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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 \(\boldsymbol{c}\) .

[6]
.

Find

  (i)     \(2\boldsymbol{a} + \boldsymbol{b}\) ;

  (ii)     \(\left| {2\boldsymbol{a} + \boldsymbol{b}} \right|\) .

[4]
a.

Find \(\boldsymbol{c}\) .

[2]
b.

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]

 

a.

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]

b.

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.

a.

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.

b.

Syllabus sections

Topic 4 - Vectors » 4.1 » Algebraic and geometric approaches to the sum and difference of two vectors; the zero vector, the vector \( - v\).

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