Video Transcript
In this video, we will learn how to
find the coordinates of a point in three dimensions. We will also calculate the distance
between two points in 3D and then midpoint. We will begin by recalling what we
know about points, midpoints, and distances in two dimensions. The two-dimensional π₯π¦-coordinate
plane is drawn below. Any point on this coordinate plane
will have an π₯- and π¦-coordinates. Letβs consider the two points π΄
and π΅ with coordinates π₯ one, π¦ one and π₯ two, π¦ two, respectively.
In order to find the midpoint of π΄
and π΅, we find the average of the π₯- and π¦-coordinates. The π₯-coordinate of the midpoint
will be equal to π₯ one plus π₯ two divided by two. And the π¦-coordinate will be equal
to π¦ one plus π¦ two over two. In order to calculate the distance
between two points on the π₯π¦-plane, we use an adaption of the Pythagorean
theorem. The distance between point π΄ and
point π΅ is the square root of π₯ two minus π₯ one squared plus π¦ two minus π¦ one
squared. We find the difference between the
π₯-coordinates and square the answer. We then find the difference between
the π¦-coordinates and square this answer. The sum of these square rooted is
the distance between the two points on the π₯π¦-plane.
We will now look at how we can
adapt these two formulas when dealing in three dimensions. The three-dimensional π₯π¦π§-plane
could be drawn in many ways on a two-dimensional surface. We know that any point will have an
π₯-, π¦-, and π§-coordinates. For example, the two points shown
have coordinates π₯ one, π¦ one, π§ one and π₯ two, π¦ two, π§ two. We can find the midpoint of π΄ and
π΅ by finding the average of the π₯-, π¦-, and π§-coordinates. The π₯-coordinate of the midpoint
will be equal to π₯ one plus π₯ two divided by two. The π¦-coordinate will be π¦ one
plus π¦ two divided by two. And the π§-coordinate will be π§
one plus π§ two divided by two.
We can extend the distance formula
in the same way. The distance between two points in
three dimensions is equal to the square root of π₯ two minus π₯ one squared plus π¦
two minus π¦ one squared plus π§ two minus π§ one squared. We simply repeat the process used
with the π₯- and π¦-coordinates with the π§-coordinate. We will now look at some questions
where we need to identify points in three dimensions.
In which of the following
coordinate planes does the point negative seven, negative eight, zero lie? Is it (A) the π₯π¦-plane, (B) the
π₯π§-plane, or (C) the π¦π§-plane?
We know that any point in three
dimensions has an π₯-, π¦-, and π§-coordinate. In this question, the π₯-coordinate
is negative seven, the π¦-coordinate is negative eight, and the π§-coordinate is
zero. As π§ is equal to zero, the point
will not move in the direction of the π§-axis. We can therefore conclude that as
π§ is equal to zero, the point will lie on the π₯π¦-plane. If our π¦-coordinate was equal to
zero but π₯ and π§ had a positive or negative value, the point would lie in the
π₯π§-plane. In a similar way, a point would lie
in the π¦π§-plane if it had coordinate zero, π¦, π§, where π¦ and π§ are positive or
negative values.
In our next question, we need to
find the coordinates of a point graphically.
Determine the coordinates of point
π΄.
Any point on the 3D plane will have
an π₯-, π¦-, and π§-coordinate. We can see from our diagram that
point π΄ has an π₯-coordinate of three. It has a π¦-coordinate of negative
three. Finally, it has a π§-coordinate of
three. We can therefore conclude that the
coordinates of point π΄ are three, negative three, three. If we werenβt able to spot this
immediately on our diagram, we could begin by considering the point π΅ in the
two-dimensional π₯π¦-plane. Point π΅ has an π₯-coordinate equal
to three and a π¦-coordinate equal to negative three. As it lies on the π₯π¦-plane, it
will have a π§-coordinate equal to zero.
Point π΄ lies directly above point
π΅. This means its π₯- and
π¦-coordinates will be the same. All we now need to work out is the
distance traveled along the π§-axis to get from point π΅ to point π΄. As this is equal to three, the
π§-coordinate of point π΄ is three. This confirms that point π΄ has
coordinates three, negative three, three.
In our next question, we need to
work out the midpoint of a line segment.
Points π΄ and π΅ have coordinates
eight, negative eight, negative 12 and negative eight, five, negative eight,
respectively. Determine the coordinates of the
midpoint of line segment π΄π΅.
We recall that in order to find the
midpoint of two points in three dimensions, we find the average of the π₯-, π¦-, and
π§-coordinates. We can begin by letting point π΄
have coordinates π₯ one, π¦ one, π§ one and point π΅: π₯ two, π¦ two, π§ two. The π₯-coordinate of our midpoint
will be equal to eight plus negative eight divided by two. Eight plus negative eight is equal
to zero and zero divided by two is equal to zero. The π¦-coordinates of π΄ and π΅ are
negative eight and five. This means that the π¦-coordinate
of the midpoint will be equal to negative eight plus five divided by two. This is equal to negative three
over two, which we could write as negative one and a half or negative 1.5. We will leave the answer as a top
heavy or improper fraction.
The π§-coordinate of our midpoint
is equal to negative 12 plus negative eight divided by two. Negative 12 plus negative eight is
equal to negative 20. Dividing this by two gives us
negative 10. The midpoint of the line segment
π΄π΅ has coordinates zero, negative three over two, negative 10. We could check this answer by
looking at the distances between these values and the corresponding values in points
π΄ and π΅. Zero is eight away from both eight
and negative eight. Negative three over two or negative
1.5 is 6.5 away from negative eight and also from five. Finally, negative 10 is two away
from negative 12 and also two away from negative eight. This confirms that the midpoint of
points π΄ and π΅ is zero, negative three over two, negative 10.
In our next question, weβll need to
find the distance between a point and one of the axes.
What is the distance between the
point 19, five, five and the π₯-axis?
Any point that lies on the π₯-axis
will have coordinates π₯, zero, zero. Both the π¦- and π§-coordinates
must be equal to zero. Weβre given the coordinates of a
point 19, five, five. The point on the π₯-axis that is
closest to this will have coordinates 19, zero, zero. The shortest distance will be to
the point where the π₯-coordinate is the same. We know that we can calculate the
distance between two points in three dimensions using an adaption of the Pythagorean
theorem. If we have two points with
coordinates π₯ one, π¦ one, π§ one and π₯ two, π¦ two, π§ two, the distance between
them is equal to the square root of π₯ two minus π₯ one squared plus π¦ two minus π¦
one squared plus π§ two minus π§ one squared.
Substituting in our two coordinates
gives us the square root of 19 minus 19 squared plus zero minus five squared plus
zero minus five squared. 19 minus 19 is equal to zero. Zero minus five is equal to
negative five. So we are left with the square root
of negative five squared plus negative five squared. Multiplying a negative number by a
negative number gives us a positive answer. Therefore, negative five squared is
equal to 25. This means that our answer
simplifies to the square root of 50.
It is worth pointing out that we
could have subtracted the coordinates in the other order as five minus zero squared
is also equal to 25. As squaring a number always gives a
positive answer, it doesnβt matter which order we subtract our coordinates in. We can actually simplify our answer
by using our laws of radicals or surds. The square root of 50 is equal to
the square root of 25 multiplied by the square root of two. As the square root of 25 equals
five, weβre left with five multiplied by the square root of two or five root
two. The square root of 50 is equal to
five root two. We can therefore conclude that the
distance between the points 19, five, five and the π₯-axis is five root two length
units.
We might actually notice a shortcut
here. To find the distance between any
point and an axis, we simply find the sum of the squares of the other two
coordinates and then square root the answer. As we want to calculate the
distance to the π₯-axis, we square the π¦- and π§-coordinates, find their sum, and
then square root our answer. If we needed to calculate the
distance between a point and the π¦-axis, we would square the π₯- and
π§-coordinates, find the sum of these, and then square root that answer. We would use the same method to
find the distance between a point and the π§-axis, this time using the π₯- and
π¦-coordinates.
In our final question, we will find
the distance between two points given their coordinates in three dimensions.
Find the distance between the two
points π΄: negative seven, 12, three and π΅: negative four, negative one, negative
eight.
We know that we can find the
distance between two points in three-dimensional space using the following
formula. The distance is equal to the square
root of π₯ two minus π₯ one squared plus π¦ two minus π¦ one squared plus π§ two
minus π§ one squared. In this question, we will let the
coordinates of point π΄ be π₯ one, π¦ one, π§ one and the coordinates of point π΅ be
π₯ two, π¦ two, π§ two. Substituting in these values gives
us the square root of negative four minus negative seven squared plus negative one
minus 12 squared plus negative eight minus three squared.
Negative four minus negative seven
is the same as negative four plus seven. This is equal to three. Negative one minus 12 is equal to
negative 13. Finally, negative eight minus three
is equal to negative 11. We know that squaring a negative
number gives a positive answer. This means that three squared is
equal to nine, negative 13 squared is 169, and negative 11 squared is 121. 169 plus 121 is equal to 290. And adding nine to this gives us
299. We can therefore conclude that the
distance between the two points negative seven, 12, three and negative four,
negative one, negative eight is the square root of 299 length units.
We will now summarize the key
points from this video. In this video, we saw that any
point in three dimensions has coordinates π₯, π¦, and π§. We saw that if our π§-coordinate is
equal to zero, the point lies on the π₯π¦-plane. If the π¦-coordinate was equal to
zero, it would lie on the π₯π§-plane. In the same way, if π₯ was equal to
zero, the point would lie on the π¦π§-plane. We also saw that if a point has two
coordinates that are equal to zero, for example, if π¦ equals zero and π§ equals
zero, it will lie on one of the axes, in this case, the π₯-axis.
If π₯ and π¦ were both equal to
zero, the point would lie on the π§-axis. And in the same way, if π₯ and π§
were equal to zero, the point would lie on the π¦-axis. We saw that the midpoint of two
points π΄ and π΅ has coordinates π₯ one plus π₯ two over two, π¦ one plus π¦ two
over two, and π§ one plus π§ two over two. We find the average of the π₯-,
π¦-, and π§-coordinates.
We also saw that we can calculate
the distance between the same two points by square rooting π₯ two minus π₯ one
squared plus π¦ two minus π¦ one squared plus π§ two minus π§ one squared. These two formulas will allow us to
solve practical problems involving coordinates in three dimensions.