Consider the vectors \(\overrightarrow {{\text{OA}}} \) = a, \(\overrightarrow {{\text{OB}}} \) = b and \(\overrightarrow {{\text{OC}}} \) = a + b. Show that if \(|\)a\(|\) = \(|\)b\(|\) then (a + b)\( \cdot \)(a − b) = 0. Comment on what this tells us about the parallelogram OACB.

(a + b)\( \cdot \)(a – b) = a\( \cdot \)a + b\( \cdot \)a – a\( \cdot \)b – b\( \cdot \)b     M1

= a\( \cdot \)a – b\( \cdot \)b     A1

= \(|\)a\({|^2}\) – \(|\)b\({|^2}\) = 0 since \(|\)a\(|\) = \(|\)b\(|\)     A1

the diagonals are perpendicular     R1

Note: Accept geometric proof, awarding M1 for recognizing OACB is a rhombus, R1 for a clear indication that (a + b) and (a – b) are the diagonals, A1 for stating that diagonals cross at right angles and A1 for “hence dot product is zero”.

Accept solutions using components in 2 or 3 dimensions.

[4 marks]

Many candidates found this more abstract question difficult. While there were some correct statements, they could not “show” the result that was asked. Some treated the vectors as scalars and notation was poor, making it difficult to follow what they were trying to do. Very few candidates realized that a – b and a + b were the diagonals of the parallelogram which prevented them from identifying the significance of the result proved. A number of candidates were clearly not aware of the difference between scalars and vectors.