three points are appearing to be on a straight line in R3 when they shouldn't be?

Question

Im given the problem:

9. Determine whether the points lie on a straight line.

(b) K(0,3,-4), L(1,2,-2), M(3,0,1)

I take the distance between KL, LM, and KM and every single time KL + LM = KM.

2.5 + 4.1 = 6.6

I use the distance formula and get for KL square root( (1-0)^2 + (1-0)^2 + (1-0)^2) = root(6) = 2.449

I use the distance formula and get for LM square root( (3-1)^2 + (0-2)^2 + (1--2)^2) = root(17) = 4.123

I use the distance formula and get for KM square root( (0-3)^2 + (3-0)^2 + (-4-1)^2) = root(43) = 6.557

KL + LM = KM.

According to the answers the points do NOT lie on a straight line.

After researching how to do this problem I was told if one vector is equal to the sum of the other two vectors, they lie on the same line.

Can someone please tell me if either I am doing my math wrong, I do not actually understand this problem completely, or if I am going insane? I don't even need to answer this problem but its driving me crazy not knowing why I'm getting the unexpected answer.

any help is appreciated, thank you.

Answer 1

If you’d used enough decimal places, then you wouldn’t have run into this (at least, not for this problem). If you take $\sqrt6\approx2.44949$, $\sqrt{17}\approx4.12311$ and $\sqrt{43}\approx6.55744$, then you have $$2.44949+4.12311=6.5726\ne6.55744$$ As JMoravitz comments, this is a good reminder to work with exact values, or at the very least with better approximations to them.

There’s another way to determine whether or not these three points in $\mathbb R^3$ are colinear besides the ones already mentioned in other answers: compute the scalar triple product $(0,3,-4)\times(1,2,-2)\cdot(3,0,1)$, which can also be written as the determinant $$\begin{vmatrix}0&3&-4\\1&2&-2\\3&0&1\end{vmatrix}=3.$$ This is the (signed) volume of a paralellepiped with edges defined by the three vectors, and will be zero if those vertices are colinear—the solid collapses into a planar figure.

Answer 2

If 3 points $K,L,M$ lie on the same straight line, then $K - L$ and $K-M$ would be the multiples of the same vector. Note that $$ K-L = (0,3,-4)-(1,2,-2) = (-1, 1, -2) $$ and $$ K-M = (0,3,-4)-(3,0,1) = (-3, 3, -5) $$ and clearly, those 2 vectors are not scales of each other, so the 3 points are not collinear.

Answer 3

Test if slope(K, L) = slope(L, M).

If yes, they are on the same straight line.

Remark: The slope formula is much easier to use than the distance formula.

Answer 4

Use vectors:

Straight line in parameter form passing through $K$ and $L$:

$\vec r = (0,3,-4) + t(1,-1,2)$;

Does not pass through $M(3,0,1)$.