1. **Problem statement:** We have triangle OAB with vectors: $\vec{OA} = 8\vec{c}$, $\vec{OB} = 4\vec{d}$, $\vec{BP} = 2\vec{d}$, and $\vec{OM} = 6\vec{c}$. Point N is the midpoint of AB. We need to: a) Find $\vec{MN}$ in terms of $\vec{c}$ and $\vec{d}$ and simplify. b) Show that points M, N, and P are collinear. 2. **Step a) Find $\vec{MN}$:** - First, find $\vec{N}$, the midpoint of AB. - Since $\vec{A} = 8\vec{c}$ and $\vec{B} = 4\vec{d}$, the midpoint $\vec{N} = \frac{\vec{A} + \vec{B}}{2} = \frac{8\vec{c} + 4\vec{d}}{2} = 4\vec{c} + 2\vec{d}$. - We know $\vec{M} = 6\vec{c}$. - Vector $\vec{MN} = \vec{N} - \vec{M} = (4\vec{c} + 2\vec{d}) - 6\vec{c} = (4 - 6)\vec{c} + 2\vec{d} = -2\vec{c} + 2\vec{d}$. 3. **Step b) Show M, N, P are collinear:** - Find $\vec{P}$ using $\vec{OB} = 4\vec{d}$ and $\vec{BP} = 2\vec{d}$. - Since $\vec{P} = \vec{B} + \vec{BP} = 4\vec{d} + 2\vec{d} = 6\vec{d}$. - Vector $\vec{NP} = \vec{P} - \vec{N} = 6\vec{d} - (4\vec{c} + 2\vec{d}) = -4\vec{c} + 4\vec{d}$. - Check if $\vec{MN}$ and $\vec{NP}$ are scalar multiples: $$\vec{NP} = 2 \times (-2\vec{c} + 2\vec{d}) = 2 \vec{MN}$$ - Since $\vec{NP} = 2 \vec{MN}$, vectors $\vec{MN}$ and $\vec{NP}$ are parallel, so points M, N, and P lie on a straight line. **Final answers:** $$\vec{MN} = -2\vec{c} + 2\vec{d}$$ M, N, and P are collinear because $\vec{NP} = 2 \vec{MN}$.