The diagram shows a triangular prism ABC, DEF - Scottish Highers Maths - Question 9 - 2018

To express vector $ightarrow{MD}$, we can start from point M, which is the midpoint of $ightarrow{BC}$. The position vector of M can be calculated by averaging the position vectors of B and C:

ightarrow{M} = rac{1}{2}( ightarrow{B} + ightarrow{C})$$ Substituting the expressions:

ightarrow{B} = ightarrow{A} + ightarrow{BA} = ightarrow{A} - t$$ and

ightarrow{C} = ightarrow{A} + ightarrow{AC} = ightarrow{A} + u$$ Thus, we find:

ightarrow{M} = rac{1}{2}(( ightarrow{A} - t) + ( ightarrow{A} + u)) = ightarrow{A} + rac{u - t}{2}$$

Now, to express vector $ightarrow{MD}$:

ightarrow{MD} = ightarrow{D} - ightarrow{M} = ( ightarrow{A} + v) - ( ightarrow{A} + rac{u - t}{2})$$ This simplifies to:

ightarrow{MD} = v - rac{u - t}{2}$$ Final expression for vector $ightarrow{MD}$ is:

ightarrow{MD} = rac{1}{2}v + rac{t - u}{2}$$.