17N.2.SL.TZ0.S_3

pestleMathematicsAISLPaper 217N· ahl-3-13-scalar-and-vector-productssource ↗

Let $\overrightarrow {{\text{AB}}} = \left( {\begin{array}{*{20}{c}} 4 \\ 1 \\ 2 \end{array}} \right)$ .

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

[2]

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

[4]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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]

Examiners’ report

[N/A]