| Date | May 2013 | Marks available | 4 | Reference code | 13M.1.sl.TZ1.1 |
| Level | SL only | Paper | 1 | Time zone | TZ1 |
| Command term | Find | Question number | 1 | Adapted from | N/A |
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 \(\boldsymbol{c}\) .
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]
Examiners report
Most candidates comfortably applied algebraic techniques to find new vectors. However, a significant number of candidates answered part (b) as the absolute numerical value of the vector components, which suggests a misunderstanding of the modulus notation. Those who understood the notation easily made the calculation.
Most candidates comfortably applied algebraic techniques to find new vectors.
Most candidates comfortably applied algebraic techniques to find new vectors. However, a significant number of candidates answered part (b) as the absolute numerical value of the vector components, which suggests a misunderstanding of the modulus notation. Those who understood the notation easily made the calculation.
