Bitesize has changed! We're updating subjects as fast as we can. Visit our new site to find Bitesize guides and clips - and tell us what you think!

Home > Maths > Geometry > Vectors



Collinear points

When you're working in three dimensions, the only way to prove that three points are in a line (collinear) involves showing that a common direction exists. For this, you need to use vectors.

You'll need to work through all the steps shown below if you want to gain full marks.

Here's how you would show that A(4, 1, 3), B(8, 4, 6) and C(20, 13, 15) are collinear.

First, choose two directed line segments with a common point

\overrightarrow {AB} = \left( \matrix{ 4 \cr 3 \cr 3 \cr} \right),\;\overrightarrow {BC} = \left( \matrix{ 12 \cr 9 \cr 9 \cr} \right)

Express one as a multiple of the other

\overrightarrow {BC} = 3\left( \matrix{ 4 \cr 3 \cr 3 \cr} \right) i.e. \overrightarrow {BC} = 3 \times \overrightarrow {AB}

and state conclusion

so \overrightarrow {AB} and \overrightarrow {BC} have a common direction.

Complete the proof

\overrightarrow {AB} and \overrightarrow {BC} have a common point. Therefore A, B and C are collinear.


Show that P(0, 2, 4), Q(10, 0, 0) and R(5, 1, 2) are collinear.


\overrightarrow {PR} = \left( \matrix{ 5 \cr - 1 \cr - 2 \cr} \right),\;\overrightarrow {QR} = \left( \matrix{ - 5 \cr 1 \cr 2 \cr} \right)

\overrightarrow {QR} = - 1\left( \matrix{ 5 \cr - 1 \cr - 2 \cr} \right) i.e. \overrightarrow {QR} = - 1 \times \overrightarrow {PR}

so \overrightarrow {QR} and \overrightarrow {PR} have a common direction

\overrightarrow {QR} and \overrightarrow {PR} have a common point R. Therefore P, Q and R are collinear.

When you have 3 collinear points you're often asked in what ratio one point divides the directed line segment. In the simpler case, in what ratio does the (middle) point divide the directed line segment?

Look again at the ABC example above. In what ratio does B divide AC? Make life easy for yourself by choosing two directed line segments with a common point - in this case, that's B.

then {{\overrightarrow {AB} } \over {\overrightarrow {BC} }} = {{\overrightarrow {AB} } \over {3 \overrightarrow {AB} }} = {1 \over 3}

i.e. AB:BC = 1:3

BBC © 2014 The BBC is not responsible for the content of external sites. Read more.

This page is best viewed in an up-to-date web browser with style sheets (CSS) enabled. While you will be able to view the content of this page in your current browser, you will not be able to get the full visual experience. Please consider upgrading your browser software or enabling style sheets (CSS) if you are able to do so.