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Properties of Parallelograms and Trapezia in Vector Form

Properties of Parallelograms and Trapezia in Vector Form

TGM Original

In the triangle \(OAB\), \(M\) is the midpoint of \(OA\). \(N\) is a point on \(OB\) such that \(ON : NB = 2 : 1\). \(MN\) is produced to \(P\) so that \(MN : NP = 1 : 2\).

Given \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\).

  1. Express, in terms of \(\mathbf{a}\) and/or \(\mathbf{b}\),
    1. \(\overrightarrow{NB}\),
    2. \(\overrightarrow{MN}\),
    3. \(\overrightarrow{NP}\).
  2. Express \(\overrightarrow{AB}\) and \(\overrightarrow{BP}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\).
  3. Write down two facts about \(\overrightarrow{AB}\) and \(\overrightarrow{BP}\).
  4. Find \(\frac{\text{area of triangle}\hspace{0.5em} PMB}{\text{area of triangle}\hspace{0.5em} PMA}\).

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Answer:(a)(i) NB = (1/3)b; (ii) MN = -(1/2)a + (2/3)b; (iii) NP = -a + (4/3)b. (b) AB = BP = -a + b. (c) AB and BP have equal magnitude and the same direction; A, B and P are collinear with B the midpoint of AP. (d) [PMB]/[PMA] = 1/2.

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