Splitting a vector along the axes so the algebra becomes coordinate arithmetic

0:00This is the hinge of the whole subject. Up to here, vectors have been arrows you draw and move about. After here, they are triples of numbers you can compute with, and everything after this runs on that. And the changeover is done in two halves of very different quality. The hard half is done properly. Three unit vectors are built rather than assumed. A position vector is taken apart into three pieces, using one law twice. And a length formula is derived, from two right triangles.

0:39The easy half is done badly. Four rules for computing with components arrive as a bare list of eleven lines, and not one of them is argued, although three of the four fall straight out of a law you already have. So we are going to do the second half properly, which takes about two minutes, and end up with one idea instead of four things to remember. Start with the three arrows everything is written in.

1:10They are not postulated. They are built. Take three points, one on each axis, each exactly one unit from the origin. Draw the position vector of each. Those three arrows are the ones that get the hats. And their length is a consequence, not a stipulation. All three come out at exactly one. That matters because it is the only reason the notation works later. If you check it: three arrows, all three of length one, three genuinely different arrows, and all six ordered pairs of different ones at right angles.

1:50And the length is being measured, not granted. Run the same test on three points two units out along the same three axes and it finds none of length one. Now take any point in space and rebuild its position vector out of those three. Here is the construction, and it is worth going slowly, because everything after this rests on it. Drop a perpendicular from the point straight down to the floor plane. Call where it lands the foot.

2:23Now walk to the point in three legs. Out along the first axis. Across to the foot. Then straight up to the point. That is the triangle law used twice: once in the floor of the box, and once going up. Two steps, not one. And that is what the foot of the perpendicular is for. Over a grid of a hundred and twenty five points, the foot lies in the floor plane every time, and at the hundred whose height is not nought it is a genuinely different point from the one above it.

2:59The walk has three legs, and each of them lies along one axis or is nothing at all, at all one hundred and twenty five. And both triangles are right-angled. Measured, at all one hundred and twenty five, not assumed from the picture. Now, that last claim sounds like it could not fail. Let us make sure it can. There are two corners in that walk: where the first leg stops, and where the second leg stops. Move either one and run exactly the same two tests.

3:35Put the floor point somewhere that is not below the point. The legs lie along the axes at twenty five of the hundred and twenty five, and the angles are square at forty five. Put the first step off the first axis instead. Twenty five and forty five again. Against a hundred and twenty five and a hundred and twenty five for the walk the construction actually makes. So both corners are doing work. The right angles are a property of that particular walk, and not of splitting a vector in general.

4:13Two names now, and they differ by one word and denote different kinds of thing. The three plain numbers are the scalar components. The three pieces they multiply are the vector components. One list is three numbers. The other is three vectors. There is also a third name floating about for the first list, rectangular components, which is the same three numbers again. Why this matters: there are questions that ask for both lists, and a student who thinks they are the same thing writes down half an answer.

4:51On a worked arrow with parts minus seven and six, the three numbers are minus seven, six and nothing. The three pieces are minus seven along the first axis, six along the second, and the zero vector. And the difference is not cosmetic. Add the three numbers and you get a number, minus one. Add the three pieces and you get the arrow back. Never say components on its own. Say which list you mean.

5:23Now the length, and this is the one derivation in the topic. You have a box. The point is at the far corner, the foot is on the floor, and there are two right triangles in there. Apply Pythagoras in the floor triangle to get the distance out to the foot. Then apply it again in the upright triangle to climb from the foot to the point. Two applications. The squared length is the sum of the three squared parts, and the length is its square root.

5:57Over the grid, two applications give the magnitude at all one hundred and twenty five points. One application gives it at twenty five. Which are exactly the twenty five points whose third part is nought, where there was nothing to climb. And the derivation is spending the right angles, not the drawing. Run the same two-step routine on the slanted split from a moment ago and it gets the length at forty five points instead of a hundred and twenty five — the same forty five where that split happened to be square.

6:35Here is something worth noticing about that formula, which you have almost certainly seen before. The magnitude of a position vector, as the square root of the sum of the squared coordinates, turns up twice, a long way apart, in two completely different roles. The first time, it is handed to you. It is the distance formula from an earlier year, quoted, with no argument. The second time, it is derived. From a box and two right triangles.

7:09The second is a proof of the first. And nothing anywhere says so. So if you have been treating that formula as something to accept, you have a proof of it, and it is three lines long. Now the payoff of the whole topic, delivered as a bare list. Two vectors in component form. Four facts follow. Their sum is got by adding corresponding components. Their difference by subtracting them. They are equal exactly when all three pairs of components agree. And a scalar multiple is got by multiplying each component.

7:49That is it. Eleven lines, four rules, no argument for any of them. They are all true, of course. Tried over a grid of twenty seven arrows: seven hundred and twenty nine pairs for the sum, the difference and equality, and a hundred and sixty two tries for scaling. Nought exceptions in each of the four. But being true is not the same as being understood, and four separate facts is three more than you need.

8:22Three of the four are the same idea, spent. Take the sum. Write each vector as its three pieces. Now you have six pieces to add. Spread the addition across all six — that is one of the distributive laws — and then gather the ones lying along each axis together. Three of the six lie along the first axis. Three along the second. Three along the third. Collect them, and out falls the componentwise rule: the first parts added, the second parts added, the third parts added.

9:00That is not a new fact. That is the distributive law being spent. Over the same seven hundred and twenty nine pairs, gathering the six pieces onto the axis each one lies along gives the plain sum every time. And the gathering is where the work is. Run the same routine but collect onto the wrong axis and it lands on the sum at forty five of the seven hundred and twenty nine.

9:30The difference rule is the same argument with a minus sign. The scaling rule is the same distributive law spread the other way. Three down. The fourth one is different, and this is the interesting part. The equality rule says: two vectors in component form are equal exactly when all three pairs of numbers agree. One direction is obvious. If the numbers agree, the vectors agree. The other direction is not obvious at all, and it is the one you lean on in every problem you solve. You have an equality of two vectors and you read off three equations. What entitles you to do that?

10:16This: a combination of the three axis arrows is the zero vector only when all three numbers are nought. And that sentence appears nowhere. So measure it. Take a hundred and twenty five triples of numbers. Exactly one of them sends the three axis arrows to the zero vector — the triple of noughts. And no two different triples write the same arrow. That is what lets one equality of arrows become three equalities of numbers.

10:50And it is a fact about those three arrows, not about arrows in general. Run the same two counts on the first axis arrow, the second, and their sum. Five triples vanish instead of one, and two hundred pairs of different triples write the same arrow. With those three, you could not read off anything. With the axis arrows, you can. That is the missing sentence, and it is worth thirty seconds.

11:21So an equality of vectors is worth three equations, and it is worth seeing that as a narrowing. Take a target vector and a hundred and twenty five candidates. Match the first component and twenty five survive. Match the second as well and five survive. Match the third and exactly one survives. A hundred and twenty five, twenty five, five, one. That is why a question that sets two component forms equal and asks for three unknowns is really three questions, and why you get all three for the price of one comparison.

12:03A short one, and it goes wrong in both directions. One-two and two-one. Both have squared length five. They are not the same vector. Equal vectors do have equal lengths — over the whole grid, not one pair of equal vectors has different lengths. The converse is nowhere near true. Of fifteen thousand six hundred and twenty five ordered pairs, two thousand two hundred and seventeen share a length. Only a hundred and twenty five of those are the same arrow.

12:38Two thousand and ninety two pairs have one length and two different arrows. A length is one number. A vector in space is three. Comparing lengths answers a much smaller question than the one you were asked. Now the coordinate version of collinearity. Two vectors are along one line when one is a number times the other. In components, that says each part of the second is the same multiple of the corresponding part of the first.

13:13So write three ratios — second over first, for each of the three parts — and set them equal. If they agree, the vectors are collinear, and their common value is the multiplier. And that last bit is worth checking rather than assuming. Search the grid independently for the number that carries the first arrow to the second — no ratios, just look. It finds one at three hundred and fifty two pairs of the fifteen thousand six hundred and twenty five and refuses the rest.

13:49At every one of the hundred and sixty pairs where the chain of ratios can be written and holds, its common value is exactly the number the search found. So the chain works. But there is a catch, and it is a big one. Look again at what the chain of ratios requires. You are dividing by each part of the first vector. If any part of the first vector is nought, that ratio is not a number, and the chain cannot be written at all.

14:22This is not a rare edge case. Of the fifteen thousand six hundred and twenty five ordered pairs on the grid, the chain can be written at eight thousand and cannot be written at seven thousand six hundred and twenty five. Nearly half. And worse: of the three hundred and fifty two genuinely collinear pairs — the exact cases the criterion exists for — the chain cannot be written at a hundred and ninety two of them. More than half.

14:55Where it can be written, it is right every time. Nought disagreements with the geometry across all eight thousand. So the chain of ratios is a correct test with a hole in its domain. If a part is zero, go back to the definition: is one a number times the other? That question always has an answer. Thirty seconds on something that is used constantly and stated nowhere. Plenty of the work here is done with two components instead of three. And nothing ever says what a two-component vector is.

15:34It is a three-component vector whose third component is nought. That is all. And that answer is worth one measurement, because the reason it works is closure. Take the twenty five arrows of the grid whose third part is nought. Add any two: the answer still has third part nought. Subtract any two: same. Scale any one by any number: same. Not one escape, in any of the three. And that is a property of that particular slice, not of slices in general. Take the twenty five arrows whose third part is one instead. Not one of their six hundred and twenty five sums stays there, and not one of their six hundred and twenty five differences either.

16:23So the plane sits inside space as a place the arithmetic cannot leave. Two dimensions is not a different subject. It is this one with a nought written in. One more, and it is the odd one out. Everywhere else, components come from two points: subtract one from the other. There is one place where they come from an angle instead. The question is: what are all the vectors of length one lying flat in the coordinate plane?

16:56And the answer is a family, not a vector. Put the cosine of an angle on the first component and the sine of the same angle on the second. Squares sum to one, for every angle. As the angle sweeps once round, the tip traces the unit circle. Check it at thirty two exact points of that circle, built from whole-number right triangles so nothing here is a decimal. All thirty two have length one, and all thirty two are different vectors.

17:30And length one is being measured, not granted: the same test, shown those same thirty two directions at twice the length, passes none of them. The sixth-of-a-right-angle case comes out the same way: root three over two, and a half, and their squares do sum to one. And notice what those two components are. In two dimensions, the components of a unit vector are the two direction cosines. Which is the bridge to the last thing.

18:04Before that, one small payoff, because it is the kind of thing that is easy to walk past. Three magnitudes to compute, and the third one is a gift. The first arrow has all three components equal to one, so its squared length is three. The second has components two, minus seven and minus three, so its squared length is sixty two. The third has one over root three in every slot, with a minus sign on the last. Its squared length is a third, plus a third, plus a third, which is one.

18:41So the third one is a unit vector, and nothing in the question says so. Noticing is the whole reward. And that is worth making a habit of. Whenever you compute a magnitude and it comes out at one, stop and say what you have just found. One sentence, easy to skip, that closes a loop opened a long way back. The three scalar components are also called the direction ratios of the vector.

19:12Now, direction ratios were defined long before components existed, and defined completely differently: the length of the vector, times each of its direction cosines. So those are two definitions of one triple, arriving from opposite directions, and one sentence quietly identifies them. Worth checking. Compute them the long way. Take each direction cosine, multiply by the length, at every one of the hundred and twenty four nonzero arrows of the grid.

19:49They come back as the components themselves. All one hundred and twenty four times. And the routine refuses the one arrow that has no direction at all. The cosines it uses are genuine direction cosines, by the way — their three squares come to one at all one hundred and twenty four. And the length in that product is doing real work. Drop it and use the cosines alone, and you get the components back at six of the hundred and twenty four — exactly the six that already had length one.

20:26So it is not a coincidence of notation. Length times direction cosine is the component, and it always was. Six things. The three axis arrows are built out of three named points, and their length one is a consequence of that construction, not an assumption. A position vector comes apart into three pieces by two applications of the triangle law, and both corners of that walk are doing work. The magnitude formula you were handed in an earlier year has a proof, and it is Pythagoras applied twice in a box.

21:08Three of the four componentwise rules are the distributive law spent once. The fourth rests on a fact nobody states: a combination of the three axis arrows is nought only when all three numbers are. Collinearity as three equal ratios is correct wherever it can be written, and it cannot be written at more than half of the collinear cases on a grid, because of a nought. And the components were the direction ratios all along.

21:41From here on, a vector is a triple of numbers, and everything is arithmetic.