A product that returns a vector, where it points, and the area it measures

0:00Two vectors. One product. And this time the answer is a vector, not a number. That single change is where all the difficulty lives, and it is worth saying why before any formula appears. A vector needs a size and a direction. The size will be easy. The direction has to be square to both of the vectors we started with — and here is the problem: there are two such directions, not one. Up out of the plane, and down out of it.

0:33So before anything can be written down, a rule is needed for which of the two to take. That rule is a convention, and there are two standard pictures of it. Picture one, a screw. Stand an ordinary screw so it points along the direction you want to name. Turn the first vector towards the second, the short way round. If the screw drives forwards, along the direction it is standing on, that direction is the one to take.

1:05Picture two, a hand. Curl the fingers of your right hand the same way — first vector turning towards the second — and your thumb points at the answer. The two pictures say the same thing and you only need one of them. Whichever one lands for you, take it and move on. Thirty seconds here is enough. Now the definition, built in three parts, in the order the parts actually matter.

1:35Part one, the size. For two vectors that are not the empty one, the size of the answer is the first length times the second length times the sine of the angle between them. Part two, the direction. Square to the first vector, and square to the second, so square to the plane they span. Part three, and this is the whole content of the definition: of the two directions square to that plane, take the one that makes the first vector, the second vector, and it, in that order, a right-handed set. The other one is the arrow drawn dashed in the picture beside it, and that dashed arrow is the reason the definition needs a third clause at all.

2:22Put together: the product is the two lengths, times the sine, times a unit arrow in the chosen square direction. That is one sentence with a choice buried in the middle of it, and everything else in this video is a consequence. Before going on, that definition is worth checking against the thing you will actually compute with, because they look nothing alike. So here it is built twice, by two routines that share no arithmetic.

2:55The first follows the words. The size comes from the two lengths and the sine — and the sine is reached from the lengths alone, since the two squared lengths multiplied, less the square of the other product, is exactly that size squared. No component is ever paired with a component in it. The direction is then FOUND by searching a net of seven hundred and twenty eight arrows for the ones square to both, and the choice between the two candidates is settled by the sign of a six-term sum over the ways of shuffling three slots, which is positive exactly for a right-handed set.

3:39The second is the component formula, three small differences, which is coming in a few minutes. Over a smaller grid, they land on the same arrow at every one of the six hundred and seventy six live pairs. And the choice in part three is load-bearing. Take the OTHER of the two square directions and the two builds agree at only fifty two pairs — exactly the pairs whose product is nothing anyway, where there is no arrow to point either way.

4:13One case the definition cannot reach, and the patch is the same one you have already seen once. If either vector is the empty arrow there is no angle between them, so there is no sine, so the definition has nothing to say. The answer is therefore SET, by convention, to the empty arrow. That is exactly what the other product of this module does with the same case: it covers the live vectors and then bolts on a convention. Two products, one gap, one patch, and the same words both times.

4:51On a grid of a hundred and twenty five arrows there are two hundred and forty nine pairs that involve the empty one, and the convention supplies the answer at all two hundred and forty nine of them. Here is the fastest way to keep the module's two products apart, and it takes one sentence each. The other product is nothing exactly when the two vectors are square to each other. This one is nothing exactly when they lie along one line.

5:25Opposite conditions. If you remember only that, you will never confuse them again. And this is worth counting rather than asserting, because vanishing is easy to get almost right. On the grid there are fifteen thousand three hundred and seventy six live pairs. The pairs where this product is the empty arrow, and the pairs that lie along one line, are not two counts that happen to match — they are the same list, disagreeing at none of its three hundred and fifty two entries.

5:59The other product is nothing at a different two thousand and sixty four pairs, and the two lists share nothing at all: a live pair cannot lie along one line and be square to itself at the same time. In particular a vector crossed with itself is nothing, and crossed with its own negative is nothing — at all hundred and twenty four live arrows each. The sine vanishes at both ends of its range, which is what those two cases are.

6:34There is a single line that ties the module's two products together, and once you have seen it the two stop being separate facts. Take the other product, square it. Take this product, take its size, square that. Add the two. You get the first squared length times the second squared length, every time. It holds at all fifteen thousand six hundred and twenty five pairs of the grid, the empty arrow included.

7:04Read it once and three things fall out. Neither product can be large where the other one is. Each is nothing exactly where the other is as large as it can get. And the size of this product can never pass the two lengths multiplied — which holds at all fifteen thousand three hundred and seventy six live pairs, and is MET at exactly the two thousand and sixty four square ones. Same list, disagreeing at none.

7:34That last one is the mirror of the bound you already met for the other product, which is met exactly on the parallel pairs. One identity, two bounds, two extreme cases. Now a right angle, on its own, because there is something to be careful about in how it is usually written. At a right angle the sine is one, so the size of the product is just the two lengths multiplied. That much is right and we have just counted it.

8:07But the statement is often written as: the product EQUALS the two lengths multiplied. And that cannot be right, for a reason that needs no arithmetic at all. The left-hand side is a vector. The right-hand side is a number. A vector is not a number, and the distinction between the two is the entire reason this module has two products in it. What has gone missing is the unit arrow. There are two ways to write it correctly. Either put size bars on the left, so a number equals a number. Or keep the unit arrow on the right, so a vector equals a vector.

8:50Both are true, they say different things, and neither of them is the version with the arrow dropped. If you ever find yourself writing a vector equal to a number, stop and put the arrow back. The three axis arrows now, because their nine products are the whole of the arithmetic and they come off one picture. Draw the three of them round a ring, in order, with arrows running one way.

9:19Any arrow crossed with itself is nothing — three of the nine gone immediately, since a vector lies along its own line. For the other six, follow the ring. Travel WITH the arrows and the product is the next one round: the first crossed with the second is the third, the second crossed with the third is the first, the third crossed with the first is the second. Travel AGAINST the arrows and you get the same answers with a minus in front. Three more.

9:52One ring, six products, and three that are nothing. That is the entire multiplication table of this product, and it is the reason the component formula looks the way it does. Swapping the two factors, and this is where a word does more damage than it should. This product is not commutative. True. But students hear that and assume the two answers are unrelated, and they are not. Swap the factors and the answer is negated. That is all that happens. On the grid it holds at all fifteen thousand six hundred and twenty five pairs.

10:34Which means the size does not change — untouched at all fifteen thousand six hundred and twenty five — and the plane does not change either. Same size, same line, opposite sense. And the reason is the definition's third clause, not some separate rule. Turning the second vector towards the first is the same angle traversed the other way, so the right-handed condition picks the OTHER of the two square directions. The dashed arrow becomes the solid one.

11:10Measured: the two answers are a genuinely different arrow at fifteen thousand and twenty four live pairs, which is exactly the pairs whose product is not nothing — same list, disagreeing at none. And at every one of those they lie along one line and point opposite ways, at all fifteen thousand and twenty four, and the same way at none. Now the reason this product is worth having at all: it measures area.

11:42Draw a parallelogram on two vectors from one corner. Its area is base times height. The base is the first length. The height is the second length times the sine of the angle between them. But the two lengths times the sine is the size of the product. So the area of the parallelogram is just the size of the product of its two sides. And the triangle on the same two sides is half of that parallelogram — the diagonal cuts it in two — so its area is half the size of the same product.

12:20Two areas, one product, and the only difference is a factor of a half. This is checked here against a base and a height that never mention this product: the height is measured as the length of what is left of the second vector once the piece lying along the first is taken off. Base times that height is the size of the product at all fifteen thousand three hundred and seventy six live pairs.

12:50And the sine is doing real work. Take the whole second length as the height — the slip of forgetting it — and you get the right answer at two thousand and sixty four pairs, which are exactly the square ones, where the two happen to coincide. Note what the area formula does NOT need: the angle. The sine is already inside the product. Computing an angle to get a sine to get an area is three steps you do not have to take.

13:23Two rules for handling this product in an expression, and one of them is what makes the component formula possible. First: it distributes over addition. Crossing with a sum is crossing with each piece and adding. On the small grid that holds at all nineteen thousand six hundred and eighty three triples, on both sides. Taking the second piece away rather than adding it is a different statement, and it still comes out right at two thousand eight hundred and thirty five of those triples — the ones where the second piece contributes nothing either way. Enough to survive one careless check.

14:07Second: a number moves freely. Scale the first vector, or scale the second, or scale the answer — all the same. Over five numbers and every live pair that is seventy six thousand eight hundred and eighty cases, and it holds at all of them. Putting the number on BOTH factors is a different statement again, and it still holds sixteen thousand seven hundred and eighty four times: the fifteen thousand three hundred and seventy six where the number is nought, and four times the three hundred and fifty two where the product is nothing whatever you do to it.

14:47One consequence worth having in your hands: a difference crossed with a sum is twice the product. The two self-products vanish and the two cross terms ADD rather than cancel, because of the minus sign in the swap. It holds at all fifteen thousand six hundred and twenty five pairs. And one thing about numbering that costs people marks. The distributive rule is the one used to get the component formula. In some presentations the step is labelled with the wrong rule number, because the same numbers get reused in different sections for different laws. Learn what the rule SAYS, not what it is numbered.

15:34The component formula. Two vectors written in components, and the product written as a three-by-three array. Top row: the three axis arrows. Middle row: the three components of the first vector. Bottom row: the three components of the second. Expand along the top row and you get the three components of the answer, with a minus in front of the middle one. That object has a name — it is a determinant — and it is worth saying out loud, because it is very easy to meet this array here without ever being told what it is called. If you have seen it before you will recognise it. If you have not, you are being handed a notation with no name attached, and that is a bad way to meet anything.

16:25The alternating sign on the middle term is not decoration. Written out, the three components are: the second times the third less the third times the second, then MINUS the first times the third less the third times the first, then the first times the second less the second times the first. Three ways of writing the same thing are run here as three separate routines: three small two-by-two differences, a first-row expansion with the alternating sign spelled out, and a six-term sum over the ways of shuffling three slots. All three agree at every one of the fifteen thousand three hundred and seventy six live pairs.

17:13And the pattern matters. Pairing slot against matching slot — first with first, second with second, third with third — is the OTHER product's pattern, and it is not this one. It happens to give the same three numbers at ninety six live pairs, which is few but not none: one worked example can fail to tell them apart. Worked, twice. Two, one, three crossed with three, five, minus two.

17:44First component: one times minus two, less three times five. Minus two less fifteen is minus seventeen. Second component, and remember the minus in front: minus, the quantity two times minus two less three times three. That is minus, minus four less nine, which is minus, minus thirteen, which is thirteen. Third component: two times five less one times three, which is seven. So the answer is minus seventeen, thirteen, seven, and its size is the square root of two hundred and eighty nine plus a hundred and sixty nine plus forty nine — the root of five hundred and seven.

18:32One more. One, minus seven, seven crossed with three, minus two, two comes out nought, nineteen, nineteen, whose size is nineteen root two. Notice the first component vanished; that costs you nothing, it just means the answer lies in a plane. Now the question this product is most often asked: give me a unit vector at right angles to these two. The recipe is two steps. Cross them, then divide by the size of what you get.

19:07Take one, one, one and one, two, three. Their sum is two, three, four and their difference is nought, minus one, minus two. Cross the sum with the difference and you get minus two, four, minus two, whose size is the root of twenty four, which is two root six. So one unit vector square to both is minus one, two, minus one, each over the root of six. And now the thing that is worth thirty seconds. That is ONE of them. There are two. The other is the same triple with all three signs flipped: one, minus two, one, over the root of six.

19:49It is not an afterthought — it is the arrow you would have got by crossing the two the other way round. The definition's third clause, made concrete on numbers. Measured on the small grid: for every live pair that does not lie along one line, the search finds exactly two shortest arrows square to both and they are each other's negatives, at all six hundred and twenty four such pairs. And crossing the two the other way round produces the other one, at all six hundred and twenty four.

20:25One more, chosen because it has no surd in it anywhere. Three, two, two and one, two, minus two. Sum four, four, nought; difference two, nought, four. Cross them and you get sixteen, minus sixteen, minus eight, whose size is exactly twenty four. So the unit vector is two, minus two, minus one, all over three. Areas, on numbers. A triangle whose two sides from one corner are nought, one, two and one, two, nought. Cross them: minus four, two, minus one. Size is the root of sixteen plus four plus one, the root of twenty one. Halve it, and the area is half the root of twenty one.

21:13A parallelogram whose adjacent sides are three, one, four and one, minus one, one. Cross them: five, one, minus four. No halving this time, so the area is the root of forty two. Now three named points, which is the form most questions take. Take one, one, two; two, three, five; and one, five, five. Build two side vectors from the first point: one, two, three, and nought, four, three. Cross them and you get minus six, minus three, four, of size the root of sixty one. The area is half of that.

21:53The move worth remembering is that first step — three points become two vectors by subtracting one of them from the other two. Which point you start from does not matter. And one last parallelogram, sides one, minus one, three and two, minus seven, one. Their product is twenty, five, minus five, of size fifteen root two. One more, because it is the only place where everything in this module is needed at once, and because it contains a small trap.

22:28A parallelogram has two neighbouring sides two, minus four, five and one, minus two, minus three. Find a unit vector along its diagonal, and find its area. The trap is in the wording: a parallelogram has TWO diagonals, and nothing in the sentence says which. One diagonal is the sum of the two sides: three, minus six, two. Its squares add to nine plus thirty six plus four, which is forty nine, so its length is exactly seven, and the unit vector along it is three, minus six, two, over seven.

23:09The other diagonal is the difference: one, minus two, eight. Its squares add to sixty nine, which is not a square of anything, so its length is the root of sixty nine. So the sum is plainly the one intended — it is the one that comes out whole. But say that rather than guessing it, because the reasoning is the answer, not the arithmetic. And the area needs the product of the two SIDES, not the diagonals: twenty two, eleven, nought, whose size is the root of six hundred and five, which is eleven root five.

23:50The two products of this module, side by side, and then one thing that is not here at all. The first returns a number; this one returns a vector. The first is nothing exactly when the two vectors are square; this one exactly when they lie along one line. The first is unchanged when you swap the factors; this one changes sign. The first measures how much of one vector points along another; this one measures the area they span.

24:23And they are tied together by the one identity: the first squared, plus the size of the second squared, is the two squared lengths multiplied. There is one angle at which the two come out the same size. It needs the sine and the cosine to agree in size, which happens half a right angle from square. On the grid the pairs where the two products have equal size, and the pairs where twice the first product squared equals the two squared lengths multiplied, are the same list of seven hundred and sixty eight, disagreeing at none.

25:02Here is a worked version. If the first length is three and the second is the root of two over three, the two lengths multiply to exactly the root of two. The product has size one exactly when the sine is one over the root of two — whose square is one half. Half a right angle, again. And finally, the thing that is not here. Two products of two vectors are defined, and no third. There is no third product of two vectors waiting in a later section, and a product of THREE vectors is not defined here either.

25:41You will meet expressions that look like one — a number paired against a product of two others. Every such expression is computable one term at a time from what you already have. Pair the first axis arrow against the second crossed with the third and you get one; the second against the first crossed with the third gives minus one; the third against the first crossed with the second gives one. Add them: one. Nothing missing.

26:11So: a size from two lengths and a sine, a direction square to both, and a choice between two. Everything else follows from those three.