### Video Transcript

In this video, we will learn how to
find the parametric equations of straight lines in space. We will begin by recalling the
different ways of writing the equation of a straight line in the π₯π¦-plane.

The form π¦ equals ππ₯ plus π is
known as slope-intercept form, where π is the slope or gradient of the line and π
is the π¦-intercept. Writing the equation of a line in
slope-point form, we have π¦ minus π¦ sub zero is equal to π multiplied by π₯ minus
π₯ sub zero. Once again, π is the slope of the
line and the point with coordinates π₯ sub zero, π¦ sub zero lies on the line. Finally, we have the general form
written ππ₯ plus ππ¦ plus π equals zero. This form is usually found by
rearranging one of the first two. The values of π, π, and π are
constants, where π is nonnegative and only one of π and π can be zero.

We can use this information to find
the vector equation of a straight line in two dimensions. The position vector π« of any point
on a line containing the point π with coordinates π₯ sub zero, π¦ sub zero with
position vector π« sub zero is given by π« is equal to π« sub zero plus π‘
multiplied by π, where π is the direction vector of the line and π‘ is any scalar
value. This equation can be written in
terms of its components, where the vector π« has components π₯ and π¦. The position vector π« sub zero is
equal to π₯ sub zero, π¦ sub zero, and we will let the direction vector π have
components π and π.

We can then write this equation in
parametric form. Looking at the first components, we
have π₯ is equal to π₯ sub zero plus π‘ multiplied by π. Equating the second components, we
have π¦ is equal to π¦ sub zero plus π‘ multiplied by π. These two equations combined give
us the equation of a straight line in two dimensions written in parametric form. We will now consider how this can
be extended to three dimensions.

The parametric equations of a line
in space are a nonunique set of three equations of the form π₯ is equal to π₯ sub
zero plus π‘π₯, π¦ is equal to π¦ sub zero plus π‘π¦, and π§ is equal to π§ sub zero
plus π‘π§, where π₯ sub zero, π¦ sub zero, π§ sub zero is a point on the line. The vector π₯, π¦, π§ is a
direction vector of the line. And π‘ is a real number that varies
from negative β to β. We will now look at an example
where we need to find the parametric equations of a line given a point that lies on
the line and its direction vector.

Give the parametric equation of the
line on point two, negative four, four with direction vector one, negative one,
five.

We begin by recalling that the
parametric equations of a line are as follows. π₯ is equal to π₯ sub zero plus
π‘π₯, π¦ is equal to π¦ sub zero plus π‘π¦, and π§ is equal to π§ sub zero plus
π‘π§, where π₯ sub zero, π¦ sub zero, π§ sub zero is a point that lies on the
line. And the vector π₯, π¦, π§ is a
direction vector of the line. We also have the parameter π‘,
which is a scalar quantity. In this question, we are told that
the point two, negative four, four lies on the line. So these will be the values of π₯
sub zero, π¦ sub zero, and π§ sub zero, respectively. We are also given a direction
vector one, negative one, five, which will be the values of π₯, π¦, and π§.

Substituting the values of π₯ sub
zero and π₯, we have π₯ is equal to two plus π‘. Next, we have π¦ is equal to
negative four minus π‘. Finally, substituting the values
four and five for π§ sub zero and π§, we have π§ is equal to four plus five π‘. The parametric equation of the line
on point two, negative four, four with direction vector one, negative one, five is
as shown.

It is important to note that this
set of equations is not unique. For example, we could multiply each
component of the direction vector by two, giving us two, negative two, 10. We could then use this as the
direction vector together with the point two, negative four, four, giving us a set
of parametric equations π₯ is equal to two plus two π‘, π¦ is equal to negative four
minus two π‘, and π§ is equal to four plus 10π‘. This would also be a valid solution
to this question. Alternatively, we could find
another point that lies on the line by choosing a value to substitute for π‘. We could then use this point
together with a direction vector to find a different set of parametric equations
that satisfy this question.

In our next question, weβll find
the parametric equation of a line that passes through two points in space.

Write the equation of the straight
line πΏ passing through the points π sub one β which is equal to four, one, five β
and π sub two β which is equal to negative two, one, three β in parametric
form. Is it option (A) π₯ is equal to
negative two minus six π‘, π¦ is equal to one, and π§ is equal to three plus two
π‘? Option (B) π₯ is equal to four
minus six π‘, π¦ is equal to one, and π§ is equal to five minus two π‘. Is it option (C) π₯ is equal to
four plus two π‘, π¦ is equal to one plus π‘, and π§ is equal to five plus three
π‘? Option (D) π₯ is equal to negative
two plus six π‘, π¦ is equal to one plus π‘, and π§ is equal to three plus two
π‘. Or is it option (E) π₯ is equal to
six plus four π‘, π¦ is equal to one, and π§ is equal to five plus two π‘? For all five options, we are told
that π‘ is greater than negative β and less than β.

We begin by recalling that the
parametric equations of a straight line are written in the form π₯ is equal to π₯
sub zero plus π‘π₯, π¦ is equal to π¦ sub zero plus π‘π¦, and π§ is equal to π§ sub
zero plus π‘π§, where π₯ sub zero, π¦ sub zero, π§ sub zero is a point that lies on
the line and the vector π₯, π¦, π§ is a direction vector of the line. We also know that π‘ is a scalar
quantity that lies between negative β and β. We are given two points that lie on
the line: four, one, five and negative two, one, three. So we can substitute either of
these for π₯ sub zero, π¦ sub zero, and π§ sub zero.

In options (A) and (D), the
coordinates of π sub two have been used, whereas in options (B) and (C), the
coordinates of π sub one have been used. π₯ sub zero equals four, π¦ sub
zero equals one, and π§ sub zero equals five. Whilst the values of π₯ sub zero,
π¦ sub zero, and π§ sub zero in option (E) do not match those coordinates in π sub
one or π sub two, this is not enough to say that the point six, one, five does not
lie on the line. This is because we can use any
point that lies on the straight line and not just one of the two that are given.

We therefore need to focus on
calculating a direction vector from the information given. One way of doing this would be to
work out the vector π sub one, π sub two. We can do this by subtracting the
vector four, one, five from the vector negative two, one, three. Subtracting the corresponding
components gives us negative six, zero, negative two. This is one possible direction
vector of the straight line πΏ. Since we are given a limited number
of options in this question, we can compare this direction vector with the direction
vectors in options (A) to (E).

In option (A), the direction vector
is equal to negative six, zero, two. In option (B), we have negative
six, zero, negative two. In options (C), (D), and (E), the
direction vectors are two, one, three; six, one, two; and four, zero, two,
respectively. The only option that has a
direction vector equal to negative six, zero, negative two is option (B).

As already mentioned, this also
passes through the point π sub one. We can therefore conclude that from
the options given, the equation of the straight line πΏ is π₯ equals four minus six
π‘, π¦ is equal to one, and π§ is equal to five minus two π‘. Option (B) is correct. It is important to note here that
the parametric equations of straight lines are not unique. However, as none of our other four
options have direction vectors that are equal to or parallel to the direction vector
negative six, zero, negative two, then these are not correct.

Before looking at one final
example, we will consider how we can write the parametric form of the equation of a
straight line from its Cartesian form.

We begin by recalling that the
Cartesian form of the equation of a line can be written π₯ minus π₯ sub zero over π₯
is equal to π¦ minus π¦ sub zero over π¦, which is equal to π§ minus π§ sub zero
over π§. Once again, the point π₯ sub zero,
π¦ sub zero, π§ sub zero lies on the line and the vector π₯, π¦, π§ is a direction
vector. These values of π₯, π¦, and π§ must
be nonzero real numbers. This is closely related to the set
of parametric equations as they simply give three values of π‘.

Firstly, by considering π‘ is equal
to π₯ minus π₯ sub zero over π₯, we can multiply both sides of this equation by π₯
and then add π₯ sub zero to both sides, giving us π₯ is equal to π₯ sub zero plus π‘
multiplied by π₯. In the same way, we have π¦ is
equal to π¦ sub zero plus π‘ multiplied by π¦ and π§ is equal to π§ sub zero plus π‘
multiplied by π§. These are the three parametric
equations that we have already seen in this video.

At this point, it is also worth
noting what happens when one component of the direction vector equals zero, for
example, if the direction vector of the line is equal to π₯, π¦, zero. This means that the third
parametric equation becomes π§ is equal to π§ sub zero. In this case, the line is
perpendicular to the π§-axis and is in a plane that is parallel to the
π₯π¦-plane. Letβs now consider what happens
when two components of the direction vector equal zero, for example, π₯, zero,
zero. If both π¦ and π§ are equal to
zero, then π¦ is equal to π¦ sub zero and π§ is equal to π§ sub zero. This means that the direction
vector is undimensional. And if this direction vector is
parallel to the π₯-axis, then π₯ equals π‘.

We will now look at an example
where we need to convert Cartesian equations into parametric equations.

Find the parametric equations of
the straight line three π₯ minus seven over negative nine is equal to eight π¦ minus
three over four, which is equal to negative eight minus six π§ over negative
nine.

We begin by recalling that the
standard form of a Cartesian equation of a straight line is π₯ minus π₯ sub zero
over π₯ is equal to π¦ minus π¦ sub zero over π¦, which is equal to π§ minus π§ sub
zero over π§. The three parts of the equation in
this question have been slightly rearranged from the general form. However, this does not matter, and
we can simply set each of the equations equal to our parameter, in this case π‘. Firstly, we have π‘ is equal to
three π₯ minus seven over negative nine. We need to rearrange this so that
it is in the form π₯ is equal to π₯ sub zero plus π‘ multiplied by π₯. This means that we need to make π₯
the subject.

Multiplying through by negative
nine and then adding seven to both sides gives us seven minus nine π‘ is equal to
three π₯. We can then divide through by
three, giving us π₯ is equal to seven-thirds minus three π‘. We will then repeat this process by
rearranging π‘ is equal to eight π¦ minus three over four so that it is in the form
π¦ is equal to π¦ sub zero plus π‘π¦. Multiplying through by four and
adding three to both sides gives us three plus four π‘ is equal to eight π¦. And then dividing through by eight,
we have π¦ is equal to three-eighths plus a half π‘. Finally, we need to rearrange the
equation π‘ is equal to negative eight minus six π§ over negative nine such that π§
is the subject. This gives us π§ is equal to
negative four-thirds plus three over two π‘.

We now have the three parametric
equations in their required form. The straight line three π₯ minus
seven over negative nine, which is equal to eight π¦ minus three over four, which is
equal to negative eight minus six π§ over negative nine, written in parametric form,
is π₯ equals seven-thirds minus three π‘, π¦ equals three-eighths plus a half π‘,
and π§ equals negative four-thirds plus three over two π‘.

We will now summarize the key
points from this video. The parametric equations of a line
are a nonunique set of equations of the form π₯ is equal to π₯ sub zero plus π‘π₯,
π¦ is equal to π¦ sub zero plus π‘π¦, and π§ is equal to π§ sub zero plus π‘π§,
where π₯ sub zero, π¦ sub zero, π§ sub zero is a point on the line. π₯, π¦, π§ is a direction vector of
the line. And π‘ is a real number that varies
from negative β to β. Where π₯, π¦, and π§ are all
nonzero real numbers, the parametric equations can be derived from the Cartesian
equations by writing π‘ is equal to π₯ minus π₯ sub zero over π₯, which is equal to
π¦ minus π¦ sub zero over π¦, which is equal to π§ minus π§ sub zero over π§. We can then rearrange each of the
three equations so theyβre written in parametric form.

When one component of the direction
vector is zero, it means that the corresponding coordinates of all the points lying
on the line are constant. For example, if the direction
vector is equal to π₯, π¦, zero, we have π₯ is equal to π₯ sub zero plus π‘π₯, π¦ is
equal to π¦ sub zero plus π‘π¦, and π§ is equal to π§ sub zero. When two components of the
direction vector are zero, the line is parallel to an axis, meaning that only one
coordinate varies, while the other two are fixed. For example, the parametric
equations for a line parallel to the π₯-axis, where the direction vector is π₯,
zero, zero, are π₯ equals π‘, π¦ is equal to π¦ sub zero, and π§ is equal to π§ sub
zero.