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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 prove that? Anyone, please help me!

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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 Сундар Oct 7 '12 at 6:15

3 Answers 3

up vote 2 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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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

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