Write down expressions for $\overrightarrow {{\text{AB}}}$ and $\overrightarrow {{\text{CB}}}$ in terms of the vectors ${\boldsymbol{a}}$ and ${\boldsymbol{b}}$ .

Hence prove that angle ${\text{A}}\hat {\rm{B}}{\text{C}}$ is a right angle.

$\overrightarrow {{\text{AB}}}  = {\boldsymbol{b}} - {\boldsymbol{a}}$     A1

$\overrightarrow {{\text{CB}}}  = {\boldsymbol{a}} + {\boldsymbol{b}}$     A1

[2 marks]

$\overrightarrow {{\text{AB}}}  \cdot \overrightarrow {{\text{CB}}}  = \left( {{\boldsymbol{b}} - {\boldsymbol{a}}} \right) \cdot \left( {{\boldsymbol{b}} + {\boldsymbol{a}}} \right)$     M1
$= {\left| {\mathbf{b}} \right|^2} - {\left| {\mathbf{a}} \right|^2}$     A1
$= 0$ since $\left| {\boldsymbol{b}} \right| = \left| {\boldsymbol{a}} \right|$     R1

Note: Only award the A1 and R1 if working indicates that they understand that they are working with vectors.

so $\overrightarrow {{\text{AB}}}$ is perpendicular to $\overrightarrow {{\text{CB}}}$ i.e. ${\text{A}}\hat {\rm{B}}{\text{C}}$ is a right angle     AG

[3 marks]

This question was poorly done with most candidates having difficulties in using appropriate notation which made unclear the distinction between scalars and vectors. A few candidates scored at least one of the marks in (a) but most candidates had problems in setting up the proof required in (b) with many using a circular argument which resulted in a very poor performance in this part.

This question was poorly done with most candidates having difficulties in using appropriate notation which made unclear the distinction between scalars and vectors. A few candidates scored at least one of the marks in (a) but most candidates had problems in setting up the proof required in (b) with many using a circular argument which resulted in a very poor performance in this part.