Projection: how much of one vector points along another

0:00Here is a picture worth carrying for the rest of this subject. A line, with a chosen direction along it. An arrow, sitting somewhere off the line, meeting it at some angle. Now shine a light straight down onto the line — square to it, not from any angle — and look at the shadow the arrow casts. That shadow is a vector. It lies along the line, it has some length, and it points either with the line or against it.

0:30That is the whole idea, and everything in the next quarter of an hour is either that picture or arithmetic for computing it. One thing to notice straight away, because it is the commonest confusion here. The shadow does not point along the original arrow. It points along the LINE. The arrow is what is being measured; the line is what the answer is reported in. How long is the shadow?

1:01Drop a perpendicular from the arrow's tip to the line and you get a right-angled triangle. The arrow is the slope, the shadow is the base, and the base is the slope times the cosine of the angle between them. So the length is the arrow's length times the cosine of that angle. Except that a cosine can be negative, and a length cannot. So the cosine gets modulus bars round it, and what the bars throw away comes back as a direction: along the line when the cosine is positive, against it when the cosine is negative.

1:42That is worth pausing on because you have met the move before. When a vector is multiplied by a number, its new length is the old length times the SIZE of the number, and the sign of the number goes into the direction instead. Same idea, second time. And it is easy to check that the bars are doing real work. Take fifteen thousand three hundred and seventy six pairs of a live arrow and a live line on a grid. Write the length without the bars and it comes out below nought at six thousand six hundred and fifty six of them — and there are exactly six thousand six hundred and fifty six pairs whose angle is obtuse. With the bars, it comes out below nought at none.

2:31Now the hazard of this topic, and it is a hazard about words rather than mathematics. The shadow is a vector, and it gets called the projection VECTOR. And there is a number that goes with it, which gets called the projection. Just that, one word shorter. One of them is a vector. The other is a number. They differ by one word, and both words are in ordinary use.

2:59So here is the rule to carry. When a question asks you to find the projection, it wants a NUMBER — and the number it wants is signed, so it can come out negative. You will see exactly where those two part company in about four minutes. Until then, keep both labels on screen. One more thing about the angle, and then we can compute. The angle here gets measured all the way round the turn, so it can be more than a straight angle. But the product we are about to use has its own angle, and that one is pinned to half a turn. Two conventions, one idea, and nothing reconciles them.

3:43Does that matter for an answer? No — and here is why, measured rather than asserted. Take any arrow and reflect it in the line, like folding the paper along the line. The reflected arrow makes the angle the other way round, so under a full-turn convention it has a different angle. But its projection is exactly the same. At all fifteen thousand three hundred and seventy six live pairs, the reflected arrow has the same projection and the same length — while being a genuinely DIFFERENT arrow at fifteen thousand and twenty four of them.

4:23The three hundred and fifty two it does not change are the ones already lying on the line, which the fold leaves alone. So the projection cannot tell you which way round the angle was measured. The wider range costs nothing here. It would cost something if you tried to carry it into the product. Now the recipe, and it is short. Take the arrow of length one pointing along the line. Then the projection of any vector onto that line is just the product of the vector with that unit arrow.

5:00That is it. One scalar product and no angle anywhere. And it is worth seeing why that works. The product of two vectors is length times length times cosine — and one of those lengths is one, so what is left is the vector's length times the cosine, which is exactly what we wanted. The unit arrow is worth a moment too. Divide the line's vector by its own length and you get length one — at all hundred and twenty four live arrows on the grid.

5:34Divide by the SQUARE of the length instead, which is the slip people make, and you get length one at six of them — and those six are exactly the arrows that already had length one, where squaring changed nothing. You will see this projection written three different ways, and they are one line rewritten twice rather than three facts to learn. First: the vector paired with the unit arrow along the line.

6:05Second: the same thing with the unit arrow spelled out — the vector paired with the line's vector divided by its own length. Third: the same thing again with the division moved outside — the product of the two vectors, all over the second length. The third is the one to use, because it does the division last and there is only one of them. Run all three as separate routines over the grid and they land on the same number at every one of the fifteen thousand three hundred and seventy six live pairs. And all three refuse outright when there is no line to project onto.

6:47So: to find a projection you need one product and one length. You never need the angle, and computing the angle and then taking its cosine is the long way round. Four angles are worth knowing by heart, and they take thirty seconds. At a zero angle the arrow already lies along the line, pointing with it. Its shadow is itself. At either right angle the arrow is square to the line and casts no shadow at all. The projection vector is nothing, and the number is nought. On the grid that is two thousand and sixty four live pairs — exactly the two thousand and sixty four whose product vanishes.

7:34Now the straight angle, and this one is worth being careful about, because it is usually paraphrased wrongly. At a straight angle the arrow lies along the line pointing against it. So it still lies on the line — and a thing that already lies on the line is its own shadow. Measured: at all three hundred and fifty two live pairs that lie along one line, both ways round, the projection vector is the ORIGINAL arrow. And it is the original reversed at none of them.

8:11And that is exactly where the two meanings of the word part company. Watch. At those hundred and seventy six pairs pointing opposite ways along one line, the projection VECTOR is the original arrow, whose length is positive. But the projection — the number — is the negative of that length, at every single one of them. One word, one situation, one positive answer and one negative answer. So how often do the two agree? On the grid, the number equals the vector's own length at eight thousand seven hundred and twenty live pairs, and equals MINUS that length at eight thousand seven hundred and twenty.

8:57The first list is the pairs at or under a right angle. The second is the pairs at or over one. They overlap only at the right angles, where both are nought. So slightly more than half the time you would never notice. That is what makes it dangerous. The working rule again: compute the signed number, and read its sign as telling you which side of a right angle the angle is on.

9:27Now a worked one, in components, where none of this shows up. Project two, three, two onto one, two, one. Pair the slots: two, plus six, plus two. The product is ten. The second arrow's squared length is one plus four plus one, which is six, so its length is root six. Divide: ten over root six. Tidy it up and that is five thirds of root six. Positive — so the angle is under a right angle, and the shadow runs the same way as the line.

10:01Notice what did not happen. No angle was found, no cosine was taken, and because the answer came out positive, nothing in this problem would have told you that a projection can be negative. Here is one that collapses if you look before you compute. Project one, minus one onto one, one. Pair the slots: one, minus one. That is nought. So the projection is nought, and you never need the second length at all. The two are square to each other, and a right angle casts no shadow.

10:39People lose time here by computing root two first and then discovering it multiplies nothing. One more, with real numbers in it. Project one, three, seven onto seven, minus one, eight. The product is seven, less three, plus fifty six — sixty. The second squared length is forty nine plus one plus sixty four, which is a hundred and fourteen. So the projection is sixty over root a hundred and fourteen, which tidies to ten root a hundred and fourteen over nineteen.

11:17Now the payoff, and it reaches back into everything you already know about components. Take any vector and project it onto the first axis. The unit arrow along that axis is the first axis arrow, so the projection is the vector paired with it — which picks out the first component and nothing else. So the projection along the first axis IS the first component. Same for the second, same for the third.

11:48The three numbers you have been writing down since components were introduced were three projections all along, and nobody says so. Measured over a hundred and twenty four live arrows against three axes — three hundred and seventy two cases — it holds at all three hundred and seventy two. And it is not something that would hold for any old pairing. Project onto the NEXT axis round instead and you still get the component you asked for at seventy two of the three hundred and seventy two, which is plenty to survive a careless spot check.

12:28The same thing arrives a second way: the vector's length times each direction cosine gives back that component, at all three hundred and seventy two. And those three cosines square-sum to one, at all hundred and twenty four. Last idea, and it is the most useful thing here for anything you meet later. Take two vectors. Split the second one into two pieces: one lying along the first, and one square to the first.

13:02You can do it without knowing anything about projections. Write the parallel piece as some multiple of the first vector, subtract it off, and choose the multiple that makes what is left pair to nought with the first vector. That forces the multiple: it is the product of the two, over the first vector's squared length. Now here is the thing nobody points out. That parallel piece is exactly the shadow. The projection vector, arrived at by a route that never mentions projections at all.

13:39Over fifteen thousand five hundred splits on the grid: the two pieces add back to the original every time, the second piece is square to the first vector every time, the first piece lies along it every time, and it is the projection vector every time. And the divisor is load-bearing. Divide by the SECOND vector's squared length instead — an easy slip, since both are right there — and the remainder is still square to the first vector at four thousand and forty of the fifteen thousand three hundred and seventy six live cases. Right often enough to fool you.

14:22The split on numbers, because it is quick. Split two, one, minus three along three, minus one, nought. The product of the two is six, minus one, plus nothing — five. The first vector's squared length is nine plus one, which is ten. So the multiplier is five over ten. One half. The piece along is half of three, minus one, nought — that is three halves, minus one half, nought.

14:52Subtract it and the piece square is one half, three halves, minus three. Two checks, both of which you should always do. Add the two pieces: two, one, minus three. The original, back. And pair the second piece with the first vector: three halves, minus three halves, plus nothing. Nought. Square, as required. What to keep. A vector projected onto a directed line casts a shadow that lies along the LINE, not along the vector.

15:27The shadow's length is the vector's length times the size of the cosine, with the sign of the cosine going into the direction instead. Modulus bars keep a length a length. One word covers two things: the projection vector, and the projection. When a question asks for the projection it wants the signed number, and that number can be negative. To compute it: one scalar product, one division by the line's length. No angle, ever.

16:01A right angle projects to nothing. An arrow already on the line is its own shadow, whichever way it points. A vector's three components are its three projections along the axes. Same three numbers, and now you know why. And the projection vector is the piece of one vector that lies along another, in the split that separates a vector into a piece along and a piece square. That is the form you will actually use.