| Date | None Specimen | Marks available | 3 | Reference code | SPNone.1.sl.TZ0.1 |
| Level | SL only | Paper | 1 | Time zone | TZ0 |
| Command term | Express | Question number | 1 | Adapted from | N/A |
Question
In the following diagram, \boldsymbol{u} = \overrightarrow {{\rm{AB}}} and \boldsymbol{v} = \overrightarrow {{\rm{BD}}} .
The midpoint of \overrightarrow {{\rm{AD}}} is E and \frac{{{\rm{BD}}}}{{{\rm{DC}}}} = \frac{1}{3} .
Express each of the following vectors in terms of \boldsymbol{u} and \boldsymbol{v} .
\overrightarrow {{\rm{AE}}}
\overrightarrow {{\rm{EC}}}
Markscheme
\overrightarrow {{\rm{AE}}} = \frac{1}{2}\overrightarrow {{\rm{AD}}} A1
attempt to find \overrightarrow {{\rm{AD}}} M1
e.g. \overrightarrow {{\rm{AB}}} + \overrightarrow {{\rm{BD}}} , {\boldsymbol{u}} + {\boldsymbol{v}}
\overrightarrow {{\rm{AE}}} = \frac{1}{2}(u + v) \left( { = \frac{1}{2}{\boldsymbol{u}} + \frac{1}{2}{\boldsymbol{v}}} \right) A1 N2
[3 marks]
\overrightarrow {{\rm{EC}}} = \overrightarrow {{\rm{AE}}} = \frac{1}{2}({\boldsymbol{u}} + {\boldsymbol{v}}) A1
\overrightarrow {{\rm{DC}}} = 3{\boldsymbol{v}} A1
attempt to find \overrightarrow {{\rm{EC}}} M1
e.g. \overrightarrow {{\rm{ED}}} + \overrightarrow {{\rm{DC}}} , \frac{1}{2}({\boldsymbol{u}} + {\boldsymbol{v}}) + 3{\boldsymbol{v}}
\overrightarrow {{\rm{EC}}} = \frac{1}{2}{\boldsymbol{u}} + \frac{7}{2}{\boldsymbol{v}} \left( { = \frac{1}{2}({\boldsymbol{u}} + 7{\boldsymbol{v}})} \right) A1 N2
[4 marks]
