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?
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$\begingroup$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.
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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.
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