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.