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.
On
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.
On
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.
On
Three non-collinear points in space form a triangle. The position vector of each vertex is given by $\vec{A}=(6,7,1)$, $\vec{B}=(2,−3,1)$ and $\vec{C}=(4,−5,0)$. You can calculate the normal vector at each vertex, lets take vertex $\vec{A}$, for example:
$$ \vec{N} = (\vec{B} - \vec{A}) \times (\vec{C} - \vec{A}) $$
If the absolute value of the normal is zero, then the three points are aligned in space. You can convince yourself of this by looking at the expression for the module of a vector:
$$ |\vec{N}| = |(\vec{B} - \vec{A}) \times (\vec{C} - \vec{A})| = |(\vec{B} - \vec{A})| \cdot |(\vec{C} - \vec{A})| \cdot sin{\space\theta} $$
where $\theta$ is the angle between the vectors $(\vec{B} - \vec{A})$ and $(\vec{C} - \vec{A})$, i.e. the angle of vertex $\vec{A}$).
This expression will be zero if $\theta = 0, \pi$. If, lets say, your points are aligned in the following order: $B - A - C$, you can see that the vertex angle at A is $\pi$ (180). If you calculated the normal at B or C then the angle would be zero. Either way, the normal module is zero.
On
Three given points $A$, $B$ and $C$ are collinear, when sum of any two distances out of $AB$, $BC$, $CA$ is equal to the remaining third otherwise the points will be the vertices of a triangle.
Find if the points joining $A=(6,7,1), B=(2,-3,1)$ and $C=(4,-5,0)$ are collinear.
D = √[(x₂ - x₁)² + (y₂ - y₁)² +(z₂ - z₁)²]
AB = √[(2 - 6)² + (-3 - 7)² +(1 - 1)²] = 10.77033
BC = √[(4 - 2)² + (-5 - -3)² +(0 - 1)²] = 3
CA = √[(6 - 4)² + (7 - -5)² +(1 - 0)²] = 12.20656
Clearly, $AB + BC \neq AC$, the given points are NOT collinear. $ABC$ is a triangle.
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:
$$\left| \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{\left|\overrightarrow{AB}\right|\cdot\left|\overrightarrow{AC}\right|} \right| = 1$$:\
$$\overrightarrow{AB}\times\overrightarrow{AC} = \overrightarrow{0}$$
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.