Find $\left| {\overrightarrow {{\text{AB}}} } \right|$.

Let $\overrightarrow {{\text{AC}}} = \left( {\begin{array}{*{20}{c}} 3 \\ 0 \\ 0 \end{array}} \right)$. Find ${\rm{B\hat AC}}$.

correct substitution     (A1)

eg$\,\,\,\,\,$$\sqrt {{4^2} + {1^2} + {2^2}}$

4.58257

$\left| {\overrightarrow {{\text{AB}}} } \right| = \sqrt {21}$ (exact), 4.58     A1     N2

[2 marks]

finding scalar product and $\left| {\overrightarrow {{\text{AC}}} } \right|$     (A1)(A1)

scalar product $= (4 \times 3) + (1 \times 0) + (2 \times 0){\text{ }}( = 12)$

$\left| {\overrightarrow {{\text{AC}}} } \right| = \sqrt {{3^2} + 0 + 0} {\text{ }}( = 3)$

substituting their values into cosine formula     (M1)

eg cos B$\hat A$C${\text{ = }}\frac{{4 \times 3 + 0 + 0}}{{\sqrt {{3^2}}  \times \sqrt {21} }},{\text{ }}\frac{4}{{\sqrt {21} }},{\text{ }}\cos \theta  = 0.873$

0.509739 (29.2059°)

${\rm{B\hat AC}} = 0.510$ (29.2°)     A1     N2

[4 marks]