Position vectors, the vector joining two points, and dividing a segment in a ratio

0:00Almost everything left in this subject starts from two named points, and almost everything left runs through one subtraction. Here is the shape of what follows. Join two points to the origin. Apply the triangle law once. Out comes the arrow between them, as a single subtraction. That is the hinge, and it is about three lines long. Then a segment gets cut in a ratio, and the two cases get very different treatment. One of them gets a derivation with a step in it that almost everybody skips. The other is usually just handed to you, with no argument at all — the one place in the whole subject where the argument is simply declined.

0:50So we will do the one that gets declined, because it is the same four lines with one sign changed, and doing it turns two formulas into one. And there is a crossing in both formulas that is where nearly all the marks go. Start with the arrow between two points, and let us find it rather than write it down. A subtraction is easy to state and easy to get backwards, so here is the arrow defined by what it does instead of by a formula.

1:24It is the arrow which, added to the first position vector, lands you on the second. That is all. No subtraction anywhere in that sentence. Now search for it. Over seven hundred and twenty nine ordered pairs of points, that arrow exists and is unique every single time. And now compare the search with the subtraction. The arrow the search finds is the second position vector minus the first, at all seven hundred and twenty nine, with the two never once disagreeing.

2:00That is the whole of the first idea, and it fell out of one closed triangle. Now the part that costs marks: which way round. The rule is terminal minus initial. The point you finish at, minus the point you start from. And the order is not a detail, it is the whole of it. Take the subtraction the other way round and compare it with the search again: it agrees at twenty seven of the seven hundred and twenty nine.

2:33Twenty seven, and there are exactly twenty seven pairs of a point with itself. So the wrong order is right only where the arrow is nothing at all. Swap the two points and every one of the three parts changes sign. At all seven hundred and twenty nine. And it is all three, not one. A rule that turns only the first part round matches the swap at eighty one — and there are exactly eighty one pairs whose arrow has nothing in its other two parts, where there was nothing else to turn.

3:08A worked case. From the point two, three, nought to the point minus one, minus two, minus four, the arrow has parts minus three, minus five and minus four. And read the other way, three, five and four. Which is why these questions say directed from. The letters alone do not tell you the order. Now its length, and this is worth thirty seconds because you have met this formula three times and probably think it is three formulas.

3:42The first time: the distance from the origin to a point, the square root of the summed squared coordinates. Handed over from an earlier year. The second time: the magnitude of a vector from its components, derived from a box and two right triangles. The third time is here: the square root of the summed squared coordinate gaps. They are one formula. The gaps version and the components version agree at all seven hundred and twenty nine ordered pairs, and the origin version is the same thing again with the first point at the origin, at all twenty seven.

4:25And the third gap is genuinely being spent. Square only two of the three and you get the right length at two hundred and forty three — exactly the pairs whose two points share their third coordinate, where the third gap was nought anyway. One formula, three appearances, and nothing anywhere says so. Now the second half of the topic. A segment, and a third point sitting on the line through it.

4:55There are two cases and they are genuinely different. In the first, the third point sits between the two ends. That is dividing the segment internally. In the second, the third point sits beyond one of the ends, out past the segment altogether. That is dividing it externally. Both are described by a ratio, and both get a formula. The formulas differ in exactly two signs. And here is the thing to hold on to: the ratio describes two parts, and which part goes with which end is not what you would guess.

5:36Take the internal case first, and let us actually derive it. You have two ends and a cutting point between them. The ratio says the first part is to the second part as one number is to another. Write that as an equation between two scaled pieces of the segment. Then write each of those pieces as a difference of position vectors — which we can now do, because that is what the first half of this topic was for.

6:08Substitute. Rearrange. And the cutting point falls out as a weighted average of the two ends. Four lines. And now the point of doing it rather than quoting it: the formula has the property it was built to have. Over six thousand three hundred and eighteen combinations of a ratio and two different points, the two parts of the cut segment stand in the stated ratio every single time. That check is worth having because it mentions no formula at all. It compares two pieces of the segment and nothing else.

6:48There is one step in that derivation that is almost always skipped, and it is usually marked with a question and then left there. It is the step where the two scaled pieces are set equal to each other. Why are you allowed to do that? Because of the definition of the ratio, two lines earlier. The ratio said the first part is to the second as one number is to the other, and an equality of ratios rearranges into exactly that equality of products.

7:22That is it. That is the whole answer. But if you do not go back for it, the derivation is four lines of symbol-pushing with a hole in the middle, and you get nothing out of it. When a derivation asks you a question, it is telling you which step is load-bearing. Now the thing that costs more marks than anything else in this topic. Look at which coefficient sits next to which endpoint.

7:50The number describing the part nearer the first end multiplies the position vector of the second end. They cross over. It is completely reasonable to expect them not to. And the formula is almost always stated without a word about it. So let us measure what happens if you get it wrong. Take the formula with the pairing undone — each number against the end it is nearer — and run the same property check.

8:21It holds at two thousand one hundred and six of the six thousand three hundred and eighteen. And there are exactly two thousand one hundred and six combinations whose two ratio numbers are equal — where there is nothing to cross, so getting the crossing wrong costs nothing. Everywhere else, which is two thirds of the time, the uncrossed version cuts the segment in the wrong place. Now the external case, and this is the one that usually arrives with no argument at all.

8:56It gets stated, and then the checking of it gets handed over to you. That is unusual enough to be worth noticing. Everything else in this subject gets derived. This one thing gets declined, out loud. So here it is, and it takes four lines, because it is the same four lines as before. Same setup: two ends, a ratio, and a cutting point. The only difference is that the cutting point is now past the second end rather than between the two.

9:31Set up the same equation between two scaled pieces. Write each as a difference of position vectors. Substitute. Rearrange. And the answer differs from the internal one in exactly two signs. Two signs. Let us be precise about where they are, because one of them is remembered and one is forgotten. The first sign is on top: a minus instead of a plus between the two weighted terms. The second sign is underneath: the two numbers subtracted instead of added.

10:08That second one is the one people miss, and it is not a small miss. Change only the sign on top and check the result against the property the external cut is supposed to have. It holds at nought of the four thousand two hundred and twelve. Not a few. None. With both signs changed it holds at all four thousand two hundred and twelve. And the two cuts really are different places: at every one of those four thousand two hundred and twelve the internal cut and the external cut are different points. The internal one lies strictly inside the segment every time, and the external one lies inside it at none of them.

10:56Now a case that looks like a problem and is actually the answer. Divide a segment externally in the ratio one to one. What do you get? The denominator is the difference of the two numbers, so the denominator is nought. And the right response to that is not to panic and not to fudge it. It is that there is no such point. A point cutting a segment externally into two equal parts does not exist.

11:26So the routine that computes this should refuse, rather than return something. Asked over seven hundred and two pairs of different points, it refuses all seven hundred and two. Overall it is asked six thousand three hundred and eighteen times, answers four thousand two hundred and twelve of them, and refuses the two thousand one hundred and six whose two numbers are equal. A formula that has no answer somewhere is not broken. It is telling you something about the geometry.

12:03Meanwhile the internal formula at one to one is the most useful thing in the topic. Set the two numbers equal and they cancel, top and bottom, and what is left is half the sum of the two position vectors. The midpoint. And it is not a separate result — it is this formula with the two numbers made equal. Checked at all seven hundred and two pairs of different points: the one-to-one cut is the average, and so is the two-to-two cut, which is the same ratio written differently.

12:39And it is the equal numbers doing it, not the formula. Over the six unequal ratios the internal cut is the average at nought of the four thousand two hundred and twelve. A worked one: the middle of the segment from two, three, four to four, one, minus two is at three, two, one. Let us put both formulas on one segment. Two points. One ratio, two to one. Two answers.

13:10From the point one, two, minus one to the point minus one, one, one. Cut internally: minus one, four and one, all over three. That point sits between the two ends. Cut externally: minus three, nought and three. That point sits out past the far end. Same two points, same two numbers, and the answers are nowhere near each other. Which is why the word internally or externally is not decoration in these questions. It is half the question.

13:45Now a worked case that makes people think they have made a mistake. The two ends are given in terms of two other vectors. The first end is three of the first vector minus two of the second. The second end is the first plus the second. Cut that in the ratio two to one, internally. And the second vector vanishes completely. The answer is five thirds of the first vector, and the second one is simply not there.

14:17That is not a coincidence and it is not an error. It is arithmetic. Tried over seven hundred and twenty nine choices of the two building vectors, the answer is five thirds of the first at every single one. The external cut in the same ratio keeps both: four times the second minus the first, again at all seven hundred and twenty nine. And the disappearance belongs to that particular ratio, not to the formula. Cut one to two instead and the answer differs from five thirds of the first at seven hundred and twenty eight of the seven hundred and twenty nine — agreeing only in the one case where both building vectors were nothing to begin with.

15:06If an answer loses one of the things you were given, check the arithmetic. Do not assume you are wrong. Now something that looks like a detour and is not. Three points are given, and the question is to show they make a right-angled triangle. The obvious tool for right angles is a product of two vectors that comes out at nought when they are perpendicular. And at this stage of the subject, that tool does not exist yet. It arrives shortly.

15:40So the right angle has to be found another way, and it is: from the three side lengths alone, using the converse of Pythagoras. Compute the three side vectors. Take their squared lengths. Forty one, six and thirty five. Six plus thirty five is forty one. That is a right angle, found without the tool for finding right angles, and the right angle is at the third of the three named points.

16:12And here is why that is worth more than the answer it gives. Shortly you will have the other tool, and this same triangle takes one line. So run both routes and compare them properly. Over seventeen thousand two hundred and fifty six genuine triangles on a small grid, the converse of Pythagoras and the product route disagree at nought, and both find eight thousand one hundred and thirty six right-angled ones.

16:42They are one fact, arrived at from two directions. And the lengths route has a trap in it. There are three pairs of sides to try, because the right angle can be at any of the three corners. A routine that tries only one pairing finds two thousand seven hundred and twelve of the eight thousand one hundred and thirty six. A third of them. So try all three. One more thing about this triangle. Relabel the three corners cyclically and it is the same triangle: the same three points, the same three squared sides in the same cyclic order, and the right angle in the same place. A different letter, the same corner.

17:30Two more, quickly, and both of them use everything so far. Three points. Show they lie on one line, and find the ratio in which the middle one cuts the outer segment. Compute the two joining arrows: four, two, six, and six, three, nine. The second is one and a half times the first, so they lie along one line, so the three points are collinear. And now the ratio just falls out. The middle point is the two-to-three cut of the outer two.

18:03And it is pinned down: of nine candidate ratios, exactly one puts the middle point where it actually is. Notice the order of work. Collinearity first, then the ratio. You cannot ask what ratio a point cuts a segment in until you know it is on the line. And the last one, which is the only place both formulas get run against each other. Two ends, given again in terms of two building vectors. Cut externally in the ratio one to two.

18:39The answer is three times the first building vector plus five times the second. At all seven hundred and twenty nine choices. Now take that answer, and average it with the second given end. And you get back the first given end. Exactly. At all seven hundred and twenty nine. So the external cut and the midpoint are consistent with each other, and that is the only item that ever checks it.

19:09Which is a nice place to stop, because it is the two halves of this topic closing on each other. Six things. The arrow between two points is terminal minus initial, and the order is the whole of it — swap the points and all three parts change sign. Its length is a formula you have now met three times, and all three are one formula with different things in the squares.

19:39The internal cut is derived in four lines, and the step people skip is the one where two scaled pieces get set equal. The coefficients cross. The number describing the part nearer one end multiplies the position vector of the other end, and getting that wrong is wrong two thirds of the time. The external cut is the same four lines with two signs changed, and the sign underneath is the one that gets forgotten. At equal numbers it has no answer, which is correct.

20:15And the midpoint is not a separate result. It is the internal formula with the two numbers equal. Two points, one subtraction, and one weighted average. That is the whole of it.