Three-dimensional coordinate systems level 4 many the formulas established for a
two-dimensional coordinate system can be extended to three dimensions, it
turns out that many of the formulas that you’re
used to working in R-squared have natural extensions in R-cubed, the first of these formulas is the all familiar midpoint formula
recall from your previous math classes that the midpoint of a line segment denoted as P sub M where M stands for midpoint, joining the
points P-one with coordinates x-one and y-one and P-two with coordinates x-two and y-two has midpoint coordinates x sub m and y sub m equal to the coordinates x-one plus x-two divided by two and y-one plus y-two divided by two respectively. In the same manner the midpoint of a line segment P sub m in space joined by the points P-one with coordinates x-one y-one and z-one and P-two with coordinates x-two, y-two, and z-two has coordinates x sub m, y sub m and z sub m equal to x sub one plus x sub two divided by two, y sub one plus y sub two divided by two and z sub one plus z sub 2 divided by two
respectively. Notice how we included a new variable, in this case we included
the variable z for the most part many of the equations
that you were introduced in your previous math classes can be extended to three dimensions, by
just including this additional variable z let’s go over an example. Find the midpoint of the line segment connecting the points (4,3,1) and (-2,5,7). Let’s go ahead and
plot these points in a three-dimensional Cartesian coordinate system just so we can get a visual
representation of where these points are located in space. The first point has coordinates (4,3,1) so we plot this point by starting at the
origin and moving four units towards the positive x-direction, then we move three units
towards the positive y-direction, and lastly move one unit
towards the positive z-direction in the same manner the second point is
plotted by starting at the origin and moving two units toward the negative
x-direction then moving five units towards the positive y-direction and lastly seven units towards the
positive z-direction this problem is asking us to find the
coordinates of the midpoint, a line segment
connecting these two points so if you imagine a line connecting these two points in space, we’re trying to find the point
that is exactly halfway between these two endpoints hence the midpoint, also keep in mind that geometrically the midpoint divides a line into two equal parts, finding the
midpoint is easy but really understanding what a
midpoint represents from a geometric point of view is much more powerful and useful when it
comes down to it the coordinates of the midpoint of a
line segment is nothing more than the averages or the arithmetic mean of the coordinates of the endpoints so the average of the x-coordinates is equal to four minus two divided by two which is equal to one. The
average of the y-coordinates is equal to five plus three divided by two
which is equal to four and the average of the z-coordinates is equal to seven plus one divided by two which is equal to four. So the coordinates of the midpoint is equal to (1,4,4). Alright let’s move along to the next formula recall that in order to find a distance between two points P-one with coordinates x-one and y-one and P-two with coordinates x-two and y-two in other words the line segment from P-one
to P-two in two dimensions we use the distance formula which was derived by using the Pythagorean theorem in the same manner the formula to find the
distance of a line segment in space can be derived
by using the Pythagorean theorem twice, let’s go over this simple exercise
so we can work out or spatial visualization skills say we have two random points in space
let’s call them point one with coordinates x-one, y-one and z-one and point two with coordinates x-two, y-two and z-two, in addition lets create a box that includes points P-one and P-two as opposite vertices where point one is located in the bottom
corner of the box and point two is located in the opposite top
corner of the box this way both points have completely
different x, y and z-coordinates since these specific vertices don’t share a
common face on the box if they did then we would be working on
a boring two-dimensional problem if both points where on the same face, okay in order to find the distance between point one and point two which is represented by the red line
segment we need to relate the length of this segment by using the coordinates of point one and point two, notice that we can actually
find the coordinates of the other vertices by using the coordinates of P-one and P-two, for example I’m going to label the vertex directly across P-one towards the positive
x-axis as vertex A and label the vertex below point two as vertex B, now in order to assign coordinates to these
vertices in terms of the coordinates of P-one and P-two we notice the following, the front face of
the box contains point two this phase is parallel to the yz-plane
this means that every single vertex in this plane has x-two as their x-coordinate including vertex A and B, similarly the faces on the sides of the box are
parallel to the xz-plane and contains point one in the left face and
point two in the right face, this means that every
single vertex on the left face will have the same y-coordinate as point one in this case y sub one, so vertex A will have y sub one as it’s y-coordinate and every single vertex
in the right face we have the same y-coordinate as point two in this case y sub two, so vertex B will
have y sub two as it’s y-coordinate, and lastly the plane that is located in the bottom of the box is parallel to the xy-plane this
means that every vertex in this plane has the same z-coordinate as point one in this case both vertex A and B will have z sub one as their z-coordinate. Now, in order to find the distance between point one and point two represented by the red line I need to
somehow relate this distance by using the coordinates of point one and point two, it turns out that we
can relate them by using right triangles notice that we can form a couple of
right triangles in this box the first right triangle contains the
line segment formed by vertex B and point two, and the line segment formed by vertex B and point one, these two segments represent the legs of this right triangle with the right angle
being formed with the bottom plane of the box and the line segment formed by point B and point two in addition, the line segment formed by point one and point two represents the hypotenuse of the right
triangle in order to find the distance we need to
know the length of each leg we can easily find the length of the leg formed by point two and vertex B this is just going to be
equal to the absolute difference between the points z-coordinates remember we need the length to be a positive value so we take the absolute value of the
difference of their coordinates the distance of the other leg is not as
straightforward and is a little bit more involved I’m going to color this line segment
with the color blue it turns out that this line segment is
also the hypotenuse of another right triangle this right
triangle has as its legs the line segment formed by
point one and vertex A, and the line segment formed by vertex A and vertex B with the right angle being
formed by these two legs in order to find the length of the blue line
segment which represents the hypotenuse of the
second right triangle we need to find the length of each of its
legs once again we can easily find these, the length of the segment formed by point one and vertex A will be equal to the
absolute difference between the points x-coordinates and the length of the segment formed by vertex A and vertex B will be equal to the absolute difference of the points y-coordinates so now we essentially have to apply the
Pythagorean theorem to find the length of the blue line the relations are as follows, the length
of the blue line segment squared has to equal to the length of the line
segment between point one and A squared, plus the length of the line
segment between point A and B squared. We know the lengths of the line
segment between point one and A, it is equal to the absolute
difference of their x-coordinates and we also know the length of the line
segment between point A and B it is equal to the absolute difference
between their y-coordinates simplifying the expression we have the following now, lets apply the Pythagorean theorem to the other right triangle that has as its hypotenuse the red
line which is ultimately what we’re trying to solve for so we have that the length of the red line
segment squared has to equal to the length of the line
segment between point one and point B squared, plus the length of the
line segment between point two and B squared. From the first application
of the Pythagorean theorem we found the line segment between point one and B this was equal to the following
expression, and the length between point two and point B is equal to the absolute difference of their z-coordinates combining them
together and simplifying we obtain the following
expression, and finally its just a matter of taking
the square root of both sides doing that we obtain the distance formula in three dimensions between the points P one with coordinates x-one, y-one and z-one and point two with coordinates x-two y-two and z-two, notice that the
equation resembles that of its two-dimensional counterpart the only difference is that we included
an extra variable in this case z, because of the fact that we
are dealing with a three dimensional coordinate system alright, in our next video we will go
over examples where we make use of this new formula
and we will also derive the equation of a sphere.