So first thing we'll do is talk about a procedure that we're going to follow to go through this.

So first thing we're going to do is we're going to pretend that the A1, A2, Z, three-axis parallel pipette within the hexagonal unit cell is actually X, Y, Z orthogonal.

We're going to sort of imagine that that's where our direction exists, and then we're going to proceed from there and determine the three-index notation for that vector.

Now we're going to call this sort of a temporary three-index system because we're not going to stop there.

We could.

In fact, a while ago, a few decades ago, people used to do that.

But the convention nowadays is to go to four-axis system to avoid any confusion with cubic systems.

So we'll take this temporary three-axis system then and then convert it to a four-axis system and then just enclose it.

So it's sort of that straightforward.

So let's go ahead and look at an example problem now.

So this first example problem, we've got a vector that originates off on the left-hand side of the base of the hexagonal unit cell and then travels up.

And so what we have to do, the first step, and this is really I think the hard step, is we have to be able to picture that direction within the A1, A2, Z three-axis system.

Our conventional origin in the middle of the unit cell there, in the middle of the base of the unit cell, is usually a place that we're comfortable visualizing this.

So what we could do is we could actually just translate that vector, translate it over, so that it now originates at our conventional origin, right in the very center of the basal plane.

And if we do that, we'll see the direction that the vector travels up in the Z direction and directly over the negative Y, sorry, the negative A3 axis.

And it travels by one step over in the horizontal direction, that is the A lattice parameter, and then during that distance, that travel, it travels up a rise of the lattice parameter.

So if we look at that vector in the three-axis system now, it would be the vector that originates at the conventional origin and then travels one over in the X, one over in the Y, and one up in the Z.

So it's in fact one of those cube diagonals, and specifically, it's the 1, 1, 1 direction in a cubic system.

So U prime, V prime, and W prime are 1, 1, and 1.

So then all we need to do is convert that to the three-axis system.

And actually, another thing we could do, instead of shifting the vector over, another thing we could do is we could just define the origin of the vector as it originally sat as our 1, 1, A2 axis originating from that origin.

It's entirely equivalent.

I either define a new origin or translate the vector over.

They're the same thing.

OK.

So anyway, we've got U prime, V prime, W prime, 1, 1, and 1.

So therefore, U is equal to 1 third times 2 times 1 minus 1.

V is, again, 1 third times 2 times 1 minus 1.

And T, the third, this new index, is, remember, it's equal to the sum of U and V made negative.

It's not U prime plus V prime made negative.

So just make sure you don't make a mistake with that.

And so we'll have that U is 1 third.

V is 1 third.

T, then, is negative 2 thirds.

And W is just W prime, which is 1.

So we're almost done.

We've just got the 3 in the denominator there, a little pesky little 3.

And so we're going to just multiply across by 3 to clear the fractions.

And we're going to end up with 1, 1, negative 2, and 1.

Sorry, and 3. 1 times 3 is 3.

So our enclosure, then, in square brackets, is going to be 1, 1, 2 bar, 3.

All right, fantastic.

Let's look at another vector now.

And this one is a little bit more challenging, perhaps.

So what we have to do is we have to translate that into our A1, A2, Z, 3-axis system, or define a new origin around there.

I think in this case, it's easier to translate the vector rather than draw the axes, but you could do either.

Either one is the same.

So if we translate it over, you can realize that, in fact, it resides entirely over the A2-axis.

So that means if you translate it back into space, back along the positive A2-axis, it'll just travel exactly over the A2-axis with no component hanging off into the A1 direction.

So when we translate it back by one A lattice parameter in the positive A2 direction, it now originates in the base, in the conventional origin, right in the very middle of the basal plane, and travels out.

But you'll notice that it travels two steps before it makes one rise in the Z direction.

Therefore, it's going to exit our unit cell at half the height.

So now we've got this vector that's going out straight along the A2 direction, with a rise for that one step in the A2 direction, a rise of 1 half.

So that vector in a three-axis notation would be, well, it's got no component in X.

It's got 1 in Y and 1 half in Z.

So we've got 0, 1, and 1 half.

Multiply across by 2, we get 0, 2, 1.

So that's the 0, 2, 1 vector.

OK.

So then we just have to convert from U prime, V prime, W prime to UVTW for an index system.

So let's do that.

And if we want to convert to U, we've got 1 third times 2 times U prime, which is 0.

So minus 1 times 1 third is negative 1 third.

And then, sorry, minus 2, minus 2.

So it's negative 2 thirds.

And then we've got V being 1 third times 2 times 2 minus 0.

So that's 4 thirds.

And then we sum U and V and make it all negative.

So we've got 4 thirds minus 2 thirds is 2 thirds made negative.

So that's negative 2 thirds.

And then W is just W prime, which is just 1.

So our indices are almost done.

Again, we've got to clear across to get rid of the 3 in the denominator.

And so we're going to end up with 2 bar, 4, 2 bar, 3 once we've multiplied across by 3.

So that was fairly challenging.

Let's look at one more.

And this one here, again, we've got to picture this somewhere within our A1, A2, Z 3-axis system somehow.

So we could translate it over, or we could just define the origin where it exists, where the origination of the vector resides right now.

And I think that's probably the easiest way to do it.

So the start of the vector, let's call that the origin.

And we can draw in our A1, A2, Z parallelopiped around that.

Hopefully, you'll be able to see how that makes, or what that would look like in our XYZ orthogonal system.

So in fact, it's traveling out along the positive A1 direction.

It rises up a little bit.

It has a little component into positive y.

But it's going to come out of that front face, essentially, of the XYZ cube or system, OK?

And so that's the first thing you need to picture is that.

It's a little difficult to see exactly where it's going to exit the unit cell.

Hopefully, if you sketch in your axes around the origin of the vector, it'll be a little more clear.

But you can see, if it actually traverses two steps along the A1 direction before it rises up 1 and Z, that means that you're going to have it exiting the front face at half the height.

OK, so that means we've got this vector now that originates at the conventional origin of our cube and travels out towards us, towards you this way, I suppose, and rises up by half of the unit cell, half the height in the Z direction, before it exits that front face.

So the point coordinates of that point where the vector exits the unit cell, starting at 0, 0, 0, would be 1 in the X, 1 half in the Y, and 1 half in the Z.

So we've got a vector that you multiply across by 2.

The vector in 3-space, or in the 3-axis system, is 2, 1, 1.

So then we just have to convert 2, 1, 1 into our 4-axis system.

So 1 third times 2 times 2 minus 1 is going to give us 4 minus 1, 3.

That's actually 1.

That's actually 1, isn't it?

And then V is going to be 1 third times 2 times 1 minus 2.

So that's 0.

OK, so that's 0.

It is 0, in fact.

And then T is 1 plus 0, made negative.

So it's negative 1.

And W, of course, is just W prime, which was 1.

So our 4-axis notation now for that vector is just 1, 0, 1 bar, 1.

OK?

But that was hard.

It looks like an easy answer, but it was a challenging question.

I hope that helped.

Thanks.