Determine whether the points lie on a straight line. Determine whether the points $A(2, 6, 2)$, $B(3, 10, 0)$, $C(1, 4, 3)$ lie on a straight line.
Is there a formula to solve this question? What is it?
 A: Yes, there is, use that $$\vec{x}=\vec{x_0}+t\vec{a}$$ where $t$ is a real number.
And the equation of the straight line is given by $$\vec{x}=[2;6;2]+t[1;4;-2]$$ so we get
$$[1;4;3]=[2;6;2]+t[1;4;-2]$$ and now compute $$t$$
$$1=2+t$$
$$4=6+4t$$ so $$t=-1$$ and $$t=-\frac{1}{2}$$ and we get no solution.
A: If they lie on a straight line, then $AC$ should be a multiple of $AB$; that is, $AC = k\cdot AB$ for some $k$.
We have
$$AB = (3,10,0) - (2,6,2) = (1,4,2)$$
and
$$AC = (1,4,3) - (2,6,2) = (-1,-2,1)$$
There is no $k$ such that $(-1,-2,1) = k(1,4,2)$, so the points do not lie on a line.
A: Consider the vectors
$$
\overset{\rightharpoonup}{AB} \quad \text{and}\quad\overset{\rightharpoonup}{AC}.
$$
That is, the vector that takes you front point $A$ to point $B$, and the one that takes you from point $A$ to point $C$. Naturally, if $A$, $B$, and $C$ are all colinear, $\overset{\rightharpoonup}{AB}$ and $\overset{\rightharpoonup}{AC}$ will be parallel. One way to test this is to exam the unit vectors in the directions of 
$\overset{\rightharpoonup}{AB}$ and $\overset{\rightharpoonup}{AC}$:
$$
u_1 = \frac{\overset{\rightharpoonup}{AB}}{\| \overset{\rightharpoonup}{AB}\|} \quad \text{and}\quad u_2 =\frac{\overset{\rightharpoonup}{AC}}{\| \overset{\rightharpoonup}{AC}\|}.
$$
If $\overset{\rightharpoonup}{AB}$ and $\overset{\rightharpoonup}{AC}$ are parallel, then either $u_1 = u_2$ or $u_1 = -u_2$. This is equivalent to testing whether $\overset{\rightharpoonup}{AB}$ and $\overset{\rightharpoonup}{AC}$ can be expressed as linear combinations of one another, i.e. $\overset{\rightharpoonup}{AB}= c\overset{\rightharpoonup}{AC}$ for some scalar $c$.
You could also exam the projections of these vectors onto one anther. If $\overset{\rightharpoonup}{AB}$ and $\overset{\rightharpoonup}{AC}$, then either
$$
u_1 \cdot u_2 = 1 \quad \text{or} \quad u_1 \cdot u_2 = -1.
$$
This is, of course, just a different way to express the same relationship.
A: As mentioned in another answer, the three points are colinear iff $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are linearly dependent, i.e., one is a scalar multiple of the other. In three dimensions, the cross product of two vectors vanishes iff they are linearly dependent, so compute $(B-A)\times(C-A)$ and see if this is the zero vector.  
Equivalently, the determinant of the matrix $[A,B,C]$ is equal to the volume of the paralellepiped with edges $\overline{OA}$, $\overline{OB}$ and $\overline{OC}$. If the three points are colinear, this paralellepiped “collapses” and has zero volume—the above determinant vanishes.
