1. Problem 1: Given parallelogram ABCD with center O and M, N midpoints of AB and CD respectively. (a) Show $\vec{DA} + \vec{DB} = 2\vec{DM}$. - Since M is midpoint of AB, $\vec{AM} = \frac{1}{2}\vec{AB}$. - Vector $\vec{DM} = \vec{DA} + \vec{AM} = \vec{DA} + \frac{1}{2}\vec{AB}$. - Note $\vec{DB} = \vec{DA} + \vec{AB}$. - So: $\vec{DA} + \vec{DB} = \vec{DA} + (\vec{DA} + \vec{AB}) = 2\vec{DA} + \vec{AB} = 2(\vec{DA} + \frac{1}{2}\vec{AB}) = 2\vec{DM}$. (b) Show $\vec{DA} + \vec{DB} + \vec{DC} = 2\vec{DO}$. - O is midpoint of diagonals, so $\vec{DO} = \frac{1}{2}(\vec{DA} + \vec{DC})$. - $\vec{DA} + \vec{DB} + \vec{DC} = \vec{DA} + (\vec{DA} + \vec{AB}) + \vec{DC} = 2\vec{DA} + \vec{AB} + \vec{DC}$. - Since $\vec{AB} + \vec{DC} = 0$ in parallelogram, - So $\vec{DA} + \vec{DB} + \vec{DC} = 2\vec{DA} = 2\vec{DO}$. (c) Show $\vec{OM} + \vec{ON} = \vec{AD}$. - $\vec{OM} = \frac{1}{2}\vec{AB}$ (as M midpoint of AB with O center). - $\vec{ON} = \frac{1}{2}\vec{DC} = -\frac{1}{2}\vec{AB}$ (opposite direction). - Sum $\vec{OM} + \vec{ON} = \frac{1}{2}\vec{AB} - \frac{1}{2}\vec{AB} = 0$, not $\vec{AD}$. - So (c) is false. (d) $DB^2 = DA^2 + DC^2 + 2DA \cdot DC \cos \angle ADC$. - By Law of Cosines in triangle ADC, this is true. 2. Problem 2: Two forces $F_1$ and $F_2$ of magnitude 50 N each act at point O with angle 60°. (a) $|F_1|=|F_2|=50$ N, true by definition. (b) $F_1 = F_2$ means vectors equal, false because direction differs. (c) Magnitude of resultant force is $$|\vec{F}| = \sqrt{50^2 + 50^2 + 2 \cdot 50 \cdot 50 \cdot \cos 60^\circ} = \sqrt{2500 + 2500 + 2500} = \sqrt{7500} = 50\sqrt{3}.$$ - So true. (d) Resultant force makes 30° with each force, not 60°. So (d) is false. 3. Problem 3: Right trapezoid ABCD with $AB = a$, $AD = 2a$, $DC = 3AB = 3a$. (a) $\vec{DC} = -3\vec{BA}$. - $\vec{BA} = -\vec{AB}$. - So $-3\vec{BA} = -3(-\vec{AB}) = 3\vec{AB} = \vec{DC}$, true. (b) $\vec{AB} + \vec{AC} = \vec{AE}$? - Point E divides DC by $\frac{2}{3}$, so $\vec{AE} \neq \vec{AB} + \vec{AC}$ generally, false. (c) $[\vec{BA} + \vec{BD}] = 2a$. - Needed context for bracket meaning unclear, likely false as a vector equality without units. (d) The set of points $M$ such that $$|\vec{MC} + 3\vec{MA}| = |\vec{MB} - \vec{MC}|$$ - corresponds to a circle centered at O with radius $\alpha \frac{\sqrt{2}}{2}$, true by given properties. 4. Problem 4: Quadrilateral ABCD with diagonals intersecting at O. (a) $\vec{OA} + \vec{OC} = \vec{AC}$, true by vector addition. (b) $DA \cdot DB < DA \cdot DB + 1$, true since $DA \cdot DB + 1$ is larger. (c) $\vec{AB} + \vec{AD} = \vec{CD} + \vec{CB}$? - By parallelogram properties, sum of adjacent sides equals sum of opposite sides, true. (d) $\vec{AB} + \vec{CD} = \vec{AD} + \vec{CB}$? - This equality holds in parallelograms; generally true. 5. Problem 5: Equilateral triangle ABC, points M, N, P with given ratios. (a) Check if $\vec{AP} + \vec{AN} = \vec{AM}$. - Using ratios and position vectors, equality does not hold generally, so false. Final answers: 1.(a) True (b) True (c) False (d) True 2.(a) True (b) False (c) True (d) False 3.(a) True (b) False (c) False (d) True 4.(a) True (b) True (c) True (d) True 5.(a) False