Show that

(i)     \({\left| {\overrightarrow {{\text{OC}}} } \right|^2} = |\)a\({|^2} + 2\)a \( \bullet \) b \( + |\)b\({|^2}\);

(ii)     \({\left| {\overrightarrow {{\text{AB}}} } \right|^2} = |\)a\({|^2} - 2\)a \( \bullet \) b \( + |\)b\({|^2}\).

Given that \(\left| {\overrightarrow {{\text{OC}}} } \right| = \left| {\overrightarrow {{\text{AB}}} } \right|\), prove that OACB is a rectangle.

METHOD 1

\({\left| {\overrightarrow {{\text{OC}}} } \right|^2} = \overrightarrow {{\text{OC}}}  \bullet \overrightarrow {{\text{OC}}} \)

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

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

= \(|\)a\({|^2}\) + 2a \( \bullet \) b + \(|\)b\({|^2}\)     AG

METHOD 2

\({\left| {\overrightarrow {{\text{OC}}} } \right|^2} = {\left| {\overrightarrow {{\text{OA}}} } \right|^2} + {\left| {\overrightarrow {{\text{OB}}} } \right|^2} - 2\left| {\overrightarrow {{\text{OA}}} } \right|\left| {\overrightarrow {{\text{OB}}} } \right|\cos ({\rm{O\hat AC}})\)    A1

\(\left| {\overrightarrow {{\text{OA}}} } \right|\left| {\overrightarrow {{\text{OB}}} } \right|\cos ({\rm{O\hat AC}}) =  - \)(a \( \bullet \) b)     A1

\({\left| {\overrightarrow {{\text{OC}}} } \right|^2} = |\)a\({|^2}\) + 2a \( \bullet \) b + \(|\)b\({|^2}\)     AG

(ii)     METHOD 1

\({\left| {\overrightarrow {{\text{AB}}} } \right|^2} = \overrightarrow {{\text{AB}}}  \bullet \overrightarrow {{\text{AB}}} \)

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

= b \( \bullet \) b − b \( \bullet \) a − a \( \bullet \) b + a \( \bullet \) a     A1

= \(|\)a\({|^2}\) – 2a \( \bullet \) b + \(|\)b\({|^2}\)     AG

METHOD 2

\({\left| {\overrightarrow {{\text{AB}}} } \right|^2} = {\left| {\overrightarrow {{\text{AC}}} } \right|^2} + {\left| {\overrightarrow {{\text{BC}}} } \right|^2} - 2\left| {\overrightarrow {{\text{AC}}} } \right|\left| {\overrightarrow {{\text{BC}}} } \right|\cos ({\rm{A\hat CB}})\)    A1

\(\left| {\overrightarrow {{\text{AC}}} } \right|\left| {\overrightarrow {{\text{BC}}} } \right|\cos ({\rm{A\hat CB}}) = \) a \( \bullet \) b     A1

\({\left| {\overrightarrow {{\text{AB}}} } \right|^2} = |\)a \({|^2} - \) 2a \( \bullet \) b + \(|\)b\({|^2}\)     AG

[4 marks]

\(\left| {\overrightarrow {{\text{OC}}} } \right| = \left| {\overrightarrow {{\text{AB}}} } \right| \Rightarrow {\left| {\overrightarrow {{\text{OC}}} } \right|^2} = {\left| {\overrightarrow {{\text{AB}}} } \right|^2} \Rightarrow |\)a\({|^2} + {\text{2}}\)a \( \bullet \) b \( + |\)b\({|^2} = |\)a\({|^2} - 2\)a \( \bullet \) b \( + |\)b\({|^2}\)     R1(M1)

Note:     Award R1 for \(\left| {\overrightarrow {{\text{OC}}} } \right| = \left| {\overrightarrow {{\text{AB}}} } \right| \Rightarrow {\left| {\overrightarrow {{\text{OC}}} } \right|^2} = {\left| {\overrightarrow {{\text{AB}}} } \right|^2}\) and (M1) for \(|\)a\({|^2} + {\text{2}}\)a \( \bullet \) b \( + |\)b\({|^2} = |\)a\({|^2} - 2\)a \( \bullet \) b \( + |\)b\({|^2}\).

a \( \bullet \) b \( = 0\)     A1

hence OACB is a rectangle (a and b both non-zero)

with adjacent sides at right angles     R1

Note:     Award R1(M1)A0R1 if the dot product has not been used.

[4 marks]

In part (a), a significant number of candidates either did not use correct vector notation or simply did not use vector notation at all. A large number of candidates who appeared to adopt a scalar product approach, did not use the scalar product ‘dot’ and represented a \( \bullet \) b as ab. A few candidates successfully used the cosine rule with correct vector notation. A small number of candidates expressed a and b in general component form. In part (a) (ii), quite a number of candidates expressed \({\overrightarrow {{\text{AB}}} }\) as a – b rather than as b – a.

In part (b), some very well structured proofs were offered by a small number of candidates.