Dividing a Segment: Get the Order Right!

A series of recent questions dealt with proportional division of a line segment. The context was vectors, and we’ll use them a lot, though the main ideas can be understood using ordinary geometry. We’ll see a mistake so easy to make that AI did it just as humans do; and how textbooks can make it easy to get it wrong.

Dividing a segment in 1:2 ratio

The first of these questions from our regular correspondent Amia came in mid-April:

Hi Dr math,

If I have a line segment in the space, and the coordinates of the beginning and the end are given, can I find the coordinates of a point that lies on the segment (such that it divides the segment to 1:2)??

Clearly there was some context to this question, so I asked for it:

Yes, you can.

I think you know how; is there a reason you haven’t yet done it?

I’d scale the vector from one point to the other. It can also be treated as a weighted average.

We’ll see both methods. For now, Amia showed the context, including a solution that goes beyond the specific question:

I have this question, and I am stuck.

How can I find N?!

Part (1) can be answered with a simple formula, which is presumably what Amia did: The midpoint M is found by just averaging the endpoints: $$M=\left(\frac{-1+-3}{2},\frac{-2+4}{2},\frac{1+-5}{2}\right)=\left(-2,1,-2\right)$$

This could have been solved using vectors, instead: If M is the midpoint of AB, then $$\overrightarrow{AM}=\frac{1}{2}\overrightarrow{AB}=\frac{1}{2}\left<(-3)-(-1),4-(-2),(-5)-1\right>=\frac{1}{2}\left<-2,6,-6\right>=\left<-1,3,-3\right>$$ so $$M=A+\overrightarrow{AM}=(-1,-2,1)+\left<-1,3,-3\right>=(-2,1,-2)$$

Part (2) generalizes this idea to find N, and then asks for the vector equation of line MN, which we’ll get to in its place.

Amia’s approach here used vectors. First, $$\overrightarrow{BC}=C-B=\left<0-(-3),(-2)-4,4-(-5)\right>=\left<3,-6,9\right>.$$ Then, if \(\overrightarrow{NC}=2\overrightarrow{BN}\), then $$\overrightarrow{NC}=\frac{2}{3}\overrightarrow{BC}=\left<2,-4,6\right>.$$

Then he subtracted, but missed a step by writing it as \(N=\overrightarrow{BC}-\overrightarrow{NC}=\left<1,-2,3\right>\).

I corrected this vector approach, and showed my formula approaches, too:

Only the last line is wrong. BC – NC = BN, not (the point) N. (I found BN as 1/3 BC, and got the same result.)

Now, N = B + BN = (-3, 4, -5) + <1, -2, 3> = (-2, 2, -2). (That is, the position vector of N is the position vector of B + the vector from B to N, ON = OB + BN.)

I mentioned using a weighted average; we could calculate N, which divides BC in the ratio 1:2, as

2/3 B + 1/3 C = 2/3 (-3, 4, -5) + 1/3 (0, -2, 4) = (-2+0, 8/3-2/3, -10/3+4/3) = (-2, 2, -2)

But you’d be expected to use vectors in your context, as you did.

Both methods give the same answer, \(N=(-2,2,-2)\).

Amia showed more of his work, using vectors:

Thank you, Dr Peterson,

Here is my work:

He has done what I suggested, finding \(\overrightarrow{BN}\) directly from \(\overrightarrow{BC}\), and then solved \(\overrightarrow{BN}=N-B\) to obtain $$N=B+\overrightarrow{BN}=(-3,4,-5)+\left<1,-2,3\right>=(-2,2,-2).$$

Then he found the equation of \(\overline{MN}\), by finding vector $$\overrightarrow{MN}=N-M=(-2,2,-2)-(-2,1,-2)=(0,1,0),$$ and writing the equation $$\vec{r}=\vec{r_1}+t\vec{v}=M+t\overrightarrow{MN}=\left<-2,1,-2\right>+t\left<0,1,0\right>=\left<-2,t+1,-2\right>.$$

You might notice that when \(t=0\), \(\vec{r}=\left<-2,0+1,-2\right>=\left<-2,1,-2\right>=M\), and when \(t=1\), \(\vec{r}=\left<-2,1+1,-2\right>=\left<-2,2,-2\right>=N\). This is a good check, and also suggests an alternative approach to these problems.

The weighted average method

Amia then asked about the weighted average, showing his attempt:

How do we get

2/3 B + 1/3 C = 2/3 (-3, 4, -5) + 1/3 (0, -2, 4) = (-2+0, 8/3-2/3, -10/3+4/3) = (-2, 2, -2) ?

He has almost the right idea for the weighted average; my work started with the second form he suggests, namely $$\frac{2}{3}B+\frac{1}{3}C=\frac{2}{3}B+\frac{1}{3}C=\frac{2}{3}(-3,4,-5)+\frac{1}{3}(0,-2,4)\\=\left(-2+0,\frac{8}{3}-\frac{2}{3},-\frac{10}{3}+\frac{4}{3}\right) = (-2, 2, -2)$$

So what is the formula, and how do you know the weights to use?

I looked for sources for the formula I used, and passed one on:

Here is an explanation of the weighted average method (among others).

That expresses it this way:

This shows the coefficient (weight) of the first point As the second term in the ratio (corresponding to the second part of the length) over the total, and gives a different way to remember this: The nearer point has the greater weight.

Here is how I think about it, pushing that idea further to make it feel almost obvious: A literal “weighted” average consists of two objects with different weights, which will balance at a point between them:

The balance point clearly will be closer to the heavier object; so the ratio of distances is opposite to the ratio of weights: $$\frac{{\color{Red}{d_1}}}{{\color{DarkGreen}{d_2}}}=\frac{{\color{DarkGreen}{W_2}}}{{\color{Red}{W_1}}}$$

This makes it almost impossible for me to forget!

And a physicist will see this as $${\color{Red}{d_1}}{\color{Red}{W_1}}={\color{DarkGreen}{d_2}}{\color{DarkGreen}{W_2}},$$ that is, the moments of the two weights about the balance point are e equal.

Note that the terms of the proportion are reversed in the weighting; the nearer point has the higher weight.

As I said,

we could calculate N, which divides BC in the ratio 1:2, as

2/3 B + 1/3 C = 2/3 (-3, 4, -5) + 1/3 (0, -2, 4) = (-2+0, 8/3-2/3, -10/3+4/3) = (-2, 2, -2)

The 1 that starts the ratio goes with the second point, not the first.

The formula can be derived using the method you used.

Here is such a derivation:

Given points \(A=(x_1,y_1,z_1)\) and \(B=(x_2,y_2,z_2)\), we want to find a point \(M\) that divides the segment \(\overline{AB}\) in the ratio m:n. So we have $$\overrightarrow{AB}=B-A=\left<x_2-x_1,y_2-y_1,z_2-z_1\right>,$$ and we want \(\left|\overrightarrow{AM}\right|:\left|\overrightarrow{MB}\right|=m:n\). Therefore, since the vectors are parallel (and in the same direction), \(m\overrightarrow{MB}=n\overrightarrow{AM}\), and \(\overrightarrow{MB}=\frac{n}{m}\overrightarrow{AM}\).

But then $$\overrightarrow{AB}=\overrightarrow{AM}+\overrightarrow{MB}=\overrightarrow{AM}+\frac{n}{m}\overrightarrow{AM}=\frac{m+n}{m}\overrightarrow{AM},$$ and $$\overrightarrow{AM}=\frac{m}{m+n}\overrightarrow{AB}.$$

Finally, expressing point M in terms of its position vector \(\overrightarrow{OM}\), we have $$\overrightarrow{OM}=\overrightarrow{OA}+\overrightarrow{AM}=\overrightarrow{OA}+\frac{m}{m+n}\overrightarrow{AB}\\=\overrightarrow{OA}+\frac{m}{m+n}\left(\overrightarrow{OB}-\overrightarrow{OA}\right)\\=\overrightarrow{OA}-\frac{m}{m+n}\overrightarrow{OA}+\frac{m}{m+n}\overrightarrow{OB}\\=\frac{n}{m+n}\overrightarrow{OA}+\frac{m}{m+n}\overrightarrow{OB}$$

In terms of coordinates, $$\overrightarrow{OM}=\frac{n}{m+n}\left<x_1,y_1,z_1\right>+\frac{m}{m+n}\left<x_2,y_2,z_2\right>\\=\left<\frac{nx_1+mx_2}{m+n},\frac{ny_1+my_2}{m+n},\frac{nz_1+mz_2}{m+n}\right>.$$

That’s our formula (in three dimensions).

I’d continued searching:

Here is another source, which writes the formula in a way that is easy to misread:

The highlights are mine. Note that this time, the terms m, n, of the ratio are written in the given order, but the coordinates are reversed, which is very easy to miss!

Searching these days can yield not only sites like those, which I wanted, but also some extra surprises:

When I did this search, for “proportional division of segment as weighted average”, Google’s AI summary got this wrong. After an incomplete explanation, it gave an incorrect example:

This divides the segment in the ratio 1:2, not 2:1!

Its explanation of the method never mentioned how the weights are related to the ratio.

Again, I added the highlights, showing that it kept both the coordinates and the ratio in the order given, missing the exact point I was emphasizing. Wherever it “learned” this method, it didn’t read carefully.

Using the formula in a larger problem

The next day, Amia had a new question, using this technique in two steps:

Hi Dr math,

I want to check if my solution is acceptable for the question below. Find m??

In his picture, numbers represent not actual lengths, but proportions: (In my version below, I used parentheses to suggest this.) The ratio AC : CB is 2:3, and the ratio AK : KD is m:3. So we are given that D is the midpoint of OB, and that C divides AB in the ratio 2:3, and want to find how K divides AD.

Read through this carefully. Do you see the two mistakes?

I answered, including a vector derivation of our weighted average formula in vector form, for these particular vectors:

Hi, Amia.

When you write \(\overrightarrow{OC}=\frac{3}{5}\vec{b}+\frac{2}{5}\vec{a}\), you make the error we discussed in the weighted average, swapping the weights. The nearer point should have the weight corresponding to the greater distance.

We can show this using vectors. We have $$\overrightarrow{OC}=\overrightarrow{OA}+\frac{2}{5}\overrightarrow{AB}=\vec{a}+\frac{2}{5}\left(\vec{b}-\vec{a}\right)=\vec{a}+\frac{2}{5}\vec{b}-\frac{2}{5}\vec{a}=\frac{3}{5}\vec{a}+\frac{2}{5}\vec{b}$$

But you made the same mistake in writing $$\overrightarrow{OK}=\frac{3}{m+3}\cdot\frac{1}{2}\vec{b}+\frac{m}{m+3}\vec{a}$$ which should be $$\overrightarrow{OK}=\overrightarrow{OD}+\overrightarrow{DK}=\frac{1}{2}\overrightarrow{OB}+\frac{3}{m+3}\overrightarrow{DA}\\=\frac{1}{2}\vec{b}+\frac{3}{m+3}\left(\vec{a}-\frac{1}{2}\vec{b}\right)=\frac{3}{m+3}\vec{a}+\frac{m}{m+3}\cdot\frac{1}{2}\vec{b}$$

These errors seem to have canceled one another.

It’s very easy to make this mistake, even just after being shown it!

But be sure the point is between the end points!

The next day, Amia had a third question, which raised a different issue:

Hi Dr math,

I want to check out my solution to the question below.

Thank you in advance.

This time, I had better news:

Yes, that is correct as long as the points are in the order X-W-Z as shown.

But if, say, they were W-X-Z, then WZ = XW + XZ = XW + 4/3 XW = 7/3 XW, so XW:WZ = 3:7.

W-----X-------Z
   3      4

Was the order specified in the problem?

Since the picture shows W in the middle, that is presumably the intent of the problem. But sometimes we can’t depend on a picture! See What Role Should a Figure Play in a Proof?

One of my sources above mentioned dividing a segment “internally or externally”; the latter is what we would do if W here were not between X and Z. That reminded me of a 4-year-old question about a deeper issue with the same formula. We’ll look at that next week.