three-dimensional coordinate systems level 5 in the last video
we started introducing common formulas associated with a
three-dimensional Cartesian coordinate system
specifically the midpoint and distance formula
we will continue covering the distance formula by going over a couple of examples
let's try the first example find the distance between the points
(4,3,1) and (-2,5,7) notice that these are the same points from the midpoint
example from the previous video
let's go ahead and use the same graph that we used from that video
the first point will be located here and the second point will be located here
this time around the problem is asking for the distance between these two
endpoints
in other words the length of the line segment connecting these two points
it's just a matter of using the distance formula, a word of caution
the common mistake that many students make is not being consistent
with the coordinates of each point, notice that it's up to you which point you
decide to designate
as point one and which one you designate as point two
as long as you're consistent everything will workout
I'm going to designate the point with coordinates (4,3,1)
as point one and designate the point with coordinates
(-2,5,7) as point two, this means that the coordinates (4,3,1)
are going to represent x-one, y-one and z-one
respectively, in the same manner the coordinates
(-2,5,7) are going to represent
x-two, y-two and z-two respectively
so I'm going to subtract the x-coordinates of the points
from each other and square them, I would do the same for the y and z-coordinates
notice that we need to make sure we subtract the correct coordinates from each
other
in this case we need to subtract the x-coordinates
y-coordinates and z-coordinates from each other, where you start getting into trouble
is when you start switching the coordinates and start subtracting the wrong coordinate
pairs
for example you might end up subtracting an x-coordinate
with a y-coordinate or an x-coordinate with a z-coordinate
the best way to avoid this mistake is by appropriately
labeling the coordinates given so you can keep track which numbers
represent the specific coordinates you're looking for, by subtracting the coordinates of
the second point with that of the first point we obtain the following
then we square each result and obtain the following
then we go ahead and add them, lastly
we take the result and take the square root we can simplify the radical
expression
by breaking it into a product of a square number and a factor of 76
in this case four times nineteen, doing that
and simplifying the expression we obtain the distance between point one and point two
which ends up being equal to 2 square root 19
alright, let's try the next example
determine whether the points PQ and R
lie on a straight line, in other words are they collinear
how can we determine if these three points lie in the same straight line
there's actually a couple of ways, consider the following
say we have three random points let's call them point A
point B and point C, let's also assume for now
that they are collinear meaning they lie on the same straight-line
if this was the case than it can easily be seen
that the distance between point A and point C has to equal to the sum
of the distance between AB and BC from elementary geometry
the sum of the length of the two smaller line segments has to equal to the length
of the line segment that includes these two smaller line segments
alternatively, we can also use the following idea
if three points are not collinear than they form a triangle
if this is the case then by using the properties of triangles
established in elementary geometry we can show
that the some of the two sides must be greater than the third side
in other words, if these three points are not collinear
than they form a triangle and we would have to show that the sum of two line
segments has to be greater than the third side of the triangle
in other words we need to show that line segment AB
plus line segment BC is greater than line segment AC
and that line segment BC plus line segment AC
is also greater than line segment AB and finally
line segment AB plus line segment AC
is greater than line segment BC in either case we need to find the length formed by each
points
and use one of these geometric relations to verify that they are collinear
let's use the first geometric relation since it's a lot easier to work with
let's first find the length of each line segment, the line segment formed by point
P and Q
is found by subtracting the coordinates from each point and then squaring them
doing that we obtain the following, then we continue simplifying the expression
and obtain the following radical expression, we can further simplify this
expression by breaking 18 into a product of a square number
and another factor of 18 in this case nine times two
then we go ahead and simplify the expression obtaining 3 root 2
next we find the length of the line segment QR
once again we use the distance formula and substitute the coordinates of
each point, doing that we obtain the following expression
then we go ahead and square the terms and simplify
doing that we obtain the following, we can simplify the radical expression by
breaking it into a product of a square number
and a factor of 72 in this case 36 times two
then we go ahead and simplify the expression obtaining 6 root two
finally, we go ahead and find the length of line segment PR
so we use the distance formula and substitute the coordinates of each point
doing that we obtain the following expression, then we go ahead and square
the terms
and simplify doing that we obtain the following
we can simplify the radical expression by breaking it into the product of a square
number
and a factor of 162 in this case 81 times two
then we go ahead and simplify the expression obtaining 9
root two, having found the length of each line segment
let's create a diagram to aid us in determining if these points are collinear
we know from the distances we found that points P and R
form the endpoints of the longest line segment
this means that point Q has to be between these two points
now let's go ahead and determine if the sum up the two smaller line segments add
up to the line segment that includes these two smaller line segments
in other words, we're determining if the length of line segment
PQ plus the length of line segment QR
is equal to the length of line segment PR let's go ahead and substitute the
values for each line segment
doing that we have the following, next let's simplify the left side of the
equation
notice how the left side is equal to the right side of the equation
meaning it checks out since this relation turned out to be true
we conclude that these points are indeed collinear
they lie in the same straight-line, if we decided to use the second geometric
relation
specifically proving that the line segments form a triangle
we would have to show that the sum of two line segments has to be greater than
a third side of the triangle
for all the following cases: the length of the line segment PQ
plus QR has to be greater than line segment PR
the length of line segment PR plus PQ has to be greater than line segment QR
and the length of line segment QR plus PR has to be greater than line
segment PQ
let's go ahead and substitute the value of the lengths for each line segment into
the relations
doing that we obtain the following, next we go ahead and simplify the left
side
of each inequality, notice that the first inequality is false
the second inequality is true and the third inequality is true
we're testing to see if these points form a triangle we need all three
inequalities to be true
but since the first inequality is false we conclude that they do not form a
triangle
meaning they must lie in the same straight line they are indeed
collinear. Alright, let's move along and continue exploring equations
in three-dimensional space in the first videos
we introduce the basic equations of coordinate planes
now we will explore the three-dimensional version of a circle
recall that the equation of a circle in R-squared with center (h,k)
and radius r is defined to be the set of all points
(x,y) such that the distance between (x,y) and (h,k)
is equal to the radius r, the distance between point P
and point C has to equal r, mathematically
we can express this as follows: the distance between the center
C and a point on the circle P is equal to r
the distance between P and C can be found by using the distance formula
in two dimensions so we substitute the distance PC
with the following expression, the square root of the quantity (x-h) squared
plus the quantity (y-k) squared equals r,
we can make the equation more user-friendly by squaring both sides of
the equation
doing that we obtain the final equation of a circle with center (h,k)
the quantity (x-h) squared plus the quantity (y-k)
squared equals r-squared, keep in mind that this equation represents a circle
in two-dimensional space we can form a surface using this curve
by rotating it about its axis of symmetry doing that
we see that we end up forming a sphere in three-dimensional space
analogous to the definition of a circle a sphere with center (h,k,l)
and radius r is defined to be the set of all-points
(x,y,z) such that the distance between (x,y,z)
and center (h,k,l) is equal to the radius r
mathematically we can express this as follows:
the distance between P and C can be found by using the distance formula for
three-dimensional space
we substitute the expression for the distance of line segment PC
with following, the square root of the quantity (x-h) squared plus
the quantity (y-k) squared plus the quantity (z-l)
squared equals r, we can make the equation more user-friendly
by squaring both sides of the equation doing that we obtain the final equation of a
sphere
with center (h,k,l) the quantity (x-h) squared plus
the quantity (y-k) squared plus the quantity (z-l)
squared equals r-squared, once again
notice how this equation is extremely similar to that of its two-dimensional
counterpart
the only difference being that we included an extra variable
in this case we included the variable z. Alright,
in our next video we will go over slightly more challenging examples that require the use
of the distance formula.