1. **Problem Statement:** Find the position vector of a point dividing a line segment in a given ratio, and prove some vector geometry theorems. 2. **Formula for internal division:** If point P divides segment AB internally in ratio $m:n$, then position vector $\vec{r}$ of P is given by: $$\vec{r} = \frac{m\vec{b} + n\vec{a}}{m+n}$$ 3. **Derivation:** - Given $|AP|:|PB| = m:n \Rightarrow n|AP| = m|PB|$ - Since $\vec{AP}$ and $\vec{PB}$ have the same direction, $n\vec{AP} = m\vec{PB}$ - Using position vectors: $\vec{AP} = \vec{r} - \vec{a}$ and $\vec{PB} = \vec{b} - \vec{r}$ - Substitute: $n(\vec{r} - \vec{a}) = m(\vec{b} - \vec{r})$ - Rearranged: $(m+n)\vec{r} = m\vec{b} + n\vec{a}$ - Hence, $\vec{r} = \frac{m\vec{b} + n\vec{a}}{m+n}$ 4. **Special case (midpoint):** If $m=n=1$, then $$\vec{r} = \frac{\vec{a} + \vec{b}}{2}$$ 5. **Formula for external division:** If P divides AB externally in ratio $m:n$, then $$\vec{r} = \frac{n\vec{a} - m\vec{b}}{n - m}$$ 6. **Proof of parallelogram diagonal bisection:** - Let ABCD be a parallelogram with position vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d}$. - Midpoint of diagonal AC: $\vec{M_1} = \frac{\vec{a} + \vec{c}}{2}$ - Midpoint of diagonal BD: $\vec{M_2} = \frac{\vec{b} + \vec{d}}{2}$ - Since $\vec{b} - \vec{a} = \vec{c} - \vec{d}$, we get $\vec{b} + \vec{d} = \vec{a} + \vec{c}$ - Thus, $\vec{M_1} = \vec{M_2}$, so diagonals bisect each other. 7. **Line joining midpoints of two sides of triangle:** - For triangle ABC with midpoints $M_1$ of CA and $M_2$ of BC, - $\vec{M_1} = \frac{\vec{a} + \vec{c}}{2}$, $\vec{M_2} = \frac{\vec{b} + \vec{c}}{2}$ - Vector $\vec{M_1M_2} = \vec{M_2} - \vec{M_1} = \frac{1}{2}(\vec{b} - \vec{a})$ - So $\vec{M_1M_2}$ is parallel to $\vec{AB}$ and half its length. 8. **Midpoints in trapezium:** - For trapezium ABCD with parallel sides AB and DC, - Midpoints $M_1$ of DA and $M_2$ of BC have position vectors: $$\vec{M_1} = \frac{\vec{d} + \vec{a}}{2}, \quad \vec{M_2} = \frac{\vec{b} + \vec{c}}{2}$$ - Vector $\vec{M_1M_2} = \frac{1}{2}[(\vec{b} - \vec{a}) + (\vec{c} - \vec{d})] = \frac{1}{2}(\vec{AB} + \vec{DC})$ - Since AB is parallel to DC, $\vec{M_1M_2}$ is parallel to both and its length is half the sum of lengths of AB and DC. **Final answers:** - Internal division: $\boxed{\vec{r} = \frac{m\vec{b} + n\vec{a}}{m+n}}$ - External division: $\boxed{\vec{r} = \frac{n\vec{a} - m\vec{b}}{n - m}}$ - Midpoint of segment: $\boxed{\vec{r} = \frac{\vec{a} + \vec{b}}{2}}$ - Diagonals of parallelogram bisect each other. - Line joining midpoints of two sides of triangle is parallel and half the third side. - Line joining midpoints of non-parallel sides of trapezium is parallel to parallel sides and half their sum in length.