All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
$\vec{OP} = \frac{1}{3} \underset{\sim}{a}$

$\vec{OQ} = \frac{3}{5} \underset{\sim}{b}$

Using Ratio Theorem in MF15, $\vec{OM} = \frac{\vec{OP} +\vec{OQ}}{2}$

$\vec{OM} = \frac{1}{6} \underset{\sim}{a} + \frac{3}{10} \underset{\sim}{b}$

Required Area

$= \frac{1}{2}|\vec{OP} \times \vec{OM}|$

$= \frac{1}{2} |\frac{1}{3} \underset{\sim}{a} \times (\frac{1}{6} \underset{\sim}{a} + \frac{3}{10} \underset{\sim}{b})|$

$= \frac{1}{2} |\frac{1}{18} \underset{\sim}{a} \times \underset{\sim}{a} + \frac{1}{10} \underset{\sim}{a} \times \underset{\sim}{b}|$

$= \frac{1}{2} |0 + \frac{1}{10} \underset{\sim}{a} \times \underset{\sim}{b}|$

$= \frac{1}{20} |\underset{\sim}{a} \times \underset{\sim}{b}|$

(ii)
(a) Since $\underset{\sim}{a}$ is a unit vector, $|\underset{\sim}{a}| = 1$

$\Rightarrow, \sqrt{4p^2 +36p^2 +9p^2} =1$

$\therefore, p= \frac{1}{7}$

(b)
It is the length of projection of $\underset{\sim}{b} \mathrm{~onto~} \underset{\sim}{a}$

(c)
$\underset{\sim}{a} \times \underset{\sim}{b} = \frac{1}{7} \begin{pmatrix}2\\{-6}\\3\end{pmatrix} \times \begin{pmatrix}1\\1\\{-2}\end{pmatrix} = \frac{1}{7}\begin{pmatrix}9\\7\\8\end{pmatrix}$

This question did not pose much of a problem. Most students were able to identify that $\underset{\sim}{a} \times \underset{\sim}{a} = \underset{\sim}{0}$ comfortably. (iib) however, was not well answered as most students overlooked that $\underset{\sim}{a}$ is the unit vector here thus the projection should be that of b onto a, instead a onto b. Lastly, students should always check their cross product using scalar product like I always advice in class.