# Exam-Style Question on Vectors

## A mathematics exam-style question with a worked solution that can be revealed gradually

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Question id: 655. This question is similar to one that appeared on an IB AA Higher paper in 2023. The use of a calculator is allowed.

The following diagram shows parallelogram OABC with $$\overrightarrow{OA}\ = \pmb{a}$$, $$\overrightarrow{OC}\ = \pmb{c}$$ and $$|\pmb{c}| = 2|\pmb{a}|$$

The angle $$ABC = \theta$$, where $$0 \lt \theta \lt \pi$$.

Point M is on [AB] such that $$\overrightarrow{AM} = k\overrightarrow{AB}$$, where $$0 \le k \le 1$$.

OM is perpendicular to MC.

(a) Express $$\overrightarrow{OM}$$ and $$\overrightarrow{MC}$$ in terms of $$\pmb{a}$$ and $$\pmb{c}$$.

(b) Hence, use a vector method to show that $$|a|^2(1-2k)(2\cos{\theta}-(1-2k))=0$$.

(c) Find the range of values for $$\theta$$ such that there are two possible positions for M.

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