Vector joining two points

If $O$ is the origin and $\overrightarrow{OA}=\bar a,\ \overrightarrow{OB}=\bar b$ are the position vectors

of $A,B$, then in $\triangle AOB$, using the triangle law,

i.e. the vector joining two points is (position vector of the end point) minus (position vector of the start point). Writing $A\equiv(x_1,y_1,z_1),\ B\equiv(x_2,y_2,z_2)$, so that

$\overrightarrow{OA}=x_1\hat\imath+y_1\hat\jmath+z_1\hat k$ and

$\overrightarrow{OB}=x_2\hat\imath+y_2\hat\jmath+z_2\hat k$, this becomes

whose magnitude $|\overrightarrow{AB}|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$ is exactly the distance

formula between two points.

Worked examples.

• For $A(2,3),B(-1,5),C(-1,1),D(-7,5)$ in a plane, $\overrightarrow{AB}=\bar b-\bar a=-3\hat\imath +2\hat\jmath$ and $\overrightarrow{CD}=\bar d-\bar c=-6\hat\imath+4\hat\jmath=2(-3\hat\imath+2\hat\jmath)$, so $\overrightarrow{CD}=2\,\overrightarrow{AB}$ shows the two are parallel.
• Given $A,B,C$ four points and needing $E=(k,l)$ with $\overrightarrow{AC}\parallel\overrightarrow{BE}$: form $\overrightarrow{AC}$ and $\overrightarrow{BE}$ (the latter in terms of the unknown $k$), set $\overrightarrow{BE}=m\,\overrightarrow{AC}$, and match components to solve simultaneously for $m$ and then $k$.
• For $A(1,-2,3),B(2,3,-4),C(0,-7,10)$: $\overrightarrow{AB}=\hat\imath+5\hat\jmath-7\hat k$ and …