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.