Magnitude, the three angles with the axes, and the cosines and ratios they give

0:00A vector, so far, has been an arrow that could be drawn anywhere. Now we pin one end down. Put the tail at the origin. Let the head sit at a point in space with coordinates x, y and z. That arrow is called the position vector of the point. And the moment you do that, something quietly collapses. The arrow's three components are the point's three coordinates. Not corresponding to them. The same three numbers.

0:32Checked at eighteen points, the arrow built from a point never once differs from the point itself. So a point and an arrow become interchangeable, and every question about one becomes a question about the other. That single choice is what makes the rest of this possible. The length comes for free, because you already know it. The distance from the origin to a point is the square root of the sum of the three squared coordinates. That was true before anybody said the word vector.

1:09So the magnitude of the position vector is that same square root. We will call it r. Nothing has been derived here. A distance has been renamed. But the renaming is worth something, because r is about to appear in every formula in this video. Now the actual content. The arrow makes an angle with each of the three axes. Three angles. And one word in that sentence is doing a great deal of work: they are measured from the positive direction of each axis.

1:46Drop the word positive and you have said nothing, because every line has two directions and the two angles differ. Hold on to that. It is the whole reason some of these angles come out obtuse, and some of these cosines come out negative, and neither is a mistake. Here is the derivation you will be shown for the cosine of each angle, and it is a good one. It is also incomplete, and almost nobody says so.

2:18Take the first axis. Drop a perpendicular from the point to it. You get a right angled triangle: the origin, the foot on the axis, and the point. The angle we want sits at the origin. The side adjacent to it runs from the origin to the foot. The hypotenuse is the arrow itself, of length r. Cosine is adjacent over hypotenuse. So the cosine is that edge, divided by r.

2:51And then the argument says: that edge is x. Cosine equals x over r. Do the same twice more and you have all three. Now. Is that edge x? It is a distance from the origin to a foot. A distance is never negative. And x is negative whenever the point is on the wrong side. So the argument gives you the size of x, not x. Let us find out how much that costs.

3:22Take eighteen points reaching into all eight octants, read each against all three axes: fifty four readings. The formula, cosine equals coordinate over r, is right at all fifty four. The triangle argument reproduces it at thirty seven. It fails at seventeen, and those seventeen are not scattered. They are exactly the readings where the coordinate is negative. Not nearly. Exactly. Counted by octant: of the eight, there is one where the triangle argument is right on all three axes at once. In each of the other seven it is wrong on at least one.

4:07The formula is still correct. The argument for it covers one eighth of space. So let us prove it properly, and the tool is older than any of this. The law of cosines. Take any two arrows from a common point, and the arrow between their tips. Then the square of that third side equals the sum of the two squares, minus twice the product of the lengths, times the cosine between them.

4:37Rearrange it and the cosine is written out of three distances. Three lengths, and nothing else. No triangle dropped to an axis. No adjacent edge. No coordinate, and therefore no sign to get wrong. Now aim it at an axis. Take our arrow and the unit arrow along the first axis, and turn the handle. The squared lengths cancel almost everything, and what is left is x over r. At all fifty four readings, in all eight octants. That is the general proof, and it is not a harder argument than the triangle. It is just a less familiar one.

5:20The triangle is a picture of the answer. This is a derivation of it. Name the three cosines l, m and n. Now read the three formulas backwards. If l is x over r, then x is r times l. Same for the other two. So the point is r times the triple of cosines. The length, and a direction. And those three products, r l, r m and r n, get a name of their own. They are the direction ratios, written a, b and c.

5:56Which means, and this is worth pausing on, the direction ratios are the coordinates over again. Checked at all fifty four places: length times cosine returns the coordinate, every time, with nothing left over. So if a vector arrives already written in components, you can read its direction ratios straight off. They are sitting there. The three cosines are not independent. Square them and add. x over r, squared, plus y over r squared, plus z over r squared. That is x squared plus y squared plus z squared, all over r squared.

6:42But the top is exactly what r squared is. So the whole thing is one. l squared plus m squared plus n squared equals one. Measured at all eighteen points, and it holds at all eighteen. That is not a formula to memorise separately. It is the length formula, divided by itself. One identity, three numbers. So how free are they? Fix two of them. The identity then fixes the square of the third, and a square fixes a number up to sign.

7:18Nine pairs were tried. Five of them leave room, and each of those admits two values for the third, one positive and one negative. Two of the nine use the identity up exactly, and admit one value: nought. And two overspend it. Their squares already come to more than one, and there is no third cosine at all. So two of the three are free and the third is decided up to sign. Which is why a question can hand you two cosines and still ask you for an angle, and expect an answer.

7:57Now the trap this whole topic exists to prevent. You have two triples. The cosines l, m, n, and the ratios a, b, c. Same direction, and they look interchangeable. They are not, and the fastest way to see it is to square and add both. The cosines come to one. Eighteen points out of eighteen. The ratios come to r squared. Eighteen out of eighteen for that too, which is a different statement entirely.

8:29How often do the ratios square to one? At none of the eighteen. Not one. Because r squared is one only when r is one, and none of these eighteen arrows happens to have length one. So when do the two triples agree? Exactly when the arrow already has length one. Then r is one, r times a cosine is the cosine, and the two triples are the same triple.

8:58Eight arrows were brought down to length one and put through both. All eight: both triples obey the identity, and the two triples agree outright. At their original lengths, only two of the eight agreed, and those two were the two that already had length one. Which gives you the cleanest sentence in the topic. Take the three cosines and read them as an arrow. Its squares add to one, so its length is one.

9:28The direction cosines are the unit vector in that direction. That is not a new fact. It is the identity, said out loud. Work one. Components one, one and minus two. The ratios are the components. One, one, minus two. Nothing to compute. The squared length is one plus one plus four, which is six, so r is the square root of six. The cosines are each component over the root of six. And rather than write a decimal for that root, check it the honest way: multiply each cosine back by r and you get its own component back. All three.

10:13Now the identity. The squares are one sixth, one sixth, and four sixths. Which is six sixths. One. And notice the third cosine is negative, because the third component is. Nobody made a mistake. Work two, and this is the item that rewards reading slowly. A vector runs from the point one, two, minus three to the point minus one, minus two, one. In that order. Subtract: the components are minus two, minus four, four. The squared length is four plus sixteen plus sixteen, which is thirty six, so r is exactly six. No root.

10:58The cosines are minus one third, minus two thirds, and two thirds. Two of the three are negative. Which is correct, and which is the whole point: two of these angles are obtuse. The arrow leans away from the positive direction of those two axes. Now travel it the other way instead. Same two points, opposite order. The length does not move. Still six. But all three cosines change sign. Every one.

11:30Which was checked by putting the reversed arrow through the very same routine, so this is a comparison, not a claim. The two points do not determine the answer. The two points and a direction of travel do. Work three. The arrow with all three components equal to one. Squared length three. All three cosines are the same number, and their squares add to one, so it is equally inclined to all three axes.

12:02But now turn the question round, which is where it gets interesting. Suppose you are told only that an arrow is equally inclined to all three axes. What are its cosines? Equal angles force equal cosines. Call the common value c. The identity says three c squared is one. So c is plus or minus one over the root of three. Two answers. And both are genuine. Both were built by the same routine with only the sign changed, both obey the identity, and each is exactly the other turned round.

12:42They are an arrow and its opposite. Keeping only the plus is not tidying up. It is losing half the answer. Work four, and this one is only answerable because of the identity. A unit vector makes a third of a half turn with the first axis, and a quarter of a half turn with the second. The angle with the third is acute. Find it. Two angles given, one asked for. The first cosine is a half, whose square is a quarter. The second is one over the root of two, whose square is a half.

13:21A quarter and a half come to three quarters. So a quarter is left for the third square. Which makes the third cosine plus or minus a half. Both were computed; both obey the identity perfectly well. So the identity alone cannot finish this. The word acute finishes it, by picking the positive one. That word was not decoration. Without it the question has two answers, and this is exactly the freedom we counted earlier: two free, the third up to sign.

13:58One last thing, and it settles which of the two triples is really about direction. Take an arrow and multiply it by seven different numbers, four positive and three negative. The cosines survive all four positive scalings unchanged. Exactly unchanged. Under the three negative ones, all three cosines turn round together. And the ratios? They survive exactly one of the seven: multiplying by one. Because the ratios carry the length, and the cosines threw it away.

14:35So the cosines describe a direction. The ratios describe a direction and a size. One caution. As defined here, the direction ratios are one specific triple. Later, working with lines, you will see any proportional triple called direction ratios too. Both usages are current, and they are not the same definition. Notice which one you are being handed. Five things. Pinning the tail to the origin makes a point and an arrow the same object, and the length is the old distance formula wearing a new name.

15:18The three angles are measured from the positive axes, and the cosine of each is that coordinate over the length. The triangle argument for it is right in one octant out of eight; the law of cosines proves it in all eight, out of three distances and no coordinates at all. The three cosines square and add to one. The three ratios square and add to r squared, and the two triples coincide exactly when r is one.

15:51Only two of the three cosines are free. The third is fixed up to sign, and something outside the identity has to choose that sign for you. And a negative cosine is not an error. It is an obtuse angle. Reverse the arrow and all three flip together, which is why the order of two points matters and why an equally inclined direction has two answers rather than one. The length is the thing that never goes negative. Everything else here is allowed to.