# Find if three points in 3-dimensional space are collinear

Find if the points joining $A=(6,7,1), B=(2,-3,1)$ and $C=(4,-5,0)$ are collinear.

How to determine collinearity in three dimensions? In two dimensions, one can compare the slopes of segments $AB$ and $BC$: if they are equal, $ABC$ are collinear. This doesn't work in 3D.

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Hint If the points lie on a straight line, then the slope between any two of the points will be the same. –  Daryl Oct 7 '12 at 6:08
Thank You Daryl –  sundar nataraj Oct 7 '12 at 6:15
No, the line through $A$ and $B$ has $z=1$. –  Yves Daoust Mar 14 at 13:35

Method 1:

Point $A$ and point $B$ ($A \ne B$) determine a line. You can find its equation. See if the coordinates of point C fits the equation. If so, A B and C are colinear, or else, no.

Method 2:

Point $A$, $B$ and $C$ determine two vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$. Suppose the latter isn't zero vector, see if there is a constant $\lambda$ that allows $\overrightarrow{AB}=\lambda \overrightarrow{AC}$.

Other properties if $A$, $B$ and $C$ are colinear:

$$\left| \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{\left|\overrightarrow{AB}\right|\cdot\left|\overrightarrow{AC}\right|} \right| = 0$$:\

$$\overrightarrow{AB}\times\overrightarrow{AC} = \overrightarrow{0}$$

Also, two ways to write the equation of a line in 3D:

$$\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$$

where $(x_0,y_0,z_0)$ is a point on the line and $(a,b,c)$ is the direction vector of the line, provided that $abc\ne 0$.

\begin{align} x&=x_0+at,\\ y&=y_0+bt,\\ z&=z_0+ct. \end{align}

All that remains is calculation.

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3rd co-ordinate of first two point says that line lies in z=1. But 3rd point has z-cord=0.
So, given points are not co-linear.

@Sundar:
How you define slope in 3D?

Please correct me it I'm wrong.

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In 3D, points are collinear if $\Delta z=(a\Delta x+b\Delta y)t$,where $a$ and $b$ are some parameters and $t$ maps out the line through 3D space. E.g $(0,0,0),\,(1,1,1),\,(2,2,2)$. –  Daryl Oct 7 '12 at 7:23

We have $\overrightarrow{AB}=(-4,-10,0)$ and $\overrightarrow{AC}=(-2,-12,-1)$ . Therefore cross product of two vectors AB and AC is $\overrightarrow{AB}\times\overrightarrow{AC}=(10, -4, 28)$ . This vector is different from vector $(0,0,0)$. So, the given points are not co-linear.

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If $A,B,C$ are collinear, then $AB$ and $AC$ are proportional.

$$(-4,-10,0)=\lambda(-2,-12,-1).$$

Obviously, no.

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the points $A$ and $B$ are on the plane $z = 1.$ that implies the line $AB$ is on the plane $z = 0.$ that is all points on the line $AB$ has $z =1$ the $z$-coordinate of the point $C$ is zero, therefore it is not on the line $AB.$

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Hint $A,B,C$ are colinear if and only if the largest of the lenghts of $AB, AC, BC$ is equal to the sum of the other two.

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can you explain why this is so? –  tmsimont Jun 12 at 2:33
@tmsimont $\Rightarrow$ is easy: the point in the middle breaks the long segment into two small pieces.... $\Leftarrow$ Assume by contradiction they are not colinear, then they form a triangle and the condition contradicts the triangle inequality.... –  N. S. Jun 12 at 2:49