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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 '15 at 13:35
up vote 4 down vote accepted

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:


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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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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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 '15 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 '15 at 2:49

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.

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

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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Here's a nice degree four polynomial predicate to test collinearity of three points in any dimension $d$. Let $a = (a_1,a_2,\ldots,a_d)$, $b = (b_1,b_2,\ldots,b_d)$, and $c=(c_1,c_2,\ldots,c_d)$ be 3 points in $\mathbb{R}^d$. The points $a$, $b$, and $c$ are collinear if and only if $c = a + \lambda (b-a)$ for some unique $\lambda \in \mathbb{R}$ that is $$ (a-c) + \lambda (b-a) = 0\\ (a_i-c_i) + \lambda (b_i-a_i) = 0, \forall i \in [d]\\ \sum_{i=1}^{d}\left[(a_i-c_i) + \lambda (b_i-a_i)\right]^2 = 0\\ \sum_{i=1}^{d}\left[(b_i-a_i)^2 \lambda^2 + 2(a_i-c_i)(b_i-a_i) \lambda + (a_i-c_i)^2\right] = 0\\ \underbrace{\left[\sum_{i=1}^{d}(b_i-a_i)^2\right]}_{A} \lambda^2 + \underbrace{\left[2\sum_{i=1}^{d}(a_i-c_i)(b_i-a_i)\right]}_{B} \lambda + \underbrace{\left[\sum_{i=1}^{d}(a_i-c_i)^2\right]}_{C} = 0\\ \lambda = \frac{-B \pm \sqrt{B^2-4AC}}{2A} $$

For $\lambda$ to exist and be unique $B^2-4AC$ must be zero. Hence, $a$, $b$, and $c$ are collinear if and only if

$$ \left[2\sum_{i=1}^{d}(a_i-c_i)(b_i-a_i)\right]^2 - 4 \left[\sum_{i=1}^{d}(a_i-c_i)^2\right]\left[\sum_{i=1}^{d}(b_i-a_i)^2\right] = 0 $$

Note that

  • $A=0$ if and only if $a=b$
  • if $A=0$ then $B=0$
  • $B^2 - 4AC$ is never positive
  • $B^2 - 4AC$ is always 0 for $d\in\{\,0,1\,\}$
  • $\frac 14 \sqrt{4AC - B^2}$ is the area of the triangle $abc$.
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If $A,B,C$ are collinear, then $AB$ and $AC$ are proportional.


Obviously, no.

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