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

\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{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}|

(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}

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

\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}

KS Comments:

This question did not pose much of a problem. Most students were able to identify that \underset{\sim}{a} \times \underset{\sim}{a} = 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.

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