O, A, B and C are distinct points such that \(\overrightarrow {{\text{OA}}}  = \) a, \(\overrightarrow {{\text{OB}}}  = \) b and \(\overrightarrow {{\text{OC}}}  = \) c.

It is given that c is perpendicular to \(\overrightarrow {{\text{AB}}} \) and b is perpendicular to \(\overrightarrow {{\text{AC}}} \).

Prove that a is perpendicular to \(\overrightarrow {{\text{BC}}} \).

c \( \bullet \) (b \( - \) a) \( = 0\)     M1

Note:     Allow c \( \bullet \) \(\overrightarrow {{\text{AB}}}  = 0\) or similar for M1.

c \( \bullet \) b \( = \) c \( \bullet \) a     A1

b \( \bullet \) (c \( - \) a) \( = 0\)

b \( \bullet \) c \( = \) b \( \bullet \) a     A1

c \( \bullet \) a \( = \) b \( \bullet \) a     M1

(c \( - \) b) \( \bullet \) a \( = 0\)     A1

hence a is perpendicular to \(\overrightarrow {{\text{BC}}} \)     AG

Note:     Only award the final A1 if a dot is used throughout to indicate scalar product.

Condone any lack of specific indication that the letters represent vectors.

[5 marks]

This was generally poorly done. The recent syllabus change refers to ‘proof of geometrical properties using vectors’ and this is clearly a topic candidates are not entirely clear with at the moment. Despite the question clearly being written as a vector question some students tried to use a geometrical approach, assuming it was two-dimensional. Many did not seem to realise that vectors being perpendicular implies that their scalar product is zero.