2020HSCShort answer

Q15 (b) (iii)

3 marks

Question

Let $OPQR$ be a parallelogram with $\overrightarrow{OP}=\mathbf p$ and $\overrightarrow{OR}=\mathbf r$ . The point $S$ is the midpoint of $QR$ and $T$ is the intersection of $PR$ and $OS$ , as shown in the diagram.

(iii) Show that
$\overrightarrow{OT}=\frac23\mathbf r+\frac13\mathbf p.$

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

Try the question before revealing the answer.

Solution

Criteria	Marks
Provides correct solution	3
Finds equations of lines $OS$ and $PR$ and attempts to solve for $T$ <br>OR<br>Writes $\overrightarrow{OT}$ in terms of $\vec p$ and $\vec r$ in two ways and attempts to solve	2
Finds equation of line $OS$ <br>OR<br>Writes $\overrightarrow{OT}$ as a multiple of $\overrightarrow{OS}=\vec r+\frac12\vec p$ , or equivalent merit	1

Sample answer:

\overrightarrow{OS}=\overrightarrow{OR}+\overrightarrow{SR}=\vec r+\frac12\vec p

Equation of line $OS$ is

\vec x=\left(\vec r+\frac12\vec p\right)\lambda.

Also equation of line $PR$ is

\vec x=\vec p+(\vec r-\vec p)\mu.

Equating, we have

\vec p\left(\frac12\lambda\right)+\vec r\lambda=\vec p(1-\mu)+\vec r\mu

\therefore\frac12\lambda=1-\mu\quad\text{and}\quad\lambda=\mu\Rightarrow\lambda=\frac23

Hence point of intersection is

\overrightarrow{OT}=\frac23\left(\vec r+\frac12\vec p\right)=\frac23\vec r+\frac13\vec p

Supplied marking guidance · human reviewed

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

Syllabus alignment

Vectors

Vectors and vector equations of linesUse coordinates and vector equations for curves, lines, line segments and spheres in two and three dimensions.

Spot a problem? Send feedback