IB DP Maths Topic 4.1 Algebraic and geometric approaches to magnitude of a vector, |v| SL Paper 2

(i) (3, \( – 4\), 0)     A1     N1

(ii) choosing velocity vector \(\left( {\begin{array}{*{20}{c}}
{ – 2}\\
3\\
1
\end{array}} \right)\)     (M1)

finding magnitude of velocity vector     (A1)

e.g. \(\sqrt {{{( – 2)}^2} + {3^2} + {1^2}} \) , \(\sqrt {4 + 9 + 1} \)

speed \(= 3.74\) \(\left( {\sqrt {14} } \right)\)     A1     N2

[4 marks]

(i) substituting \(p = 7\)     (M1)

\({\text{B}} = ( – 11{\text{, }}17{\text{, }}7)\)     A1     N2

(ii) METHOD 1

appropriate method to find \(\overrightarrow {{\rm{AB}}} \) or \(\overrightarrow {{\rm{BA}}} \)     (M1)

e.g. \(\overrightarrow {{\rm{AO}}}  + \overrightarrow {{\rm{OB}}} \) , \({\rm{A}} – {\rm{B}}\)

\(\overrightarrow {{\rm{AB}}}  = \left( {\begin{array}{*{20}{c}}
{ – 14}\\
{21}\\
7
\end{array}} \right)\) or \(\overrightarrow {{\rm{BA}}}  = \left( {\begin{array}{*{20}{c}}
{14}\\
{ – 21}\\
{ – 7}
\end{array}} \right)\)     (A1)

distance \(= 26.2\) \(\left( {7\sqrt {14} } \right)\)     A1     N3

METHOD 2

evidence of applying distance is speed × time     (M2)

e.g. \(3.74 \times 7\)

distance \(= 26.2\) \(\left( {7\sqrt {14} } \right)\)     A1     N3

METHOD 3

attempt to find AB² , AB     (M1)

e.g. \({(3 – ( – 11))^2} + {( – 4 – 17)^2} + (0 – 7){)^2}\) , \(\sqrt {{{(3 – ( – 11))}^2} + {{( – 4 – 17)}^2} + (0 – 7){)^2}} \)

AB² \(= 686\), AB \(= \sqrt {686} \)     (A1)

distance AB \(= 26.2\) \(\left( {7\sqrt {14} } \right)\)     A1     N3

[5 marks]

correct direction vectors \(\left( {\begin{array}{*{20}{c}}
{ – 2}\\
3\\
1
\end{array}} \right)\) and \(\left( {\begin{array}{*{20}{c}}
{ – 1}\\
2\\
a
\end{array}} \right)\)     (A1)(A1)

\(\left| {\begin{array}{*{20}{c}}
{ – 1}\\
2\\
a
\end{array}} \right| = \sqrt {{a^2} + 5} \) , \(\left( {\begin{array}{*{20}{c}}
{ – 2}\\
3\\
1
\end{array}} \right) \bullet \left( {\begin{array}{*{20}{c}}
{ – 1}\\
2\\
a
\end{array}} \right) = a + 8\)      (A1)(A1)

substituting     M1

e.g. \(\cos {40^ \circ } = \frac{{a + 8}}{{\sqrt {14} \sqrt {{a^2} + 5} }}\)

\(a = 3.21\) , \(a = – 0.990\)     A1A1     N3

[7 marks]

Many candidates demonstrated a good understanding of the vector equation of a line and its application to a kinematics problem by correctly answering the first two parts of this question.

Many candidates demonstrated a good understanding of the vector equation of a line and its application to a kinematics problem by correctly answering the first two parts of this question.

Some knew that speed and distance were magnitudes of vectors but chose the wrong vectors to calculate magnitudes.

Very few candidates were able to get the two correct answers in (c) even if they set up the equation correctly. Much contorted algebra was seen and little evidence of using the GDC to solve the equation. Many made simple algebraic errors by combining unlike terms in working with the scalar product (often writing \(8a\) rather than \(8 + a\) ) or the magnitude (often writing \(5{a^2}\) rather than \(5 + {a^2}\) ).