A product that returns a number, and the angle you can extract from it

0:00Before anything gets defined, here is a question worth asking. Why would anyone want a second way to multiply two arrows? Look at multiplication elsewhere. Two numbers multiply to a number. Two matrices multiply to a matrix. You put in two of a thing and get back one of the same thing. But two functions do not behave like that. Multiply them value by value and you get a function. Feed one into the other and you get a function that way too — a completely different one, built by a completely different operation.

0:40So multiplication is not always a single operation. Sometimes there are two, and they are not variations of each other. That is the case with arrows. Two products, genuinely different, and the giveaway is what comes out. One of them hands you back a plain number. That is the one we are doing. Here is the definition. Take two arrows, neither of them nothing. Put their tails together so there is an angle between them.

1:13The product is the length of the first, times the length of the second, times the cosine of that angle. Three ingredients, and only one of them is new — the cosine. And notice what has just happened. Two arrows went in. A length is a number, a cosine is a number, and a number times a number times a number is a number. So the answer is not an arrow. No direction, no parts. One number, positive, negative or nought — and almost every mistake in this topic is downstream of forgetting that sentence.

1:54There is a clause in that definition doing more work than it looks. The angle runs from nothing up to a straight angle. Half a turn, and no further. Why does that matter? Two arrows meeting at a point do not subtend one angle, they subtend two — one going round each way, adding to a full turn. And those two have different cosines. So without the restriction the definition would not name a single number. The first half turn picks exactly one of the two, and that is the whole reason the clause is there.

2:31It also pins the cosine between minus one and one. Over fifteen thousand three hundred and seventy six pairs of arrows on a grid, it never once leaves that range. It reaches one at exactly a hundred and seventy six of them — the pairs lying along one line and pointing the same way. And it reaches minus one at exactly a hundred and seventy six — the pairs along one line pointing opposite ways.

3:04Now the case the definition cannot reach. Suppose one of the two arrows is nothing at all. There is no arrow to lay alongside the other, so there is no angle, and the definition is asking for the cosine of something that does not exist. So the value gets written in by hand. If either arrow is nothing, the product is nought. That is a decision, not a deduction. But it is not an arbitrary decision, and here is the argument nobody makes out loud.

3:38On the grid, two hundred and forty nine pairs carry the empty arrow, and the angle routine refuses at exactly those two hundred and forty nine — as it should, since there is nothing to measure. Every other route to the answer still works there, and every one returns nought, at all two hundred and forty nine. So the value chosen is the only one that agrees with everything else.

4:06Keep this shape in mind. The other product of two arrows is defined the same way — a definition for the live case, and a separate convention for the empty one. The answer is a number, so it has a sign, and the sign carries information. The two lengths are never negative, so the sign of the whole thing is the sign of the cosine — and the cosine changes sign at a right angle.

4:36Under a right angle, positive. At a right angle, nought. Over a right angle, negative. So one number, read for its sign alone, tells you what kind of angle you are looking at. You never need to find the angle itself. And that can be checked against a reading that never mentions the product — the side facing the angle, compared against the other two. The two readings agree at every one of the fifteen thousand three hundred and seventy six live pairs: six thousand six hundred and fifty six under a right angle, two thousand and sixty four at one, and six thousand six hundred and fifty six over.

5:21The middle case deserves its own line, because it is the one you will use most. Two live arrows are perpendicular exactly when their product is nought. Read left to right, that is a test: compute the number, get nought, conclude a right angle. Read right to left, it is a condition: demand a right angle, and you have one equation to solve. Here is the same fact as a picture. Lay the two arrows out as the sides of a parallelogram; the diagonals are the sum and the difference. Those two have equal length exactly when the sides are perpendicular — a statement about lengths, with no product in it anywhere.

6:07And it picks out precisely the same pairs: at all fifteen thousand three hundred and seventy six, the vanishing product and the equal diagonals agree. Now put an arrow against itself. The angle between an arrow and itself is nothing, and the cosine of nothing is one. So the product is the length times the length — the square of the length. Small, and used constantly. Every squared magnitude you meet in the working of these problems came from here.

6:43Now put an arrow against its own negative. The angle is a straight one, the cosine is minus one, and the answer is that square with the sign turned — the genuine particular case of the straight-angle rule, and the one that usually gets skipped. Across a hundred and twenty five arrows both hold every time, while the control — asking for the square unturned against the negative — survives at exactly one arrow, the empty one, where there was no sign to turn.

7:20Now the single most-used fact in everything that follows. Take the three axis arrows, the unit ones pointing along the three directions. There are nine products you can form, taking two at a time in every order — and you do not have to remember nine things, you have to remember two. Each has length one, so each against itself gives one times one times the cosine of nothing, which is one. And different axes are perpendicular, so each against a different one gives nought.

7:56So the nine products form a three-by-three table with ones down the diagonal and noughts everywhere else. Six of the nine vanish, three survive. That table is built here from the geometric definition, not from any formula, and then the formula is asked the same nine questions and gives the same nine answers. It is worth seeing what a wrong table looks like. Pair each slot with the next one round instead, and you get a table that is just as tidy — three ones, six noughts — with its ones off the diagonal. Tidy is not the same as right.

8:40The definition runs one way: give me the angle and I will give you the number. Now run it backwards. Divide both sides by the two lengths, and the cosine of the angle is the product over the two lengths multiplied. That is the working formula for the angle, and it is why the product is worth having at all: it turns an arithmetic quantity into a geometric one. Here it is on a case. Take one, one, minus one, and one, minus one, one. Pair the slots — one, minus one, minus one — and the product is minus one.

9:20Each arrow has squared length three, so each has length root three, and the two multiplied give three. The cosine is minus one third. Negative, so the angle is over a right angle, and nothing has gone wrong. And it is not one of the tidy ones. Double it and the cosine becomes minus seven ninths, neither nought nor minus one, so twice the angle is neither a right angle nor a straight one. The inverse cosine of minus one third is the answer, and that is that.

9:57Compare a tidy one. Lengths one and two with a product of one give a cosine of one half, and three of that angle make a straight angle exactly — so it is a third of a half turn. Two properties, and then we can derive the formula everybody actually uses. The first: the product spreads across a sum. An arrow against a sum of two arrows is the same as the two products added.

10:27That is checked here over nineteen thousand six hundred and eighty three triples, and it holds at every one. The second: a scalar can sit anywhere. Multiply the first arrow by a number, or the second, or multiply the answer by it afterwards, and you get the same thing. Three placements, one value, at every one of seventy eight thousand one hundred and twenty five cases tried. And note what the second property does not say. It does not say you may put the scalar on both arrows. Do that and you have multiplied by the number twice.

11:08That is a real error and not a harmless one — it still gives the right answer at twenty four thousand eight hundred and seventy seven of those cases, which are exactly the ones where the scalar is nought or the product already is. Both of these will get used in the next ninety seconds, and then almost never stated again. Here is the piece worth staying awake for. A definition is about to turn into an algorithm.

11:41Write each arrow in components: some number of the first axis arrow, plus some number of the second, plus some number of the third. Same for the other. Now multiply. Spread across the first sum, spread across the second, and move all six scalars outside. That is both properties, used once each. What is left is nine terms, each a number, times a number, times a product of two axis arrows.

12:12And we have those nine. Six are nought, so six terms disappear. Three are one, so three terms lose their factor entirely. What survives is the first parts multiplied, plus the second parts multiplied, plus the third. That is the formula — pair the slots and add. Six lines, every step justified, and nothing in it that was not already on the board. This is the one place in this subject where you watch a definition become a rule you could hand to a machine.

12:49And it is worth checking rather than believing. The nine-term expansion, run as code with those nine values plugged in, lands on the paired-slot formula at every one of fifteen thousand six hundred and twenty five pairs. Now the formula earns its keep, because everything from here is arithmetic. First case. Take five, minus one, minus three, and one, three, minus five. Show that their sum and their difference are perpendicular.

13:24The sum is six, two, minus eight. The difference is four, minus four, two. Pair the slots: twenty four, less eight, less sixteen. Nought. Perpendicular. Second case, and this is the shape most of the questions take: three arrows, and you want the one multiplier that makes a combination of the first two perpendicular to the third. Take two, two, three; minus one, two, one; and three, one, nought. The first pairs with the third to give eight, the second to give minus one.

14:00Spread the product across the combination and the condition becomes eight, minus the multiplier, equals nought. So the multiplier is eight, and sweeping every whole number from minus twenty to twenty finds that one and no other. The formula also runs the other way, giving you lengths from products. Take the difference of two arrows. Its squared length is the first squared length, plus the second, minus twice the product — a rule that holds at every one of the fifteen thousand six hundred and twenty five pairs, so it is safe to use blind.

14:43So: lengths two and three with a product of four. Four, less eight, plus nine — five. The difference has length root five, somewhere between two and three. Here is a neater one. Take an unknown arrow and a unit arrow, form the sum and the difference, and multiply those together. It is a difference of two squares, and the squares are squared lengths, so the answer is the unknown squared length minus one. That holds at all seven hundred and fifty cases the grid offers.

15:19So if the product comes out eight, the squared length is nine and the length is three. If it comes out twelve, the length is root thirteen. And in each case you throw away the negative root, because a length is never negative — which the machinery here does by refusing a negative input outright rather than returning one. Two named inequalities now, and they are one argument in two halves. The first bounds the product: the size of the product of two arrows is never more than the two lengths multiplied.

15:59You can see why immediately. The product is those two lengths times a cosine, and a cosine is never bigger than one in size. Multiply by at most one and nothing grows. That is the Cauchy–Schwarz inequality, and the whole of its proof is the sentence you just heard, plus a line disposing of the empty arrow, because the argument divides by both lengths. Over the grid it holds at all fifteen thousand six hundred and twenty five pairs.

16:35Now, when is it tight? When the cosine is one in size — which is when the two arrows lie along one line, either way round. Exactly six hundred and one pairs meet the bound, and there are exactly six hundred and one pairs that lie along one line or carry the empty arrow. Insist on strict inequality everywhere instead and you are wrong at those six hundred and one.

17:05The second inequality bounds a sum. The length of the sum of two arrows is never more than the two lengths added. Obvious if you draw it — two sides of a triangle against the third — and the proof is where the first inequality gets spent. Square the sum: the first squared length, plus twice the product, plus the second squared length. The middle term is the only one you cannot control, so cap it. By the first inequality, twice the product is at most twice the two lengths multiplied.

17:42Put that in, and the right-hand side is the first squared length, plus twice the lengths multiplied, plus the second — which is exactly the two lengths added and then squared. So the square of the left is at most the square of the right, and both are lengths, so the left is at most the right. Every step runs cleanly at all fifteen thousand six hundred and twenty five pairs. Skip the capping step and you are right at only four hundred and twenty five.

18:15Which brings us to those four hundred and twenty five. Equality here is stricter than before: not along one line, but along one line pointing the SAME way. Loosen it to either direction and you are wrong at a hundred and seventy six pairs. Now run that equality case backwards, and something rather nice falls out. Take three points. Join the first to the second, the second to the third, and the first to the third. The first two arrows add to the third.

18:51By the inequality, that third length is at most the other two added — and equality means the two arrows point the same way, which from a common point means one straight line. So: if the two shorter joining lengths add exactly to the longest, the three points lie on a line, with the middle one between the other two. A collinearity test made of nothing but three lengths. Over all nineteen thousand six hundred and eighty three triples of points on a small grid, it picks out precisely the triples whose middle point sits on the way from the first to the third. One thousand five hundred and twenty nine do.

19:36And it matters which two you add. Add the two sharing the first point instead and you still agree eighteen thousand and eighty three times — but you are wrong at sixteen hundred. Here it is on numbers. Three points whose joining arrows come out three, minus one, minus two; then six, minus two, minus four; and then nine, minus three, minus six. The squared lengths are fourteen, fifty six and a hundred and twenty six, so the lengths are root fourteen, twice it, and three times it.

20:12One plus two is three. The points are on a line, and the middle one sits a third of the way along. And notice — that argument never touched the product. Only lengths. Now four points, and two joining arrows: one, four, minus one, and minus two, minus eight, two. Pair the slots — minus two, minus thirty two, minus two — and the product is minus thirty six. The squared lengths are eighteen and seventy two, so the lengths are three root two and six root two, and multiplied they give exactly thirty six.

20:53So the cosine is minus thirty six over thirty six. Minus one exactly, and the angle is a straight one. The formula did not break at the extreme; it returned the extreme. Now here is the one line that made all of that unnecessary. The second arrow is minus two times the first. One is a scalar multiple of the other, so they lie along one line, and the multiplier is negative, so they point opposite ways. Straight angle. No product, no lengths, no division.

21:31Both routes are worth having, but the second one is the one you should be looking for first. Two problems now that are the same idea run forwards and backwards, and they are much easier as a pair than separately. Forwards. Three arrows of lengths three, four and five, each one perpendicular to whatever the other two add up to. How long is the sum of all three? Square it: the three squared lengths, plus twice each of the three pairwise products. Group those cross terms by arrow, and each group is exactly one of the three given conditions. All of them vanish.

22:15So the squared length is nine plus sixteen plus twenty five, which is fifty, and the answer is five root two. On the small grid, eight hundred and forty seven triples have that property, and the splitting holds at every one — while asking it of every triple regardless holds at only three thousand three hundred and sixty seven of nineteen thousand six hundred and eighty three. Now backwards. Three arrows of lengths three, four and two, and this time they add to nothing. Find the three pairwise products added together.

22:53Same move. The sum is nothing, so its square is nothing: the three squared lengths, plus twice the thing you want. So twice what you want is minus nine minus sixteen minus four — minus twenty nine — and the answer is minus twenty nine halves. Three hundred and forty three such triples on the grid, and the rule reads the products off the three lengths correctly at every one. Do it with three unit arrows instead and you get minus three halves.

23:28Last idea, and it is the one that catches people. A product comes out nought — what may you conclude? Two quite different reasons it could be nought: either one of the arrows is the empty one, or both are alive and genuinely perpendicular. On the grid, two thousand three hundred and thirteen pairs have a vanishing product. Two hundred and forty nine of those carry the empty arrow. Only the remaining two thousand and sixty four are actually perpendicular.

24:03So a vanishing product means perpendicular only if you already know both arrows are there. Here is the trap built on that gap. You are told an arrow pairs to nought with every arrow there is. It has to be the empty one — and on the grid exactly one arrow does that. Now someone hands you a second condition about that same arrow. It is automatically satisfied, for every partner, all hundred and twenty five of them. It tells you nothing whatever.

24:38Answer "perpendicular" there and you have fallen for it. The honest answer is that nothing can be concluded. One more thing you may not do: cancel. If an arrow pairs to the same number with two others, those two others need not be equal. The three axis arrows already break it — the first pairs to nought with each of the other two, and those two are not the same arrow. Across the grid, thirteen thousand two hundred and forty eight such cases.

25:12So. What is worth keeping. The product of two arrows is a number. Not an arrow. Everything else is downstream of that. It is a length, a length and a cosine, with the angle pinned to the first half turn so that there is only one of it. And where there is no angle to be had, the value is chosen rather than deduced — chosen as the only one that agrees with everything else.

25:40Its sign tells you the kind of angle. Positive, acute. Nought, right. Negative, obtuse, and nothing has gone wrong. Nine products of three axis arrows collapse to ones on a diagonal, and those nine turn the definition into the rule you actually use: pair the slots and add. Not a separate fact — six lines of consequence, and you have watched all six. Two inequalities, and the first is used inside the second. The equality case of the second is a collinearity test made of three lengths.

26:20And a product of nought means one of two things, so before you say perpendicular, check that both arrows are actually there.