Collinearity

Video lesson · from 11:572 worked examples

One lesson video covers all of Vectors, so it opens at 11:57 for this topic — not from the beginning.

Theory

Points A, B and C are collinear if $\vec{AB}$ and $\vec{BC}$ are parallel, with B a common point.

NB: vectors are parallel if they are scalar multiples of the same vector:

e.g. $\vec{u} = \begin{pmatrix} 8 \\ -2 \\ 0 \end{pmatrix}$ , $\vec{v} = \begin{pmatrix} 16 \\ -4 \\ 0 \end{pmatrix} = 2\begin{pmatrix} 8 \\ -2 \\ 0 \end{pmatrix} = 2\vec{u} \implies \vec{u}$ and $\vec{v}$ are parallel.
e.g. $\vec{a} = \begin{pmatrix} 12 \\ -3 \\ 9 \end{pmatrix} = 3\begin{pmatrix} 4 \\ -1 \\ 3 \end{pmatrix}$ , $\vec{b} = \begin{pmatrix} 20 \\ -5 \\ 15 \end{pmatrix} = 5\begin{pmatrix} 4 \\ -1 \\ 3 \end{pmatrix} \implies \vec{a}$ and $\vec{b}$ are parallel.

⚠️ Common Examiner Traps

Parallel is not enough: you must also state that the two vectors share a common point. Without that, they could be parallel lines that never meet.
Show one is a multiple of the other: $\vec{AB} = k\vec{BC}$ . State the value of $k$ explicitly.
Write the conclusion: say that the vectors are parallel, that the point is common, and therefore that the points are collinear. Conclusions here are very often left unstated.
Check every component gives the same $k$ : if one component disagrees, the vectors are not parallel at all.

Worked examples

Example 1

A is the point $(1, -2, 5)$ , B is $(8, -5, 9)$ and C is the point $(22, -11, 17)$ .

Show that A, B and C are collinear.
Find the ratio in which B divides AC.

1. Find $\vec{AB}$ and $\vec{BC}$ :

\vec{AB} = \vec{b} - \vec{a} = \begin{pmatrix} 8 \\ -5 \\ 9 \end{pmatrix} - \begin{pmatrix} 1 \\ -2 \\ 5 \end{pmatrix} = \begin{pmatrix} 7 \\ -3 \\ 4 \end{pmatrix}

\vec{BC} = \vec{c} - \vec{b} = \begin{pmatrix} 22 \\ -11 \\ 17 \end{pmatrix} - \begin{pmatrix} 8 \\ -5 \\ 9 \end{pmatrix} = \begin{pmatrix} 14 \\ -6 \\ 8 \end{pmatrix}

Notice that $\vec{BC} = 2\vec{AB}$ . Since $\vec{BC}$ and $\vec{AB}$ are scalar multiples, they are parallel. Since B is a common point, A, B and C are collinear.

2. The ratio $AB:BC$ is $1:2$ , because $\vec{BC}$ is twice as long as $\vec{AB}$ .

Example 2

D, E and F have coordinates $(10, -8, -15)$ , $(1, -2, -3)$ and $(-14, 8, 17)$ respectively.

Show that D, E and F are collinear.
Find the ratio in which E divides DF.

1. Find $\vec{DE}$ and $\vec{EF}$ :

\vec{DE} = \vec{e} - \vec{d} = \begin{pmatrix} 1 \\ -2 \\ -3 \end{pmatrix} - \begin{pmatrix} 10 \\ -8 \\ -15 \end{pmatrix} = \begin{pmatrix} -9 \\ 6 \\ 12 \end{pmatrix} = 3\begin{pmatrix} -3 \\ 2 \\ 4 \end{pmatrix}

\vec{EF} = \vec{f} - \vec{e} = \begin{pmatrix} -14 \\ 8 \\ 17 \end{pmatrix} - \begin{pmatrix} 1 \\ -2 \\ -3 \end{pmatrix} = \begin{pmatrix} -15 \\ 10 \\ 20 \end{pmatrix} = 5\begin{pmatrix} -3 \\ 2 \\ 4 \end{pmatrix}

Since both $\vec{DE}$ and $\vec{EF}$ are scalar multiples of $\begin{pmatrix} -3 \\ 2 \\ 4 \end{pmatrix}$ , they are parallel. Since E is a common point, D, E and F are collinear.

2. The ratio in which E divides DF ( $DE:EF$ ) is $3:5$ .