Three-Dimensional Coordinate systems Level 6 In this video we are going to go over 3 slightly
more complicated problems that require the use of the distance formula in three dimensions.
let's try first example. Find the lengths of the sides of the triangle
PQR with the given vertices and determine whether the triangle
is a right triangle, an isosceles triangle, equilateral triangle or neither.
How do we determine what type of triangle these 3 points form? Will it depends, say
we have triangle ABC if this triangle was an equilateral triangle then the length of
each of its sides would be the same meaning that the length of line segment AB is equal
to line segment AC which is also equal to line segment BC. If the triangle where to
be an isosceles triangle then two of its sides would have to be the same meaning that the
length of line segment AB equals line segment AC or line segment AB equals line segment
BC or line segment AC equals line segment BC, if the triangle were to be a right triangle
then we can use the converse of the Pythagorean theorem let's assume that in this diagram
line segment BC is the hypotenuse meaning it's the longest side of the triangle, if
the sum of the squares of the other two line segment in this case AB and AC are equal to
the square of line segment BC then the triangle is a right triangle. In any case we need to
find the lengths of each line segment, so let's go ahead and use the distance formula
to find the length of each of three line segments. The length of Line segment PR can be found
by using the coordinates of point P and R as follows, we go ahead and square the difference
and simplify it by adding the sum doing that we obtain the square root of 14, notice that
we cannot simplify this radical since the factors of 14 are 7 and 2 none of these are
a square number. In the same manner we find the length of line segment QR by using the
coordinates of point Q and R as follows, we go ahead and square the difference and simplify
it by adding the sum, doing that we obtain the square root of 45, we can simplify this
radical by breaking apart 45 into a product of a square number and one of its factors
in this case 9 times 5 simplifying the expression we obtain 3 root 5. Finally we find the length
of line segment PQ by using the coordinates of point P and Q as follows, we go ahead and
square the difference and simplify it by adding the sum doing that we obtain the square root
of 59, we cannot simplify the expression because 59 is a prime number so we leave the expression
as it is. Having found the length of all the line segments of triangle PQR we go ahead
and start comparing the sides to determine what type of triangle the points form. Let's
check to see if it's an equilateral triangle, meaning is the length of all 3 sides equal?
in this case no. Every single line segment has a different length, so we can eliminate
this triangle from being an equilateral triangle or isosceles triangle. lets test to see if it's a right triangle,
we first need to identify the hypotenuse and the legs of the triangle, remember the hypotenuse
is always the longest side of a right triangle in this case line segment PQ is the hypotenuse
and line segment QR and PR are the legs, next let's verify that the Pythagorean theorem
is true, meaning that the length of line segment PQ squared is equal to the length of line
segment QR squared plus the length of line segment PR squared, so we go ahead and substitute
the values of each line segment into the equation, next we go ahead and square the values, simplifying
this expression and adding the values on the right side of the equation we obtain 59 is equal to
59 so it checks out, hence by the converse of the Pythagorean theorem triangle PQR is
a right triangle with the right angle located at vertex R.
Alright, let's try the second example: If line L passes through point A(1,2,3) and
is perpendicular to the xy-plane, what are the coordinates of the points that lie on
line L that are at a distance of 7 from point B(3,-1,5)?
Alright, let's first draw a diagram so we have an idea of what we are dealing with, let's
first plot point A with coordinates (1,2,3). Now, the problem states that line L passes
through point A, in addition we also know that this line is perpendicular to the xy
plane meaning it forms a right angle with this plane, as follows, this is a visual representation
of line L that is perpendicular to the xy plane and passes through point A. The problem
is asking us to find the points that lie on this line L, that are at a distance of 7 from
point B with coordinates (3,-1,5), so essentially we are looking for all the points that lie
on this line that are at a distance of 7 from point B. Since we are dealing with the distance
between two points it makes sense to use the distance formula in three dimensions, in order
to use this formula we need two points, since we are trying to find the distance from point
B to a point on the line, we will assign point B as point 1 and the point located on the
line as point 2, lets label this point as point P, now notice that we know the coordinates
of point B, they are 3,-1 and 5, but how about the coordinates of point P? well lets take
a closer look at the diagram we know that line L passes through the point (1,2,3) and
that it's also perpendicular to the xy-plane, this means that any point on this line will
have its x and y coordinates fixed they will not change. So every point that lies in this
line will be guaranteed to have an x-coordinate of 1 and a y-coordinate of 2, so point P should
have 1 and 2 as its x-coordinate and y-coordinate respectively. In addition, since this line
is perpendicular to the xy plane, the z-coordinates of every single point that lie in this line
will be distinct since the points are free to move along this line in both the positive
z-direction and negative z-direction. Since the values of z can change we will assign
the variable z as the z-coordinate for point P, now that we have values for the coordinates
of point P lets go ahead and apply the distance formula. Substituting the coordinates for
point P and B we have the following, we also know that the distance from point B to point
P has to equal 7 so we set this expression equal to 7. Notice that this problem reduces
to finding the value of the z coordinate of point P, so let's go ahead and solve for z,
first we simplify the difference of the x and y coordinates, then we go ahead and square
them and collect like terms leaving the quantity (z-5) squared intact, next we go ahead and square
both sides of the equation as follows, doing that we obtain the following expression. Next
we subtract 13 from both sides, then we go ahead and take the square root of both sides,
here we have to be careful recall that when you take the square root of a quantity squared
this expression reduces to the absolute value of the quantity. Alternatively we can take
the square root of both sides, with the square root of 36 being equal to positive and negative
6, in either case we have two distinct equations for which we need to solve for z. So solving
for z on both equations we obtain z = 11 and z = -1, Notice that we obtained two answers
this means that there are actually two distinct points that lie on line L for which they are
at a distance of 7 from point B, the first of these is the point with coordinates (1,2,11)
and the second point has coordinates (1,2,-1). Alright, let try the final example.
Find the point on the y-axis equidistant from the point (2,5,-3) and (-3,6,1)
Alright lets go ahead and plot the first point and label it as point A, lets also plot the
second point and label it as point B. This time around the question is asking us to find
the point that lies in the y-axis that is equidistant to these two points. Lets label
this point as point P, mathematically we are trying to find the point for which the distance
between point P and A is equal to the distance between point P and B. We can find the distance
between each points by using the distance formula in three dimensions but first we need
the coordinates of point P, from the problem we know that point P lies on the y axis, this
means that the x and z coordinate remain constant specifically both the x and z coordinate are
equal to zero, leaving the y coordinate of point P free to change, so we will assign
this coordinate with the variable y, now that we have the coordinates of point P lets go
ahead and apply the distance formula to both line segments, the distance from point P to
point A is represented as follows, and the distance from point P to point B is represented
as follows, next we go ahead and simplify both sides by squaring the difference of the
x and z coordinates leaving the difference of the y-coordinates intact for now. Next
we go ahead and square both sides to eliminate the square root from both sides, and then
we go ahead and simplify the expressions, now let's go ahead and expand the binomial
from each side of the equation. Doing that we have the following, we can eliminate the
y squared term by adding it to both sides, next is just a matter of solving for y by
collecting like terms, and isolating the variable doing that we have y = 4, so the point that
is equidistant to both point A and point B that lies on the y axis is the point with
coordinates (0,4,0) Alright, in our next video we will start solving
basic problems that require the use of the equation of a sphere.