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202210141016 Solution to 1975-CE-AMATH-II-XX

In the Figure below, OABC is a square. D and E are the mid-points of AB and BC respectively. Let the vector \overrightarrow{OA} be represented by \mathbf{i} and \overrightarrow{OC} by \mathbf{j}.

(a) Express \overrightarrow{OD} in terms of \mathbf{i} and \mathbf{j}.
(b) Express \overrightarrow{OE} in terms of \mathbf{i} and \mathbf{j}.
(c) Evaluate \displaystyle{\frac{\overrightarrow{OD}\cdot\overrightarrow{OE}}{\left|\overrightarrow{OD}\right|\left|\overrightarrow{OE}\right|}}, where \left|\overrightarrow{OD}\right| and \left|\overrightarrow{OE}\right| are the magnitudes of \overrightarrow{OD} and \overrightarrow{OE} respectively.
(d) Hence calculate \angle DOE.

Notation.

In what follow vectors will be typographically boldfaced, i.e.,

\begin{aligned} \mathbf{OA} & := \overrightarrow{OA} \\ \mathbf{OC} & := \overrightarrow{OC} \\ \mathbf{OD} & := \overrightarrow{OD} \\ \mathbf{OE} & := \overrightarrow{OE} \\ \end{aligned}

etc., in place of an overhead arrow \overrightarrow{[\cdot ]} more than usually handwritten, the direction of which (e.g. \overrightarrow{AB}) is intended an initial point (e.g. A) make to a terminal point (e.g. B).

(a)

\begin{aligned} \mathbf{OD} & = \mathbf{OA} + \mathbf{AD} \\ & = \mathbf{OA} + \frac{1}{2}\mathbf{AB} \\ \dots\enspace\because\enspace & \mathbf{AB}=\mathbf{OC}\enspace\dots \\ & = \mathbf{OA} + \frac{1}{2}\mathbf{OC} \\ & = \mathbf{i} + \frac{1}{2}\mathbf{j} \\ \end{aligned}

(b)

\begin{aligned} \mathbf{OE} & = \mathbf{OC} + \mathbf{CE} \\ & = \mathbf{OC} + \frac{1}{2}\mathbf{CB} \\ \dots\enspace\because\enspace & \mathbf{CB}=\mathbf{OA}\enspace\dots \\ & = \mathbf{OC} + \frac{1}{2}\mathbf{OA} \\ & = \mathbf{j} + \frac{1}{2}\mathbf{i} \\ \end{aligned}

(c) and (d) are not chosen.