2020HSCShort answer

Q15 (b) (i)

2 marks

Question

The point $C$ divides the interval $AB$ so that
$\frac{CB}{AC}=\frac mn.$
The position vectors of $A$ and $B$ are $\mathbf a$ and $\mathbf b$ , respectively, as shown in the diagram.

(i) Show that
$\overrightarrow{AC}=\frac n{m+n}(\mathbf b-\mathbf a).$

Uncropped source page 12 for question 15.b.i; requires human review and cropping — Reviewed visual · source page 12

Try the question before revealing the answer.

Solution

Criteria	Marks
Provides correct solution	2
Recognises that $\overrightarrow{CB}=\frac mn\overrightarrow{AC}$ , or equivalent merit	1

Sample answer:

\overrightarrow{OA}+\overrightarrow{AC}+\overrightarrow{CB}=\overrightarrow{OB}

\therefore \vec a+\overrightarrow{AC}+\frac mn\overrightarrow{AC}=\vec b

\therefore\overrightarrow{AC}\left(1+\frac mn\right)=\vec b-\vec a

\therefore\overrightarrow{AC}=\frac n{m+n}(\vec b-\vec a)

Supplied marking guidance · human reviewed

Compare this with your working, not only your final sentence.

Syllabus alignment

Vectors

Introduction to three-dimensional vectorsRepresent and operate with vectors in three dimensions using unit, component, ordered-triple and column-vector forms.

Spot a problem? Send feedback