# How to tell if a point is between two others in $\mathbb{R}^3$

Find the coordinates $c$ of the point $C$ $st$ $C$ is on the line from $A (2, -3, 1)$ to $B (8, 9, -5)$, it is between A and B, and $\vec{AC} = 2\vec{CB}$.

So, I found the vector $\vec{AB}$ to be $(6, 12, -6)$ and from the equality, I set up the following:

$\left(\begin{array}{cc} c_1 - 2\\ c_2 +3\\ c_3 - 1\\ \end{array}\right)= \left(\begin{array}{cc} 16 - 2c_1\\ 18 - 2c_2\\ -10-2c_3\\ \end{array}\right)$

to get $c_1 = 6$, $c_2 = 7$, $c_3 = 3$

but I wasn't sure how to tell if $(6, 7, 3)$ was in between A and B and I also had no use for $\vec{AB}$ which seemed strange.

How do I solve this problem? and how do I know if a point is between two others generally in dimensions greater than $\mathbb{R}^2$?

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Your solution would have been fine if you’d solved correctly for $c_1,c_2$, and $c_3$: you should have got $c_2=5$ and $c_3=-3$.

Yours is the more straightforward approach, but here’s another way to think about it.

Points on the line through $A$ and $B$ are precisely those that you can reach by going from the origin to $A$ and then some distance, possibly negative, in the direction $\overrightarrow{AB}$. In other words, they correspond to vectors of the form $\vec a+\lambda\vec v$, where $\vec a=\langle 2,-3,1\rangle$ and $\vec v=\langle 6,12,-6\rangle$. You want the point that is two-thirds of the way from $A$ to $B$, so you want $$\vec a+\frac23\vec v=\langle 2,-3,1\rangle+\frac23\langle 6,12,-6\rangle=\langle 6,5,-3\rangle\;.$$ Thus, $C$ should be the point $\langle 6,5,-3\rangle$.

As a check, $|AB|=\sqrt{6^2+12^2+(-6)^2}=\sqrt{216}=6\sqrt6$, $|AC|=\sqrt{4^2+8^2+(-4)^2}=\sqrt{96}=4\sqrt6$, and $CB=\sqrt{2^2+4^2+(-2)^2}=\sqrt{24}=2\sqrt6$, exactly as desired.

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If I used my method, assuming I don't make any mistakes next time, would I ever come across a situation where the point satisfies the equality, but is not on the line AB? i.e. would it be prudent to always proceed with your check? –  stariz77 May 21 '12 at 9:11
@stariz77: No, you’d be fine. You solved $\vec v-\vec a=2\vec b-\vec v$, which is equivalent to $3\vec v=2\vec b+\vec a$, or $\vec v=\frac23\vec b+\frac13\vec a$. A combination $\lambda\vec a+(1-\lambda)\vec b$ always gives you a point on the line through $A$ and $B$, and it’s on the segment $\overline{AB}$ if and only if $0\le\lambda\le 1$. In that case it’s a kind of weighted average of $\vec a$ and $vec b$: here $\vec b$ gets $2/3$ of the weight, and $\vec a$ only $1/3$, so $\vec v$ ends up twice as close to $B$ as to $A$. –  Brian M. Scott May 21 '12 at 9:27
+1. I usually find it helpful to write the parameterization explicitly as $A + \lambda (B-A)$. That way, one can see easily that $\lambda=0$ gives $A + 0 = A$, the start point; and that $\lambda=1$ gives $A+(B-A)=B$, the end point; and that $\lambda = 1/2$ gives $A+(1/2)(B-A) = (A+B)/2$, the midpoint. This reinforces the intuition that, say, $\lambda = 1/3$ and $\lambda=2/3$ give the "third point" and "two-third point", and that $\lambda$ in general is just a ratio. –  Blue May 21 '12 at 9:29

The points $C$ between $A$ and $B$ can be parametrized by $$C=At+B(1-t)\tag{1}$$ where $0< t< 1$

Therefore, we wish to solve for $t\in(0,1)$ so that \begin{align} |C-A|&=2|C-B|\\ |At+B(1-t)-A|&=2|At+B(1-t)-B|\\ |(B-A)(1-t)|&=2|(A-B)t|\\ 1-t&=2t\\ &t=\tfrac13\tag{2} \end{align} Thus, plugging $(2)$ into $(1)$ yields \begin{align} C &=\tfrac13A+\tfrac23B\\ &=\tfrac13(2, -3, 1)+\tfrac23(8, 9, -5)\\ &=(6,5,-3)\tag{3} \end{align}

Determining if $C$ is between $A$ and $B$:

In general, $C$ is between $A$ and $B$ when $$|B-A|=|B-C|+|C-A|\tag{4}$$ Equation $(4)$ is the extreme case of the triangle inequality, which says $$|B-A|\le|B-C|+|C-A|\tag{5}$$

Check:

Let's check if $C=(6, 7, 3)$ is between $A=(2, -3, 1)$ and $B=(8, 9, -5)$: $$|A-B|=|(-6,-12,6)|=6\sqrt{6}$$ but $$|A-C|+|C-B|=|(-4,-10,-2)|+|(-2,-2,8)|=2\sqrt{30}+6\sqrt{2}$$ Numerically, $6\sqrt{6}\approx14.70$ and $2\sqrt{30}+6\sqrt{2}\approx19.44$. Thus, $$(6, 7, 3)\text{ is not between }(2, -3, 1)\text{ and }(8, 9, -5)\tag{6}$$ Let's check if $C=(6, 5, -3)$ is between $A=(2, -3, 1)$ and $B=(8, 9, -5)$: $$|A-B|=|(-6,-12,6)|=6\sqrt{6}$$ and $$|A-C|+|C-B|=|(-4,-8,4)|+|(-2,-4,2)|=4\sqrt{6}+2\sqrt{6}=6\sqrt{6}$$ Thus, $$(6, 5, -3)\text{ is between }(2, -3, 1)\text{ and }(8, 9, -5)\tag{7}$$

Extension:

Given this question, one might wonder what is the set of all points $C$ so that $$|C-A|=2|C-B|\tag{8}$$ Squaring $(8)$ yields $$C\cdot C-2A\cdot C+A\cdot A=4C\cdot C-8B\cdot C+4B\cdot B$$ $$0=3C\cdot C-2(4B-A)\cdot C+4B\cdot B-A\cdot A$$ Therefore, \begin{align} 0 &=C\cdot C-2\left(\tfrac43B-\tfrac13A\right)\cdot C+\tfrac43B\cdot B-\tfrac13A\cdot A\\ &=\left|C-\left(\tfrac43B-\tfrac13A\right)\right|^2+\left(\tfrac43B\cdot B-\tfrac13A\cdot A\right)-\left(\tfrac43B-\tfrac13A\right)\cdot\left(\tfrac43B-\tfrac13A\right)\\ &=\left|C-\left(\tfrac43B-\tfrac13A\right)\right|^2-\tfrac49|A-B|^2\tag{9} \end{align} So the set of points that satisfy $(8)$ is the sphere centered at $\tfrac43B-\tfrac13A$ with radius $\frac23|A-B|$.

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