Another common error was to set the vector equations equal using the same parameter, from which the candidates did not recognize a system to solve. Furthermore, it is interesting to note that while only one parameter value is needed to answer the question, most candidates find or attempt to find both, presumably out of habit in the algorithm.
| Marks available | 6 |
| Reference code | 18M.1.sl.TZ1.6 |
Question
Six equilateral triangles, each with side length 3 cm, are arranged to form a hexagon.
This is shown in the following diagram.
The vectors p , q and r are shown on the diagram.
Find p•(p + q + r).
Markscheme
METHOD 1 (using |p| |2q| cosθ)
finding p + q + r (A1)
eg 2q, 
| p + q + r | = 2 × 3 (= 6) (seen anywhere) A1
correct angle between p and q (seen anywhere) (A1)
\(\frac{\pi }{3}\) (accept 60°)
substitution of their values (M1)
eg 3 × 6 × cos\(\left( {\frac{\pi }{3}} \right)\)
correct value for cos\(\left( {\frac{\pi }{3}} \right)\) (seen anywhere) (A1)
eg \(\frac{1}{2},\,\,\,3 \times 6 \times \frac{1}{2}\)
p•(p + q + r) = 9 A1 N3
METHOD 2 (scalar product using distributive law)
correct expression for scalar distribution (A1)
eg p• p + p•q + p•r
three correct angles between the vector pairs (seen anywhere) (A2)
eg 0° between p and p, \(\frac{\pi }{3}\) between p and q, \(\frac{{2\pi }}{3}\) between p and r
Note: Award A1 for only two correct angles.
substitution of their values (M1)
eg 3.3.cos0 +3.3.cos\(\frac{\pi }{3}\) + 3.3.cos120
one correct value for cos0, cos\(\left( {\frac{\pi }{3}} \right)\) or cos\(\left( {\frac{2\pi }{3}} \right)\) (seen anywhere) A1
eg \(\frac{1}{2},\,\,\,3 \times 6 \times \frac{1}{2}\)
p•(p + q + r) = 9 A1 N3
METHOD 3 (scalar product using relative position vectors)
valid attempt to find one component of p or r (M1)
eg sin 60 = \(\frac{x}{3}\), cos 60 = \(\frac{x}{3}\), one correct value \(\frac{3}{2},\,\,\frac{{3\sqrt 3 }}{2},\,\,\frac{{ – 3\sqrt 3 }}{2}\)
one correct vector (two or three dimensions) (seen anywhere) A1
eg \(p = \left( \begin{gathered}
\,\,\,\frac{3}{2} \hfill \\
\frac{{3\sqrt 3 }}{2} \hfill \\
\end{gathered} \right),\,\,q = \left( \begin{gathered}
3 \hfill \\
0 \hfill \\
\end{gathered} \right),\,\,r = \left( \begin{gathered}
\,\,\,\,\frac{3}{2} \hfill \\
– \frac{{3\sqrt 3 }}{2} \hfill \\
\,\,\,\,0 \hfill \\
\end{gathered} \right)\)
three correct vectors p + q + r = 2q (A1)
p + q + r = \(\left( \begin{gathered}
6 \hfill \\
0 \hfill \\
\end{gathered} \right)\) or \(\left( \begin{gathered}
6 \hfill \\
0 \hfill \\
0 \hfill \\
\end{gathered} \right)\) (seen anywhere, including scalar product) (A1)
correct working (A1)
eg \(\left( {\frac{3}{2} \times 6} \right) + \left( {\frac{{3\sqrt 3 }}{2} \times 0} \right),\,\,9 + 0 + 0\)
p•(p + q + r) = 9 A1 N3
[6 marks]
Examiners report
| Marks available | 1 |
| Reference code | 18M.1.sl.TZ1.9 |
Question
Point A has coordinates (−4, −12, 1) and point B has coordinates (2, −4, −4).
The line L passes through A and B.
Show that \(\mathop {{\text{AB}}}\limits^ \to = \left( \begin{gathered}
\,6 \hfill \\
\,8 \hfill \\
– 5 \hfill \\
\end{gathered} \right)\)
Find a vector equation for L.
Point C (k , 12 , −k) is on L. Show that k = 14.
Find \(\mathop {{\text{OB}}}\limits^ \to \, \bullet \mathop {{\text{AB}}}\limits^ \to \).
Write down the value of angle OBA.
Point D is also on L and has coordinates (8, 4, −9).
Find the area of triangle OCD.
Markscheme
correct approach A1
eg \(\mathop {{\text{AO}}}\limits^ \to \,\, + \,\,\mathop {{\text{OB}}}\limits^ \to ,\,\,\,{\text{B}} – {\text{A}}\,{\text{, }}\,\left( \begin{gathered}
\,\,2 \hfill \\
– 4 \hfill \\
– 4 \hfill \\
\end{gathered} \right) – \left( \begin{gathered}
\, – 4 \hfill \\
– 12 \hfill \\
\,\,\,1 \hfill \\
\end{gathered} \right)\)
\(\mathop {{\text{AB}}}\limits^ \to = \left( \begin{gathered}
\,6 \hfill \\
\,8 \hfill \\
– 5 \hfill \\
\end{gathered} \right)\) AG N0
[1 mark]
any correct equation in the form r = a + tb (any parameter for t) A2 N2
where a is \(\left( \begin{gathered}
\,\,2 \hfill \\
– 4 \hfill \\
– 4 \hfill \\
\end{gathered} \right)\) or \(\left( \begin{gathered}
\, – 4 \hfill \\
– 12 \hfill \\
\,\,\,1 \hfill \\
\end{gathered} \right)\) and b is a scalar multiple of \(\left( \begin{gathered}
\,6 \hfill \\
\,8 \hfill \\
– 5 \hfill \\
\end{gathered} \right)\)
eg r \( = \left( \begin{gathered}
\, – 4 \hfill \\
– 12 \hfill \\
\,\,\,1 \hfill \\
\end{gathered} \right) + t\left( \begin{gathered}
\,6 \hfill \\
\,8 \hfill \\
– 5 \hfill \\
\end{gathered} \right),\,\,\left( {x,\,\,y,\,\,z} \right) = \left( {2,\,\, – 4,\,\, – 4} \right) + t\left( {6,\,\,8,\,\, – 5} \right),\) r \( = \left( \begin{gathered}
\, – 4 + 6t \hfill \\
– 12 + 8t \hfill \\
\,\,\,1 – 5t \hfill \\
\end{gathered} \right)\)
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 (solving for t)
valid approach (M1)
eg \(\left( \begin{gathered}
\,k \hfill \\
12 \hfill \\
– k \hfill \\
\end{gathered} \right) = \left( \begin{gathered}
\,\,2 \hfill \\
– 4 \hfill \\
– 4 \hfill \\
\end{gathered} \right) + t\left( \begin{gathered}
\,6 \hfill \\
\,8 \hfill \\
– 5 \hfill \\
\end{gathered} \right),\,\,\left( \begin{gathered}
\,k \hfill \\
12 \hfill \\
– k \hfill \\
\end{gathered} \right) = \left( \begin{gathered}
\, – 4 \hfill \\
– 12 \hfill \\
\,\,\,1 \hfill \\
\end{gathered} \right) + t\left( \begin{gathered}
\,6 \hfill \\
\,8 \hfill \\
– 5 \hfill \\
\end{gathered} \right)\)
one correct equation A1
eg −4 + 8t = 12, −12 + 8t = 12
correct value for t (A1)
eg t = 2 or 3
correct substitution A1
eg 2 + 6(2), −4 + 6(3), −[1 + 3(−5)]
k = 14 AG N0
METHOD 2 (solving simultaneously)
valid approach (M1)
eg \(\left( \begin{gathered}
\,k \hfill \\
12 \hfill \\
– k \hfill \\
\end{gathered} \right) = \left( \begin{gathered}
\,\,2 \hfill \\
– 4 \hfill \\
– 4 \hfill \\
\end{gathered} \right) + t\left( \begin{gathered}
\,6 \hfill \\
\,8 \hfill \\
– 5 \hfill \\
\end{gathered} \right),\,\,\left( \begin{gathered}
\,k \hfill \\
12 \hfill \\
– k \hfill \\
\end{gathered} \right) = \left( \begin{gathered}
\, – 4 \hfill \\
– 12 \hfill \\
\,\,\,1 \hfill \\
\end{gathered} \right) + t\left( \begin{gathered}
\,6 \hfill \\
\,8 \hfill \\
– 5 \hfill \\
\end{gathered} \right)\)
two correct equations in A1
eg k = −4 + 6t, −k = 1 −5t
EITHER (eliminating k)
correct value for t (A1)
eg t = 2 or 3
correct substitution A1
eg 2 + 6(2), −4 + 6(3)
OR (eliminating t)
correct equation(s) (A1)
eg 5k + 20 = 30t and −6k − 6 = 30t, −k = 1 − 5\(\left( {\frac{{k + 4}}{6}} \right)\)
correct working clearly leading to k = 14 A1
eg −k + 14 = 0, −6k = 6 −5k − 20, 5k = −20 + 6(1 + k)
THEN
k = 14 AG N0
[4 marks]
correct substitution into scalar product A1
eg (2)(6) − (4)(8) − (4)(−5), 12 − 32 + 20
\(\mathop {{\text{OB}}}\limits^ \to \, \bullet \mathop {{\text{AB}}}\limits^ \to \) = 0 A1 N0
[2 marks]
\({\text{O}}\mathop {\text{B}}\limits^ \wedge {\text{A}} = \frac{\pi }{2},\,\,90^\circ \,\,\,\,\,\left( {{\text{accept}}\,\frac{{3\pi }}{2},\,\,270^\circ } \right)\,\) A1 N1
[1 marks]
METHOD 1 (\(\frac{1}{2}\) × height × CD)
recognizing that OB is altitude of triangle with base CD (seen anywhere) M1
eg \(\frac{1}{2} \times \left| {\mathop {{\text{OB}}}\limits^ \to } \right| \times \left| {\mathop {{\text{CD}}}\limits^ \to } \right|,\,\,{\text{OB}} \bot {\text{CD}},\) sketch showing right angle at B
\(\mathop {{\text{CD}}}\limits^ \to = \left( \begin{gathered}
– 6 \hfill \\
– 8 \hfill \\
\,5 \hfill \\
\end{gathered} \right)\) or \(\mathop {{\text{DC}}}\limits^ \to = \left( \begin{gathered}
\,6 \hfill \\
\,8 \hfill \\
– 5 \hfill \\
\end{gathered} \right)\) (seen anywhere) (A1)
correct magnitudes (seen anywhere) (A1)(A1)
\(\left| {\mathop {{\text{OB}}}\limits^ \to } \right| = \sqrt {{{\left( 2 \right)}^2} + {{\left( { – 4} \right)}^2} + {{\left( { – 4} \right)}^2}} = \left( {\sqrt {36} } \right)\)
\(\left| {\mathop {{\text{CD}}}\limits^ \to } \right| = \sqrt {{{\left( { – 6} \right)}^2} + {{\left( { – 8} \right)}^2} + {{\left( 5 \right)}^2}} = \left( {\sqrt {125} } \right)\)
correct substitution into \(\frac{1}{2}bh\) A1
eg \(\frac{1}{2} \times 6 \times \sqrt {125} \)
area \( = 3\sqrt {125} ,\,\,15\sqrt 5 \) A1 N3
METHOD 2 (subtracting triangles)
recognizing that OB is altitude of either ΔOBD or ΔOBC(seen anywhere) M1
eg \(\frac{1}{2} \times \left| {\mathop {{\text{OB}}}\limits^ \to } \right| \times \left| {\mathop {{\text{BD}}}\limits^ \to } \right|,\,\,{\text{OB}} \bot {\text{BC}},\) sketch of triangle showing right angle at B
one correct vector \(\mathop {{\text{BD}}}\limits^ \to \) or \(\mathop {{\text{DB}}}\limits^ \to \) or \(\mathop {{\text{BC}}}\limits^ \to \) or \(\mathop {{\text{CB}}}\limits^ \to \) (seen anywhere) (A1)
eg \(\mathop {{\text{BD}}}\limits^ \to = \left( \begin{gathered}
\,6 \hfill \\
\,8 \hfill \\
– 5 \hfill \\
\end{gathered} \right)\), \(\mathop {{\text{CB}}}\limits^ \to = \left( \begin{gathered}
– 12 \hfill \\
– 16 \hfill \\
\,10 \hfill \\
\end{gathered} \right)\)
\(\left| {\mathop {{\text{OB}}}\limits^ \to } \right| = \sqrt {{{\left( 2 \right)}^2} + {{\left( { – 4} \right)}^2} + {{\left( { – 4} \right)}^2}} = \left( {\sqrt {36} } \right)\) (seen anywhere) (A1)
one correct magnitude of a base (seen anywhere) (A1)
\(\left| {\mathop {{\text{BD}}}\limits^ \to } \right| = \sqrt {{{\left( 6 \right)}^2} + {{\left( 8 \right)}^2} + {{\left( 5 \right)}^2}} = \left( {\sqrt {125} } \right),\,\,\left| {\mathop {{\text{BC}}}\limits^ \to } \right| = \sqrt {144 + 256 + 100} = \left( {\sqrt {500} } \right)\)
correct working A1
eg \(\frac{1}{2} \times 6 \times \sqrt {500} – \frac{1}{2} \times 6 \times 5\sqrt 5 ,\,\,\frac{1}{2} \times 6 \times \sqrt {500} \times {\text{sin}}90 – \frac{1}{2} \times 6 \times 5\sqrt 5 \times {\text{sin}}90\)
area \( = 3\sqrt {125} ,\,\,15\sqrt 5 \) A1 N3
METHOD 3 (using \(\frac{1}{2}\)ab sin C with ΔOCD)
two correct side lengths (seen anywhere) (A1)(A1)
\(\left| {\mathop {{\text{OD}}}\limits^ \to } \right| = \sqrt {{{\left( 8 \right)}^2} + {{\left( 4 \right)}^2} + {{\left( { – 9} \right)}^2}} = \left( {\sqrt {161} } \right),\,\,\left| {\mathop {{\text{CD}}}\limits^ \to } \right| = \sqrt {{{\left( { – 6} \right)}^2} + {{\left( { – 8} \right)}^2} + {{\left( 5 \right)}^2}} = \left( {\sqrt {125} } \right),\,\) \(\left| {\mathop {{\text{OC}}}\limits^ \to } \right| = \sqrt {{{\left( {14} \right)}^2} + {{\left( {12} \right)}^2} + {{\left( { – 14} \right)}^2}} = \left( {\sqrt {536} } \right)\)
attempt to find cosine ratio (seen anywhere) M1
eg \(\frac{{536 – 286}}{{ – 2\sqrt {161} \sqrt {125} }},\,\,\frac{{{\text{OD}} \bullet {\text{DC}}}}{{\left| {OD} \right|\left| {DC} \right|}}\)
correct working for sine ratio A1
eg \(\frac{{{{\left( {125} \right)}^2}}}{{161 \times 125}} + {\text{si}}{{\text{n}}^2}\,D = 1\)
correct substitution into \(\frac{1}{2}ab\,\,{\text{sin}}\,C\) A1
eg \(0.5 \times \sqrt {161} \times \sqrt {125} \times \frac{6}{{\sqrt {161} }}\)
area \( = 3\sqrt {125} ,\,\,15\sqrt 5 \) A1 N3
[6 marks]