Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Cartesian coordinate system in space. This is considered a right-handed system. There is a difference between right-handed and left-handed systems and ours is a right-handed system. For the Cartesian coordinate system, ordered triplets are going to be of the form XY and Z. The X is going to be how far out we come. Y is how for right and left. Z is how for up and down. The only standard consistent labeling for the octants because there will be eight of them. If we think about the Z going up is positive, so the Z coming down would be negative. The X coming out of this is positive. So the X going backwards is negative. The Y going to the right is positive, the Y going to the left is negative. So there are 8 octants depending on the application. The other seven labels are not consistent, but it is consistent that the first octant is the plus plus plus distance between two points. The distance between two points is going to look very similar to the distance formula when it was in a Cartesian coordinate system in a plane. So the absolute value of P1P2 equal the square root of the quantity X2 minus X 1 ^2 + y two minus Y 1 ^2 + C two minus C1 squared. Let's look at this proof. If we have a picture and we have two given points P1P2, we're trying to find the distance here. This distance is going to be based off of looking at the Pythagorean theorem several times. So if I thought about what's the distance between .1 and a right here, Well, if we can look at .1 and A, we realize that the Y ones and Z ones are both the same. So really the only distance is occurring in the X direction. So P1A has got a distance of the absolute value of X 2 -, X one. Now if we wanted to think about AB, that's this distance here. And if we look at this, the X's and the Z's have the same. So now we're really only talking about the change in the Y direction. So the absolute value of Y 2 -, y one, if we wanted to think about BP two, that would be this location here. And the only change now is going in the Z direction. So we're going to start by looking at this triangle, that is P1A and B. So we can see that by Pythagorean theorem, P1B quantity squared equals P1A squared plus AB squared. Well, the P1B we don't know right now, but the P1A we said a minute ago was just the difference of the XS and the AB was just the difference of the YS. So now if we look at the new triangle that's being used by P1, BP 2B and then the distance between P1P2, that is indeed a right triangle. It's just standing up on its side. It's in three dimensions. So P1P2 squared is P one b ^2 + P 2B squared. The P1B squared is this location. Here the BP 2 ^2 was just the Z 2 -, C one. If we put it all together, we can see that we do indeed get the distance between the two points. Formula equation for a sphere in space is going to be very similar. It's just going to have taken this last equation and squared each side. So instead of solving for P1P2, it's really just going to be this distance formula in the form X -, X not squared plus y -, y not squared plus Z -, Z not squared equaling a ^2. Now the difference is that X not Y not Z not is a given point and the center is A and the radius. Where as before we knew X1Y1Z1 and X2Y2Z2, now we know different information. So in this equation for the sphere, we're saying what are all the points XY and Z that are a certain distance from a given point. So let's look at some examples. We want to describe the set of points in space. So if X is -1 and Z is 0. If we thought about our coordinate system here, if X is -1, we're going to go back one, but the Z is 0, so we didn't go up or down any. So if we think about we went back one, but we didn't go up or down any, but we don't have any restrictions on Y. So we know that Y could go left and right as far as we wanted. So this is really going to be a line through -1 is what our X is that's required. Our Y we don't know and we know our Z is 0. So it's a line through -1 Y zero. And this line is parallel to the Y axis. So we went back one on the axe. We didn't go up or down any, but we didn't know how far right or left we went. Oh, that was not a straight line. So a line through -1 Y zero parallel to the Y axis. Our next example, X ^2 + y ^2 + Z ^2 equal 25. Well, we know that this is a sphere, and this sphere is now getting intersected by Y equaling -4. Y equaling -4 is really a plane. So what we're going to do is we're going to say X ^2 plus, because we know what the Y is, we're going to put it in Z ^2 equaling 25. So we end up with X ^2 + Z ^2. This is 16, and if I subtract it over, we get 9. So we get a circle in the Y equal -4 plane. This one's a little harder to draw. So X ^2 + y ^2 + e ^2 equal 25. It means that we're going to go out five in each direction. Sometimes it's easiest to draw it in the YZ plane first because that's going to be, I'm going to just draw the top half to start with, and then we can connect the YX portion. And then we could put some lines trying to show the three dimensionality. Oh, that looks bad, sorry. OK, let's not do that part. So, but we wanted Y equaling -4 So if we came this way, we know that our plane Y equal -4 is going to be right here. So if we thought about cutting that sphere with a plane, that's how we get the circle. How about if we had X ^2 + y ^2 less than or equal to one when Z = 0? So this part says we're now on or inside because of the less than or equal portion. So this is a circle that's intersected by a plane. So this is the circle in the XY plane. Why is it in the XY plane? Because our Z was zero, we didn't go up or down any. This next one's going to be the circle on or inside the circle in the plane Z equal 3. So all it did was it took the circle and it raised it up three or we're in a parallel plane to the XY plane. This last one, if there isn't any restrictions, what's going to happen is this is going to turn into a solid cylinder because we want X ^2 + y ^2 equaling 1. So we know that X ^2 + y ^2 equaling one is going to be a circle. And it's not going to look like a circle in three dimensions because it's we're looking at a three-dimensional object in two. But because there wasn't any restriction on the Z, it's going to go all the way up and down. So it's going to be a bunch of circles basically stacked one on top of another on top of another, and that's going to make us, oh, that's going to make us the solid cylinder. So all these circles stacked on top of each other. What about the plane through three negative 1/2? That's perpendicular to the X axis. So if we look at the plane through three negative 1/2, let's start by figuring out where three negative 1/2 is. We're going to come out three, we're going to go to the left one, and then we're going to go up two. So we came out three to the left one and up two. So this is our .3 negative 1/2, and if we want it to be perpendicular to the X axis, perpendicular to here, we'd have our plane coming this direction if it's perpendicular, so it would be represented by X equal 3:00. What if we wanted to do it perpendicular to the Y axis this way? So perpendicular here would be Y equaling -1 because basically we have to go through that point, and if we're perpendicular, we're going to have the Y equal or the X equal the actual value of the point. How about another example? How about a distance between the origin and the .0 to 0? So let's start by understanding that XY and Z is any given point, and we want the point that's the equal distance between the origin and this point. So we're going to use the distance formula that says X - 0 ^2 + y - 0 ^2 + C - 0 ^2. That's the distance between some generic point in the origin, and we want it to equal the distance between some generic point and zero to 0. When we solve this out, we're going to square each side. We're going to see that the X squareds and Z squareds cancel, and we're going to get y ^2 on the left. And if I go ahead and foil out the right side, I get y ^2 -, 4 Y +4. If I solve that, I can see that Y equal 1. So the thing that is the points that are equal distance between the origin and a given .020 are all of the points on the plane Y equal 1. We thought about looking at that pictorially. Here's the origin zero to the right two and not up or down any. So if we wanted equal distance between there, wouldn't it make sense that the Y had to equal 1? So we wanted all of those points on that plane. The next example we're going to do is we're given an equation and we actually want to find the center and the radius. So the first thing I would recommend doing is to get A1 coefficient on your X ^2, y ^2, and Z ^2 by dividing everything through by three. I'm going to also group my like variables. So I have y ^2 + 2/3 Y, I'm going to have Z ^2 -, 2/3 Z, and then that's all going to equal 3. Now we're going to complete the square, so we're going to take half of the linear coefficient, IE half of 2/3. Remember, when I take half of something, it's just really multiplication. So half a 2/3 is 1/3, and then we're going to square it. So I'm going to add 1 ninth. But if I add 1 ninth to one side, we've got to add 1 ninth to the other. I'm going to do the same thing for the Z half. A negative 2/3 is negative 1/3, but when I square it, it still turns out positive. So we're going to add that to each side. So now when we complete our square, we get X ^2 + y + 1/3 quantity squared plus Z -, 1/3 quantity squared equaling. Let's see, three is going to be 27 ninths, 2829 ninths. So our Center for the sphere is going to be zero negative one third, one third with the radius of sqrt 29 / 3, actually 3 dimensions. We call it a instead of R. Thank you and have a wonderful day.