1. **Problem:** Find the position vector of point Q dividing the line segment \(\overline{AB}\) externally in the ratio 3 : 2. 2. **Formula:** If a point Q divides the line segment joining points A and B externally in the ratio \(m:n\), then the position vector of Q is given by: $$\vec{Q} = \frac{m\vec{B} - n\vec{A}}{m - n}$$ 3. **Explanation:** For external division, the ratio is taken with a minus sign in the denominator because the point lies outside the segment. 4. **Intermediate work:** Let \(\vec{A}\) and \(\vec{B}\) be the position vectors of points A and B respectively. Then, $$\vec{Q} = \frac{3\vec{B} - 2\vec{A}}{3 - 2} = 3\vec{B} - 2\vec{A}$$ 5. **Interpretation:** The position vector of Q is a linear combination of \(\vec{A}\) and \(\vec{B}\) weighted by the ratio 3 and 2, with subtraction indicating external division. 6. **Final answer:** $$\boxed{\vec{Q} = 3\vec{B} - 2\vec{A}}$$ --- 1. **Problem:** Find the fourth vertex D of parallelogram ABCD with vertices \(A(3, -4)\), \(B(-1, -3)\), and \(C(-6, 2)\). 2. **Formula:** In a parallelogram, the position vector of D is given by: $$\vec{D} = \vec{A} + \vec{C} - \vec{B}$$ 3. **Explanation:** Because \(\vec{AB} + \vec{AD} = \vec{AC}\), rearranged to find \(\vec{D}\). 4. **Intermediate work:** $$D_x = A_x + C_x - B_x = 3 + (-6) - (-1) = 3 - 6 + 1 = -2$$ $$D_y = A_y + C_y - B_y = -4 + 2 - (-3) = -4 + 2 + 3 = 1$$ 5. **Final answer:** $$\boxed{D(-2, 1)}$$ --- 1. **Problem:** Find values of \(x\) and \(y\) such that \(A(1, 2)\), \(B(4, y)\), \(C(x, 6)\), and \(D(3, 5)\) form a parallelogram. 2. **Formula:** For parallelogram ABCD, $$\vec{A} + \vec{C} = \vec{B} + \vec{D}$$ 3. **Intermediate work:** Equate components: $$1 + x = 4 + 3 = 7 \implies x = 6$$ $$2 + 6 = y + 5 \implies 8 = y + 5 \implies y = 3$$ 4. **Final answer:** $$\boxed{x = 6, y = 3}$$ --- 1. **Problem:** Show that the line segments joining the midpoints of the sides of a quadrilateral consecutively form a parallelogram. 2. **Explanation:** Let the quadrilateral have vertices \(A, B, C, D\) and midpoints \(E, F, G, H\) of sides \(AB, BC, CD, DA\) respectively. 3. **Key fact:** The segment joining midpoints of two sides is parallel to the diagonal and half its length. 4. **Result:** Joining \(E, F, G, H\) forms a parallelogram by the midpoint theorem. --- 1. **Problem:** Show that line segments joining midpoints of diagonals and midpoints of any two opposite sides of a quadrilateral consecutively form a parallelogram. 2. **Explanation:** Using midpoint theorem and vector addition, these segments form a closed figure with opposite sides parallel. --- 1. **Problem:** Prove that the segment joining midpoints of diagonals of a trapezium is parallel to the parallel sides and its length is half the difference of the lengths of the parallel sides. 2. **Explanation:** Using vector properties and midpoint theorem, the segment is parallel and length equals half the difference of parallel sides. --- 1. **Problem:** In trapezium OPQR with \(\overline{OP} = \vec{r}\), \(\overline{OS} = \vec{s}\), and M midpoint of \(\overline{OR}\), (i) Find \(\overline{PS}\) in terms of \(\vec{r}\) and \(\vec{s}\). (ii) Show \(\overline{OQ}\) is parallel to \(\overline{SM}\). 2. **Solution:** (i) \(\overline{PS} = \vec{s} - \vec{r}\) (ii) Using midpoint and vector addition, \(\overline{OQ}\) and \(\overline{SM}\) are parallel. --- 1. **Problem:** In trapezium ABCD with \(\overline{AB} \parallel \overline{DC}\), E lies on diagonal \(\overline{DB}\) such that \(DE = \frac{1}{3} DB\). Show \(\overline{BC} \parallel \overline{AE}\). 2. **Explanation:** Using section formula and vector properties, \(\overline{BC}\) and \(\overline{AE}\) are parallel. --- 1. **Problem:** In regular hexagon ABCDEF with \(\overline{AB} = \vec{a}\) and \(\overline{BC} = \vec{b}\), express \(\overline{AC}, \overline{CD}, \overline{EF}, \overline{DA}, \overline{EB}, \overline{FA}, \overline{FC}\) in terms of \(\vec{a}\) and \(\vec{b}\). 2. **Expressions:** $$\overline{AC} = \vec{a} + \vec{b}$$ $$\overline{CD} = \vec{b} - \vec{a}$$ $$\overline{EF} = -\vec{a}$$ $$\overline{DA} = -\vec{a} - \vec{b}$$ $$\overline{EB} = \vec{b} - \vec{a}$$ $$\overline{FA} = -\vec{b}$$ $$\overline{FC} = -\vec{a} + \vec{b}$$