11a.4. Position Vectors, Distance Between Points, Collinear Points

Position vectors

Position vectors vs general vectors

A vector comprises of magnitude and direction: the starting or ending position of the vector is not important. For example, the vector $\mathbf{a} = \mathbf{i} + 2 \mathbf{j}$ involves a movement of 1 unit along the $x$-axis and 2 units along the $y$-axis. It doesn't matter where it starts from: any such movement are all considered $\mathbf{a}$. In fact, vectors can be "moved around" as long as we keep it parallel and maintain its magnitude.

When we study geometry (in particular, 3-dimensional geometry for our syllabus) it is often useful to work with points. We have already worked with coordinates in curve sketching and can write, for example, the coordinates of a point $A$ to be $(1,2,0)$. We let the origin $O$ be $(0,0,0)$ and thus have a vector $\overrightarrow{OA} = \mathbf{i}+2\mathbf{j}$.

However, we will "lose track" of the point $A$ if we allow the vector $\overrightarrow{OA}$ to "move". To handle this discrepancy, we call the vectors that originate from the origin and end at a point the position vector of the point. This special vector is often handled slightly differently from a general (direction) vector. A position vector is generally considered to be "fixed" while direction vectors can be "moved around".

Notationally for our syllabus, we write coordinates horizontally, for example, $B (1,-3,2)$ and represent position vectors in column vector form ($\overrightarrow{OB}=\mathbf{b}=\begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix}$). Using $\mathbf{i},\mathbf{j}$ and $\mathbf{k}$ will work for the latter as well.

We are often interested to form a vector starting from a point $A$ and ending at a point $B$. Using concepts from vector addtion and subtraction, we get $\overrightarrow{AB}=\overrightarrow{AO} + \overrightarrow{OB} = \overrightarrow{OB}-\overrightarrow{OA}$. This is illustated in the diagram below.

$$ \overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}$$

This seemingly trivial formula is especially useful and will be used very often in our syllabus since it allows us to go from coordinates/position vectors to a vector starting from one point and ending at another.

Distance between points

We now know how to calculate $\overrightarrow{AB}$. To find the the distance between two points, $A$ and $B$, we use the magnitude of $\overrightarrow{AB}$, $\left | \overrightarrow{AB} \right |$.

Collinear points

We are sometimes interested in whether three points lie on a straight line. If they do, we call them collinear . To check whether 3 points $A,B$ and $C$ are collinear, we can form the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ (other combinations like $\overrightarrow{BC}$ and $\overrightarrow{CB}$ are allowed too). If $A,B$ and $C$ are collinear, then $\overrightarrow{AB}$ and $\overrightarrow{AC}$ will be parallel so $\overrightarrow{AB}=k\overrightarrow{AC}$.

Conversely, if $\overrightarrow{AB}=k\overrightarrow{AC}$ for some $k \neq 0$, then $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are parallel. Moreover, they share a common point $A$ so $A,B$ and $C$ must collinear.

$ A, B \textrm{ and } C \textrm{ are collinear} \Leftrightarrow \overrightarrow{AB}=k\overrightarrow{AC} \textrm { for some } k \neq 0$

Solutions to examples

1. Given points $A(5,-1,3)$ and $B$ with position vector $\overrightarrow{OB} = \mathbf{i} + 2 \mathbf{j} - 4 \mathbf{k}$, find the vector $\overrightarrow{AB}$.

$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} =$ $-4 \mathbf{i} + 3 \mathbf{j} -7\mathbf{k}$.

2. Find the distance between $A$ and $B$.

Distance required $=|\overrightarrow{AB}| =$ $\sqrt{74}$ units.

3. It is given further that $C$ has coordinates $(13,y,17)$. Given that $A,B$ and $C$ are collinear, find $y$.

$\overrightarrow{AC} = \overrightarrow{OC}-\overrightarrow{OA} = \begin{pmatrix} 8 \\ y+1 \\ 14 \end{pmatrix}$. Since $A,B$ and $C$ are collinear, $\overrightarrow{AC} = k \overrightarrow{AB}$ for some $k$. $\begin{pmatrix} 8 \\ y+1 \\ 14 \end{pmatrix} = k\begin{pmatrix} -4 \\ 3 \\ -7 \end{pmatrix}$. Hence $k = -2$ and $y+1 = (-2)(-4)$. This gives $y=7$.

4. If $D$ has coordinates $(-3,5,-11)$, show that $A,B$ and $D$ lie on the same line.

$\overrightarrow{AD} = \overrightarrow{OD} - \overrightarrow{OA} = -8\mathbf{i} + 6 \mathbf{j} -14 \mathbf{k}$. Hence $\overrightarrow{AD} = 2 \overrightarrow{AB}$, so $AD$ is parallel to $AB$. Moreover, they share a common point $A$. Thus $A,B$ and $D$ lie on the same line.