Q.Find the vector equation of the line which passes through the origin and the point $(5, -2, 3)$.

When a line is described not by a point-plus-direction but by two of its own points, $A(\bar{a})$ and $B(\bar{b})$, its direction is obtained simply by subtracting the two position vectors: $\overline{AB} = \bar{b}-\bar{a}$. For any other point $P(\bar{r})$ on the line, $\overline{AP}$ and $\overline{AB}$ are collinear, so $\overline{AP} = \lambda\overline{AB}$ for some scalar $\lambda$, giving the vector equation $\bar{r} = \bar{a} + \lambda(\bar{b}-\bar{a}).$ This is really the point-and-direction-vector form applied with $\bar{b}-\bar{a}$ playing the role of the direction vector — so once the difference $\bar{b}-\bar{a}$ is computed, the two situations are handled identically. As with the point-and-vector form, this can also be written without a parameter as $(\bar{r}-\bar{a})\times(\bar{b}-\bar{a}) = \bar{0}$. A useful self-check after finding such an equation is to substitute $\lambda = 1$ …