Q.In the figure 5.34 (triangle PQR with $\bar a=\overrightarrow{QP}$ along the left side, $\bar b=\overrightarrow{PR}$ along the right side, and S the midpoint of QR with $\bar d=\overrightarrow{SR}$, $-\bar d=\overrightarrow{SQ}$) express $\bar c=\overrightarrow{SP}$ and $\bar d$ in terms of $\bar a$ and $\bar b$.

Take $Q$ as the origin for position vectors. Since $\bar a=\overrightarrow{QP}$, $P$ has position vector

$\bar a$. Since $\bar b=\overrightarrow{PR}$, $R$ has position vector $\overrightarrow{QR}=\overrightarrow{QP} +\overrightarrow{PR}=\bar a+\bar b$.

The figure marks $\overrightarrow{SR}=\bar d$ and $\overrightarrow{SQ}=-\bar d$ with equal magnitude and

opposite sign, which means $S$ is equidistant from $Q$ and $R$ along the segment $QR$, i.e. $S$ is the

midpoint of $QR$. So $S$ has position vector

$\bar s=\frac{\bar 0+(\bar a+\bar b)}{2}=\frac{\bar a+\bar b}{2}.$ …

Assign Q as origin, build position vectors of P and R from a and b via the triangle law, identify S as the midpoint of QR from the matching +/- d lab …

Treating S as an arbitrary point on QR rather than recognising, from the equal-and-opposite d …