Determine even if it is the clues $A(2, 6, 2)$, $B(3, 10, 0)$, $C(1, 4, 3)$ lie on a directly line.

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Is there a formula to solve this question? What is it?

Yes, there is, use that $$\vecx=\vecx_0+t\veca$$ wherein $t$ is a genuine number.And the equation the the directly line is given by $$\vecx=<2;6;2>+t<1;4;-2>$$ so us get$$<1;4;3>=<2;6;2>+t<1;4;-2>$$ and also now compute $$t$$

$$1=2+t$$$$4=6+4t$$ therefore $$t=-1$$ and $$t=-\frac12$$ and we obtain no solution.

If they lie ~ above a straight line, then $AC$ must be a multiple of $AB$; the 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 $(-1,-2,1) = k(1,4,2)$, for this reason the points execute not lied on a line.

Consider the vectors$$\overset\rightharpoonupAB \quad \textand\quad\overset\rightharpoonupAC.$$That is, the vector the takes girlfriend front allude $A$ to point $B$, and also the one the takes girlfriend from allude $A$ to suggest $C$. Naturally, if $A$, $B$, and $C$ room all colinear, $\overset\rightharpoonupAB$ and also $\overset\rightharpoonupAC$ will certainly be parallel. One method to test this is to exam the unit vectors in the direction of $\overset\rightharpoonupAB$ and $\overset\rightharpoonupAC$:

$$u_1 = \frac\overset\rightharpoonupAB \overset\rightharpoonupAB\ \quad \textand\quad u_2 =\frac\overset\rightharpoonupAC \overset\rightharpoonupAC\.$$

If $\overset\rightharpoonupAB$ and also $\overset\rightharpoonupAC$ room parallel, then either $u_1 = u_2$ or $u_1 = -u_2$. This is indistinguishable to experimentation whether $\overset\rightharpoonupAB$ and $\overset\rightharpoonupAC$ can be to express as linear combinations the one another, i.e. $\overset\rightharpoonupAB= c\overset\rightharpoonupAC$ for some scalar $c$.

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You could also exam the projections of these vectors top top one anther. If $\overset\rightharpoonupAB$ and also $\overset\rightharpoonupAC$, climate either$$u_1 \cdot u_2 = 1 \quad \textor \quad u_1 \cdot u_2 = -1.$$This is, that course, just a different means to refer the very same relationship.