Three-Dimensional Coordinate systems (Level 9) In the previous video we covered intermediate

level examples that required the use of the equation of a sphere. Now we are going to

go over 2 more challenging examples. Let’s go over the first example,

Find an equation of the largest sphere with center (5,4,9) that is contained in the first

octant. This examples builds upon the problems from

the previous video, the problem is asking us to find the largest sphere with center

(5,4,9) that is contained in the first octant. How can we figure this one out? Let’s first think

about an easier problem that is related to this one, say we have a random circle that

is placed in the first quadrant of a two dimensional coordinate system. If we were trying to find

the largest circle that is contained in this quadrant, one way we can solve this problem

is by expanding the circle until it literally hits one of the two axis if we allow this

to happen the circle will actually be tangent to one of the coordinate axis, it turns out

that we will actually hit the coordinate axis this is closest to the center. In the three

dimensional case if we allow this sphere to expand as if we were inflating a ball or a

balloon it will end up hitting one of the coordinate planes, similar to the two dimensional

case the sphere will end up touching the coordinate plane that is closest to its center. Lets

actually find the distance from the center of the sphere to the each of the 3 coordinate

planes. Recall from the previous video that since all 3 coordinate planes are perpendicular

to one of the coordinate axis, finding the length reduces to just measuring the distance

from the center to each of the coordinate planes along the direction of a single coordinate

axis. This means that the xy-plane is 9 units away from the center, the yz-plane is 5 units

away from the center and the xz-plane is 4 units away from the center, If we were to

let this sphere expand it will end up hitting the xz coordinate plane, if the sphere is

allowed to expand beyond this plane it will seize to be contained in the first octant,

so the radius of the sphere is equal to 4, the largest sphere that is contained in the

first octant with the given center is represented by the equation the quantity (x-5) squared plus the quantity (y-4) squared plus the quantity (z-9) squared equals 16. The hard part of this problem was to understand the idea

of containing the sphere in the first octant it turns out that the shortest distance from

the center to the coordinate plane limits how much the sphere will be able to expand

and at the same time be constrained in the first octant.

Alright, lets try the next example. Find an equation of the sphere with center

in the xz-plane and passing through the points P, Q, and R

Notice that in this problem we are not provided with the center or the radius of

the sphere, nor are we provided with the end points of the diameter, we are dealing with

a completely different problem not addressed in the previous videos. We are asked to find

the equation of a sphere with its center located in the xz-plane, all we know from this specific

information is that the center of the sphere will have coordinates equal to (h,0,l). Since

we don’t know what the x and z coordinates of the center are equal to we leave them as

h and l, in addition, we also know that k is equal to zero because any point located

in the xz plane has 0 as its y coordinate. Next we are given 3 distinct points that are

located on the surface of this sphere, we need to somehow use these points to find the

value for h, l, and the radius of the sphere. How do we accomplish this? What kind of geometric

relations can we construct with the given information? We have an expression for the

center of the sphere and we also have 3 points that are located on the surface of

the sphere, with this information we can generate a relation for the radius of the sphere by

using the center and one of these points, it turns out that we can actually generate

3 relations that represent the radius of the sphere, it can be seen that the radius is

equal to the length between point P and C which is also equal to the length between

point Q and C which in turn is also equal to the length between point R and C. Since

all of these relations represent the radius of the sphere we can equate all three expressions

together, this is similar to the law of sines that you learned in your precaluclus or math

analysis recall that you can equate any two relations with each other in order to solve a specific

expression. By using the distance formula in three dimensions and using point P as point

1 and the center C as point 2 we obtain the following expression for the radius, in the

same manner we can find an expression for the distance between point Q and point C as

follows, simplifying the expression we obtain the following, and finally the distance between

point R and the center C is expressed as follows, simplifying the expression we obtain the following,

since all of these expressions represent the radius of the sphere we can equate all 3 expressions

with each other. We can make the expressions more user friendly by getting rid of the radical

so we go ahead and square all three expressions as follows. Next it’s just a matter of solving the system

of equations, we need to solve for two distinct variables in this case h and l, and we can

also generate three equations from the given relations, this is more than enough to solve

the system of equations. Let’s start by equating the first expression with the third expression,

Lets avoid using the second expression for now since it contains two binomials squared

we want to work with the simpler expressions first, notice that when we equate the first

and third expression we can eliminate the h squared term since it appears on both sides

of the equation, next lets go ahead and expand the binomial square, doing that we have the

following, notice how we can also eliminate the l squared term since it appears on both

sides of the equation, next we go ahead and add like terms moving the constant on one

side of the equation and the terms with variables on the other side, doing that we have the following,

finally we go ahead and solve for l, doing that we obtain l=12, now we need equate

two distinct expressions and substitute this value for l and solve for h, lets equate the

first and second equations together, next lets go ahead replace the value of l with

12 and simplify the expressions, after that lets go ahead and expand the binomial square,

notice that the h squared terms cancel out since it appears on both sides, then it’s

just a matter of getting all the terms with the variables to one side and the constants

to the other side and simplify the expression, lastly we go ahead and solve for h, doing

that we obtain h is equal to -7, we now have the center of our sphere, to find the radius

its just a matter of substituting the values for h and l into any of the three line segments

since all three of them represent the radius of the sphere, let’s go ahead and substitute

them into the expression representing line segment PC, simplifying the expression we

obtain the square root of 257, so the equation of the sphere that is located in the xz-plane

and that passes through points P,Q, and R is represented by the equation the quantity (x+7) squared

plus y-squared plus the quantity (z-12) squared equals 257

Alright, in our final video on three dimensional coordinate systems we will go over how to

graph equations whose domain are restricted to a given interval.

Hey Mathfortress, thanks for all these videos. I've definitely learned a lot from watching these videos. I'll make sure to spread the word to my friends about your channel! 🙂