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.