# How do I determine if 3 vectors are collinear?

I only know how to show that $$2$$ vectors are collinear, but for $$3$$ vectors I only know how to prove coplanarity.

• if vectors are 2 by 2 collinear than they are all three collinear. Jan 12, 2014 at 15:31
• Ah, yes, of course. For the front two and back to two share a line, this must mean both the first and third vector are collinear with the middle vector. Which only leaves the option that they are all 3 lying along the same line. Thanks! Jan 12, 2014 at 15:41
• @Threethumb Watch out! What about $[1,0,0], [0,0,0], [0,1,0]$? The first two are collinear, so are the last two, but not all three of them! Jan 12, 2014 at 15:54
• Oh, that's a good point. But something like [2, 4, 6], [4, 8, 12] and [8, 16, 24] would be collinear, right? I just wrote a random first vector, then a second with every component being twice as big, and did the same for the third. Jan 12, 2014 at 19:42
• @HagenvonEitzen this occurs mainly due to the unique property of the null vector to be collinear with any vector... right? Oct 8, 2022 at 5:07

A similar problem is the determining if three points are collinear within a plane.

Given points a, b and c form the line segments ab, bc and ac. If ab + bc = ac then the three points are collinear.

The line segments can be translated to vectors ab, bc and ac where the magnitude of the vectors are equal to the length of the respective line segments mentioned.

By example of the points you've given in response to Naveen.

a(2, 4, 6) b(4, 8, 12) c(8, 16, 24)

$$\overline{ab} = \sqrt[]{56}$$ $$\overline{bc} = \sqrt[]{224}$$ $$\overline{ac} = \sqrt[]{504}$$

$$\overline{ab} + \overline{bc} = \overline{ac}$$

• You can use matrix also, | 2 4 8 | A = | 4 8 16 | | 6 12 24 | If determinant of A is zero, then it is collinear, else, not. Refer wolframalpha.com/input/… Mar 31, 2017 at 5:50

For any vectors x,y,z $||x-y||\leq ||x-z||+||y-z||$ and equality hold iff all three are collinear.

• So [2, 4, 6], [4, 8, 12] and [8, 16, 24] would be collinear? Jan 12, 2014 at 15:52

Ok see here..... Since 'use vectors to' is pretty vauge, you can do this a bunch of ways.

For example, you could use the dot product ((b⃗ −c⃗ )⋅(a⃗ −b⃗ )) / (|b⃗ −c⃗ | * |a⃗ −b⃗ |) = ± 1

or the cross product (b⃗ − c⃗ ) × (a⃗ −b⃗ )=0⃗

Actually you may proceed in a general co-ordinate geometry process also...... let us consider that the 3 vectors if we poit them into co-ordinate system we will get 3 points.....

let us consider they are: (a,b) , (c,d) , (e,f).. So, you will have 3 points on the co-ordinates......ok.

now you, find the equations of the straight line joining the points (a,b) and (c,d) and do the same for the points (c,d) , (e,f) ----- if these 3 points are co-linear then you will get the same equation of straight line!!

OR, You can proceed in this way: first measure the SLOPE of the line joining the points (a,b) and (e,f) that is : (f - b) / (e - a) .

and then measure the SLOPE of the line joining the points (a,b) and (c,d) that is : (d - b) / (c - a) .

          If the pints  A(a,b) , B(c,d) , C(e,f) are co-linear then the measured slope will be equal


OR, You can find the equation of the st. line joining the points A and B and then you can find the equation of the st. line joining the points B and C then you find the angle between AB st.line and BC st.line...by applying the co-ordinate geometry formula (that you can easily get from your text book) and see whether the angle between the straight lines is 180deg. or not....if it is 180deg. you can easily say that the points A , B , C are co-linear.......

• As in question vectors has been mentioned, it is a specific and the relevant answer to this question. In other way we can say, if three vectors a, b, c are Collinear, the area of the triangle constructed by those vectors will become zero. i.e. the cross product (b - a) × (c - a) = 0. Trying to find the determinant value i.e. scalar triple product [a b c] to be zero will lead to satisfy Coplanarity NOT Collinearity. Feb 10, 2020 at 16:01