Collinear vectors and collinear points – Collinear Points and Orthogonal Vectors in 2D (Senior 3 Mathematics) | Ganzaa

Collinear Points and Orthogonal Vectors in 2D

Lessons

Senior 3 · Mathematics

Collinear Points and Orthogonal Vectors in 2D

3 lessons

Lesson 1 of 3

Collinear vectors and collinear points

25 min

Is the tap on the pipe line?

A water pipe must run in one straight line from a tank at $A$ , through a valve at $B$ , to a tap at $C$ . On the site plan (1 unit = 10 m) the three points are $A (1, 2)$ , $B (4, 4)$ and $C (10, 8)$ . Before the trench is dug, the plumber wants to be sure that $C$ really is on the line through $A$ and $B$ . A drawing can mislead by a few millimetres; two vectors cannot.

Vectors that point along one line

In Senior 2 you wrote vectors as columns, such as $\overrightarrow{AB} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$ , and multiplied a vector by a scalar $k$ : $k\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} kx \\ ky \end{pmatrix}$ . The vector $k\,\vec{u}$ lies along the same line as $\vec{u}$ : the same way if $k > 0$ , the opposite way if $k < 0$ .

Definition

Two vectors $\vec{u}$ and $\vec{v}$ are collinear (or parallel) if one is a scalar multiple of the other: $\vec{v} = k\,\vec{u}$ for some real number $k$ . The zero vector is collinear with every vector.

Three or more points are collinear if they lie on one straight line.

Example: $\begin{pmatrix} -6 \\ 9 \end{pmatrix} = -3\begin{pmatrix} 2 \\ -3 \end{pmatrix}$ , so these two vectors are collinear; they point in opposite directions because $k = - 3 < 0$ .

Important

The determinant test. For $\vec{u} = \begin{pmatrix} x_1 \\ y_1 \end{pmatrix}$ and $\vec{v} = \begin{pmatrix} x_2 \\ y_2 \end{pmatrix}$ ,

\vec{u} \text{ and } \vec{v} \text{ are collinear} \iff \det(\vec{u}, \vec{v}) = \begin{vmatrix} x_1 & x_2 \\ y_1 & y_2 \end{vmatrix} = x_1y_2 - x_2y_1 = 0 .

Why. If $\vec{v} = k\vec{u}$ , then $x_{2} = k x_{1}$ and $y_{2} = k y_{1}$ , so $x_{1} y_{2} - x_{2} y_{1} = k x_{1} y_{1} - k x_{1} y_{1} = 0$ . Conversely, if the determinant is 0 and $x_1 \neq 0$ , put $k = \frac{x_2}{x_1}$ : then $x_{2} = k x_{1}$ and $y_2 = \frac{x_2y_1}{x_1} = ky_1$ .

Three points. $A$ , $B$ and $C$ are collinear exactly when $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are collinear. The two vectors are parallel and share the point $A$ , so they lie on one line.

Worked example 1: testing two vectors

Is $\vec{u} = \begin{pmatrix} 2 \\ -3 \end{pmatrix}$ collinear with (a) $\vec{v} = \begin{pmatrix} -6 \\ 9 \end{pmatrix}$ ; (b) $\vec{w} = \begin{pmatrix} 4 \\ -5 \end{pmatrix}$ ?

Solution (a) $\det(\vec{u}, \vec{v}) = 2 \times 9 - (-6) \times (-3) = 18 - 18 = 0$ : collinear. Indeed $\vec{v} = -3\vec{u}$ .

(b) $\det(\vec{u}, \vec{w}) = 2 \times (-5) - 4 \times (-3) = -10 + 12 = 2 \neq 0$ : not collinear.

Worked example 2: the pipe line

Show that $A (1, 2)$ , $B (4, 4)$ and $C (10, 8)$ are collinear, and find the ratio $A B : B C$ . Is $D (7, 7)$ on the same line?

Solution $\overrightarrow{AB} = \begin{pmatrix} 4 - 1 \\ 4 - 2 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$ and $\overrightarrow{AC} = \begin{pmatrix} 10 - 1 \\ 8 - 2 \end{pmatrix} = \begin{pmatrix} 9 \\ 6 \end{pmatrix} = 3\overrightarrow{AB}$ .

The vectors are collinear and share the point $A$ , so $A$ , $B$ and $C$ are collinear: the tap is on the line. Since $\overrightarrow{BC} = \begin{pmatrix} 6 \\ 4 \end{pmatrix} = 2\overrightarrow{AB}$ , $A B : B C = 1 : 2$ .

For $D$ : $\overrightarrow{AD} = \begin{pmatrix} 6 \\ 5 \end{pmatrix}$ and $\det(\overrightarrow{AB}, \overrightarrow{AD}) = 3 \times 5 - 6 \times 2 = 3 \neq 0$ . $D$ is not on the line.

Worked example 3: a missing coordinate

Find $k$ so that $P (- 1, 3)$ , $Q (2, k)$ and $R (8, - 3)$ are collinear.

Solution $\overrightarrow{PQ} = \begin{pmatrix} 3 \\ k - 3 \end{pmatrix}$ and $\overrightarrow{PR} = \begin{pmatrix} 9 \\ -6 \end{pmatrix}$ .

\det(\overrightarrow{PQ}, \overrightarrow{PR}) = 3 \times (-6) - 9(k - 3) = -18 - 9k + 27 = 9 - 9k .

Collinear means $9 - 9 k = 0$ , so $k = 1$ . Check: $\overrightarrow{PQ} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}$ and $\overrightarrow{PR} = 3\overrightarrow{PQ}$ . ✓

Worked example 4: where a line crosses an axis

The line through $A (1, 2)$ and $B (4, 4)$ crosses the $x$ -axis at $P (x, 0)$ . Find $x$ .

Solution $\overrightarrow{AB} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$ and $\overrightarrow{AP} = \begin{pmatrix} x - 1 \\ -2 \end{pmatrix}$ . Collinear: $3 \times (-2) - (x - 1) \times 2 = 0$ , so $- 6 - 2 x + 2 = 0$ and $x = - 2$ .

Check: $\overrightarrow{AP} = \begin{pmatrix} -3 \\ -2 \end{pmatrix} = -\overrightarrow{AB}$ , so $P$ is on the line, on the other side of $A$ from $B$ .

Worked example 5: collinear without coordinates

$O$ is the origin, $\overrightarrow{OA} = \vec{a}$ , $\overrightarrow{OB} = \vec{b}$ and $\overrightarrow{OC} = 3\vec{b} - 2\vec{a}$ . Prove that $A$ , $B$ and $C$ are collinear and find $A B : B C$ .

Solution $\overrightarrow{AB} = \vec{b} - \vec{a}$ and $\overrightarrow{AC} = (3\vec{b} - 2\vec{a}) - \vec{a} = 3\vec{b} - 3\vec{a} = 3(\vec{b} - \vec{a}) = 3\overrightarrow{AB}$ .

So $\overrightarrow{AC}$ is a multiple of $\overrightarrow{AB}$ and they share $A$ : the points are collinear. $\overrightarrow{BC} = \overrightarrow{AC} - \overrightarrow{AB} = 2\overrightarrow{AB}$ , so $A B : B C = 1 : 2$ . The figure shows one case, $\vec{a} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}$ and $\vec{b} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$ , which gives $C (- 2, 7)$ .

Warning

Parallel is not yet collinear. If $\overrightarrow{AB} = 2\overrightarrow{CD}$ , the lines $A B$ and $C D$ are parallel, but they may be two different lines. Three points are collinear only when the two vectors share a point, such as $\overrightarrow{AB}$ and $\overrightarrow{AC}$ .

Mind the order in the determinant: $x_{1} y_{2} - x_{2} y_{1}$ , top of the first times bottom of the second, minus top of the second times bottom of the first.

Tip

Before using the determinant, look at the numbers: if one column is an obvious multiple of the other, such as $\begin{pmatrix} 9 \\ 6 \end{pmatrix}$ and $\begin{pmatrix} 3 \\ 2 \end{pmatrix}$ , say so. It is quicker and shows the ratio at once.

My Maths Guy: Showing collinear points using vectors. Watch for the two things every proof must state: one vector is a multiple of the other, and they share a point.

✦

Practice

The last two are harder.

Are $\begin{pmatrix} 4 \\ -6 \end{pmatrix}$ and $\begin{pmatrix} -2 \\ 3 \end{pmatrix}$ collinear? If so, find $k$ with $\begin{pmatrix} -2 \\ 3 \end{pmatrix} = k\begin{pmatrix} 4 \\ -6 \end{pmatrix}$ .
Are $\begin{pmatrix} 5 \\ 2 \end{pmatrix}$ and $\begin{pmatrix} 15 \\ 7 \end{pmatrix}$ collinear?
Find $m$ so that $\begin{pmatrix} m \\ 6 \end{pmatrix}$ and $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$ are collinear.
Are $A (0, - 1)$ , $B (2, 3)$ and $C (5, 9)$ collinear?
Are $P (- 3, 4)$ , $Q (1, 2)$ and $R (6, 0)$ collinear?
Find $t$ so that $A (2, 1)$ , $B (5, t)$ and $C (11, 7)$ are collinear.
Find where the line through $A (- 2, 5)$ and $B (4, 2)$ crosses the $x$ -axis.
$\overrightarrow{OA} = \vec{a}$ , $\overrightarrow{OB} = \vec{b}$ and $\overrightarrow{OC} = 4\vec{b} - 3\vec{a}$ . Show that $A$ , $B$ and $C$ are collinear and find $A B : B C$ .

Show answersHide answers

$4 \times 3 - (-2) \times (-6) = 12 - 12 = 0$ : yes, with $k = -\frac{1}{2}$ .
$5 \times 7 - 15 \times 2 = 35 - 30 = 5 \neq 0$ : no.
$3m - 2 \times 6 = 0$ , so $m = 4$ .
$\overrightarrow{AB} = \begin{pmatrix} 2 \\ 4 \end{pmatrix}$ , $\overrightarrow{AC} = \begin{pmatrix} 5 \\ 10 \end{pmatrix} = \frac{5}{2}\overrightarrow{AB}$ : yes.
$\overrightarrow{PQ} = \begin{pmatrix} 4 \\ -2 \end{pmatrix}$ , $\overrightarrow{PR} = \begin{pmatrix} 9 \\ -4 \end{pmatrix}$ : $4 \times (-4) - 9 \times (-2) = 2 \neq 0$ , so no.
$\overrightarrow{AB} = \begin{pmatrix} 3 \\ t - 1 \end{pmatrix}$ , $\overrightarrow{AC} = \begin{pmatrix} 9 \\ 6 \end{pmatrix}$ : $18 - 9 (t - 1) = 0$ , so $t = 3$ .
$\overrightarrow{AB} = \begin{pmatrix} 6 \\ -3 \end{pmatrix}$ , $\overrightarrow{AP} = \begin{pmatrix} x + 2 \\ -5 \end{pmatrix}$ : $6 \times (-5) - (x + 2)(-3) = 0$ , so $3 x - 24 = 0$ , $x = 8$ . The point is $(8, 0)$ .
$\overrightarrow{AC} = 4\vec{b} - 4\vec{a} = 4\overrightarrow{AB}$ , with $A$ shared: collinear. $\overrightarrow{BC} = 3\overrightarrow{AB}$ , so $A B : B C = 1 : 3$ .

Sources and credits

Written by Ganzaa.
Video: Showing Collinear Points Using Vectors, My Maths Guy, embedded from YouTube.

The diagrams, worked examples and practice items were written by Ganzaa; the site plan is illustrative.

1Is the tap on the pipe line?

2Vectors that point along one line

Is the tap on the pipe line?

Vectors that point along one line