IB DP Maths Topic 4.4 Finding the point of intersection of two lines. SL Paper 2

appropriate approach     (M1)

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

\(\overrightarrow {{\rm{AB}}} = \left( {\begin{array}{*{20}{c}}
1\\
{ – 1}\\
1
\end{array}} \right)\)     A1     N2

[2 marks]

any correct equation in the form \({\boldsymbol{r}} = {\boldsymbol{a}} + t{\boldsymbol{b}}\)     A2 N2

where \({\boldsymbol{b}}\) is a scalar multiple of \(\left( {\begin{array}{*{20}{c}}
1\\
{ – 1}\\
1
\end{array}} \right)\)

e.g. \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
1\\
{ – 1}\\
4
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
1\\
{ – 1}\\
1
\end{array}} \right)\) , \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
{2 + t}\\
{ – 2 – t}\\
{5 + t}
\end{array}} \right)\) , \({\boldsymbol{r}} = 2{\boldsymbol{i}} – 2{\boldsymbol{j}} + 5{\boldsymbol{k}} + t({\boldsymbol{i}} – {\boldsymbol{j}} + {\boldsymbol{k}})\)

 [2 marks]

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

finding scalar product and magnitudes     (A1)(A1)(A1)

scalar product \( = 1 \times 2 + – 1 \times 1 + 1 \times 3\) \(\left( { = 4} \right)\)

magnitudes \(\sqrt {{1^2} + {{( – 1)}^2} + {1^2}}{\text{  }}( = 1.73 \ldots )\) , \(\sqrt {4 + 1 + 9}{\text{  }}( = 3.74 \ldots )\)

substitution into \(\frac{{{\boldsymbol{u}} \bullet {\boldsymbol{v}}}}{{|{\boldsymbol{u}}||{\boldsymbol{v}}|}}\) (accept \(\theta = \frac{{{\boldsymbol{u}} \bullet {\boldsymbol{v}}}}{{|{\boldsymbol{u}}||{\boldsymbol{v}}|}}\) , but not \(\sin \theta = \frac{{{\boldsymbol{u}} \bullet {\boldsymbol{v}}}}{{|{\boldsymbol{u}}||{\boldsymbol{v}}|}}\) )     M1

e.g. \(\cos \theta = \frac{{1 \times 2 + – 1 \times 1 + 1 \times 3}}{{\sqrt {{1^2} + {{( – 1)}^2} + {1^2}} \sqrt {{2^2} + {1^2} + {3^2}} }}\) , \(\cos \theta = \frac{4}{{\sqrt {42} }}\)

\(\theta = 0.906\) \(({51.9^ \circ })\)     A1      N5

[7 marks]

METHOD 1 (from \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
  1 \\
  { – 1} \\
  4
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
  1 \\
  { – 1} \\
  1
\end{array}} \right)\) )

appropriate approach     (M1)

e.g. \({\boldsymbol{p}} = {\boldsymbol{r}}\) , \(\left( {\begin{array}{*{20}{c}}
1\\
{ – 1}\\
4
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
1\\
{ – 1}\\
1
\end{array}} \right)t = \left( {\begin{array}{*{20}{c}}
2\\
4\\
7
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
2\\
1\\
3
\end{array}} \right)s\)

two correct equations    A1A1

e.g. \(1 + t = 2 + 2s\) , \( – 1 – t = 4 + s\) , \(4 + t = 7 + 3s\)

attempt to solve     (M1)

one correct parameter     A1

e.g. \(t = – 3\) , \(s = – 2\)

C is \(( – 2{\text{, }}2{\text{, }}1)\)     A1     N3

METHOD 2 (from \({\boldsymbol{r}} = \left( {\begin{array}{*{20}{c}}
  2 \\
  { – 2} \\
  5
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
  1 \\
  { – 1} \\
  1
\end{array}} \right)\) )

appropriate approach     (M1)

e.g. \({\boldsymbol{p}} = {\boldsymbol{r}}\) , \(\left( {\begin{array}{*{20}{c}}
2\\
{ – 2}\\
5
\end{array}} \right) + t\left( {\begin{array}{*{20}{c}}
1\\
{ – 1}\\
1
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
2\\
4\\
7
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
2\\
1\\
3
\end{array}} \right)s\)

two correct equations     A1A1

e.g. \(2 + t = 2 + 2s\) , \( – 2 – t = 4 + s\) , \(5 + t = 7 + 3s\)

attempt to solve     (M1)

one correct parameter     A1

e.g. \(t = – 4\) , \(s = – 2\)

C is \(( – 2{\text{, }}2{\text{, }}1)\)     A1     N3

[6 marks]