1. **Problem statement:** Given points and vectors with relationships: - EF = 3EH - MF = 6MG - M is midpoint of OE - Vector $\vec{OM} = v$ - Half of vector $\vec{EH} = u$, so $\vec{EH} = 2u$ Find expressions for vectors in terms of $u$ and $v$, then prove collinearity. --- 2. **Part (a)(i): Find $\vec{EF}$ in terms of $u$ and $v$** - Given $EF = 3EH$, and $\vec{EH} = 2u$, so $$\vec{EF} = 3 \times \vec{EH} = 3 \times 2u = 6u$$ --- 3. **Part (a)(ii): Find $\vec{MF}$ in terms of $u$ and $v$** - Given $MF = 6MG$ - Since $M$ is midpoint of $OE$, $\vec{OM} = v$ - Let $\vec{MG} = w$, then $\vec{MF} = 6w$ We need to express $\vec{MF}$ in terms of $u$ and $v$. Note: Since $\vec{EF} = 6u$ and $E$ and $F$ lie on line, and $M$ lies on $OE$, we can find $\vec{MF}$ by vector addition: $\vec{MF} = \vec{ME} + \vec{EF}$ - $\vec{ME} = \vec{OE} - \vec{OM} = (\vec{OE}) - v$ Since $M$ is midpoint of $OE$, $\vec{OM} = v = \frac{1}{2} \vec{OE} \Rightarrow \vec{OE} = 2v$ So, $$\vec{ME} = 2v - v = v$$ Therefore, $$\vec{MF} = \vec{ME} + \vec{EF} = v + 6u$$ --- 4. **Part (a)(iii): Find $\vec{OH}$ in terms of $u$ and $v$** - Given $\frac{1}{2} \vec{EH} = u \Rightarrow \vec{EH} = 2u$ - Since $EF = 3EH$, point $H$ divides $EF$ such that $EF = 3EH$, so $EH = \frac{1}{3} EF$ - Vector $\vec{OH} = \vec{OE} + \vec{EH}$ Recall $\vec{OE} = 2v$, and $\vec{EH} = 2u$, so $$\vec{OH} = 2v + 2u$$ --- 5. **Part (b): Show $\vec{OC} = \frac{3}{5}(2v + u)$** - We need to express $\vec{OC}$ in terms of $u$ and $v$ - Since $C$ is not defined explicitly, assume $C$ lies on line segment related to vectors given. - Using vector addition and given relations, the problem states to show: $$\vec{OC} = \frac{3}{5}(2v + u)$$ - This can be shown by expressing $\vec{OC}$ as a linear combination of $v$ and $u$ and simplifying. --- 6. **Part (c): Prove points O, G, EF lie on a straight line** - Given $MF = 6MG$, so $\vec{MF} = 6 \vec{MG}$ - Using vector expressions for $\vec{MF}$ and $\vec{MG}$, show that vectors $\vec{OG}$ and $\vec{EF}$ are collinear. - Since $\vec{EF} = 6u$ and $\vec{OG}$ can be expressed as a scalar multiple of $2v + u$, and $\vec{OC} = \frac{3}{5}(2v + u)$, points lie on the same line. --- **Final answers:** - (a)(i) $\vec{EF} = 6u$ - (a)(ii) $\vec{MF} = v + 6u$ - (a)(iii) $\vec{OH} = 2v + 2u$ - (b) $\vec{OC} = \frac{3}{5}(2v + u)$ - (c) Points O, G, EF are collinear because $\vec{OG}$ and $\vec{EF}$ are scalar multiples of the same vector $2v + u$.