### Video Transcript

In this video, we will learn how to
find the equation of a straight line in vector form. There are many different ways that
we can write the equation of a straight line in the π₯π¦-plane. Letβs recall a few of these.

Firstly, in slope intercept form,
we have π¦ equals ππ₯ plus π, where π is the slope or gradient and π is the
π¦-intercept. In point slope form, we have π¦
minus π¦ sub zero is equal to π multiplied by π₯ minus π₯ sub zero, where once
again π is the slope and the line passes through the point π₯ sub zero, π¦ sub
zero. Thirdly, we have the standard form,
which is π΄π₯ plus π΅π¦ equals πΆ, where π΄, π΅, and πΆ are integers and π΄ is
nonnegative. It is also important to note that
only one of π΄ and π΅ can be zero.

All three of these methods have
advantages and disadvantages. As both slope intercept and point
slope form do not allow for vertical lines, weβll use the vector notation for a
straight line to overcome this. We can find the position vector of
any point on a straight line by using a known point π with coordinates π₯ sub zero,
π¦ sub zero on the line that has a position vector π« sub zero together with any
nonzero vector π that is parallel to the line. Vector π is called the direction
vector of the line.

This leads us to the following
definition. The position vector π« of any point
on a line containing the point π 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 is because if the point π
lies on the line, which is parallel to the nonzero vector π, then we can find the
position vector of any point on the line by adding a scalar multiple of π to the
position vector of π.

Letβs consider the line with
direction vector one, two that passes through the point with coordinates one, one
such that π« sub zero is equal to one, one. The vector equation of this line
will therefore be equal to one, one plus π‘ multiplied by one, two. We can represent this on the
coordinate plane. We know that the line passes
through the point with coordinates one, one. The direction vector is one,
two. And we recall that the first
component tells us the horizontal displacement and the second component, the
vertical displacement. This means that the vector one, two
represents movement one unit to the right and two units upwards. We move along the line from the
point one, one in scalar multiples of the direction vector one, two.

If the scalar multiple was two,
then our direction vector becomes two, four. In fact, any nonzero vector
parallel to the line is an equivalent direction vector of the line since weβre
allowed to take any scalar multiples of the direction vector. We will now consider how we can
find the π₯- and π¦-intercepts using the vector form of the equation of a line.

To find the π₯- and π¦-intercepts,
we set each component of the vector equation equal to zero and then solve for
π‘. Recalling the vector equation of a
line that we saw earlier, we know that this vector will have π₯- and
π¦-components. When the line intersects the
π¦-axis, we know that π₯ is equal to zero. Simplifying the right-hand side of
our equation, we have the components one plus π‘ and one plus two π‘. We can then equate the
π₯-components. And solving this gives us π‘ is
equal to negative one. As π¦ is equal to one plus two π‘,
π¦ is also equal to negative one. We can therefore conclude that the
π¦-intercept occurs at π¦ equals negative one.

Using the same method and setting
the π¦-component equal to zero, we see that the line intersects the π₯-axis at π₯
equals one-half. We noted earlier that there is one
big advantage for using the vector equation of a line; that is, we can find the
vector equation of any line, including vertical ones.

Before moving on to some specific
examples, letβs consider this. The line π₯ equals negative one
passes through the point negative one, zero and its direction is vertical. This means that it has no
horizontal direction at all. This means that the first component
of the direction vector will be equal to zero. Using the vector form of the
equation of a line, we see that π« is equal to negative one, zero plus π‘ multiplied
by zero, one. This can be seen on the π₯π¦-plane
as shown. We have a position vector negative
one, zero and a direction vector zero, one. Letβs now look at some specific
questions involving the vector equation of a straight line.

Write the vector equation of the
straight line that passes through the point six, negative nine with direction vector
nine, negative two. Is it (A) π« is equal to nine,
negative two plus π multiplied by six, negative nine? (B) π« is equal to six, negative
nine plus π multiplied by nine, negative two. Is it option (C) π is equal to
six, negative nine plus vector π« multiplied by nine, negative two? Or option (D) π is equal to nine,
negative two plus vector π« multiplied by six, negative nine.

We begin by recalling that the
vector equation of a straight line is written in the form π« is equal to π« sub zero
plus π‘ multiplied by π, where π« sub zero is a position vector of a point on the
line and π is the direction vector of the line. In this question, we are given both
of these. As the line passes through the
point six, negative nine, π« sub zero is equal to six, negative nine. We are also told that the direction
vector is nine, negative two. Therefore, π« is equal to six,
negative nine plus π‘ multiplied by nine, negative two. Noting that the value of π‘ can be
any scalar quantity, to match the options, we will let this equal π. The correct answer is therefore
option (B).

Before moving on to our next
example, we will consider how we can write the equation of a line given its
slope. We know that the slope or gradient
of any line is equal to the change in π¦ over the change in π₯. Letβs begin by considering the line
drawn with slope negative eight over three. This means that for every three
units we move horizontally to the right, we need to move eight units vertically
down. This can also be written as the
vector three, negative eight.

It is important to note that this
is equivalent to the direction vector negative three, eight. This demonstrates a useful property
for finding the direction vector when given the slope of a line. If a line has slope π, then the
line will have direction vector one, π. When the slope is equal to a
fraction π over π as in this case, the line will have direction vector π, π. We will now look at an example
where we will use this information.

Find the vector equation of the
straight line passing through the points six, negative seven and negative four,
six. Is it (A) π« is equal to six,
negative seven plus π multiplied by 10, negative 13? (B) π« is equal to negative four,
six plus π multiplied by negative 13, 10. (C) π« is equal to six, negative
four plus π multiplied by negative seven, six. Or (D) π« is equal to negative
four, six plus π multiplied by 10, 13.

We begin by recalling that the
vector equation of a straight line is written in the form π« is equal to π« sub zero
plus π multiplied by π, where π« sub zero is the position vector of any point that
lies on the line, π is the direction vector of the line, and π is any scalar. We are given the coordinates of two
points, six, negative seven and negative four, six, that both lie on the line. And we can use either of these as
the position vector to help find the vector equation of the line. In option (A), the position vector
is six, negative seven. And in options (B) and (D), the
position vector is negative four, six.

Letβs begin by letting the position
vector π« sub zero be six, negative seven. We now need to find the direction
vector given the two points that lie on the line. We recall that the slope of any
line is equal to the change in π¦ over the change in π₯. Substituting in the points given,
we see that the slope is equal to negative seven minus six over six minus negative
four. This is equal to negative 13 over
10. Recalling that any line with slope
π equal to π over π has direction vector π, π, we see that the direction vector
here is equal to 10, negative 13. As this is one possible direction
vector of our straight line, we have π« is equal to six, negative seven plus π
multiplied by 10 negative 13.

We noticed that this corresponds to
option (A), proving that this is the vector equation of the straight line passing
through the points six, negative seven and negative four, six. Looking at the other options, we
recall that options (B) and (D) do indeed pass through the point negative four,
six. However, they do not have a
direction vector which is equal to or parallel to 10, negative 13. We can therefore rule out options
(B) and (D). The direction vector of option (C)
is negative seven, six. And this too is not parallel to the
direction vector 10, negative 13.

At this point, it is worth
recalling that the vector equation of a straight line is not unique. Other possible vector equations of
the straight line from the information given are π« is equal to negative four, six
plus π multiplied by 10, negative 13; π« is equal to six, negative seven plus π
multiplied by negative 10, 13; and π« is equal to negative four, six plus π
multiplied by negative 10, 13. The direction vectors in the last
two options move in the opposite direction. Any of these four solutions are
valid. However, the only one that was
given as one of the options is π« is equal to six, negative seven plus π multiplied
by 10, negative 13.

This leads us to a useful result
about finding the direction vector of a line when we are given two distinct points
on the line. Given two distinct points π΄ and π΅
with coordinates π₯ sub one, π¦ sub one and π₯ sub two, π¦ sub two, the vector
equation of the line that passes through them is equal to π₯ sub one, π¦ sub one
plus π multiplied by π₯ sub two minus π₯ sub one, π¦ sub two minus π¦ sub one.

We can also use this information to
determine whether three points in the π₯π¦-plane are collinear. We will see an example of this in
our final question.

Using the vector form of the
equation of a straight line, identify whether the points negative seven, five;
negative one, two; and five, negative one are collinear.

We recall that a set of points are
said to be collinear if all of the points lie on the same straight line. There are several ways of checking
this. One way is to begin by finding the
equation between a pair of points and then checking whether the third point
satisfies this. We will begin by finding the vector
equation of the line which passes through negative seven, five and negative one,
two. We recall that this line has
equation π« is equal to π₯ sub one, π¦ sub one plus π multiplied by π₯ sub two
minus π₯ sub one, π¦ sub two minus π¦ sub one.

Substituting in the given values,
the right-hand side becomes negative seven, five plus π multiplied by negative one
minus negative seven, two minus five. This, in turn, simplifies to
negative seven, five plus π multiplied by six, negative three. Grouping the corresponding
components on the right-hand side, we have π« is equal to negative seven plus six
π, five minus three π. If the three points are collinear,
our third point five, negative one will lie on this line. We can check this by substituting
its position vector five, negative one into the vector equation of the line. Equating the components, we have
five is equal to negative seven plus six π and negative one is equal to five minus
three π.

We can solve the first equation by
adding seven to both sides and then dividing through by six. This gives us π is equal to
two. The second equation also gives us a
solution of π equals two. As this value of π is the same, we
can therefore conclude that the point five, negative one lies on the line. The correct answer is therefore
yes, the three points are collinear. This can also be shown graphically
on the π₯π¦-plane.

We will now finish this video by
summarizing the key points. The vector equation of a line can
be written in the form π« is equal to π« sub zero plus π‘ multiplied by π, where π«
sub zero is the position vector of any point that lies on the line, π is the
direction vector of the line, and π‘ is any scalar. If we have two distinct points π΄
and π΅ that lie on the line, we can find its direction vector as shown, where the
coordinates of the two points are π₯ sub one, π¦ sub one and π₯ sub two, π¦ sub
two.

We also saw in this video that a
line of slope π has direction vector one, π. And if this value of π is equal to
the fraction π over π, then the direction vector can be written π, π. We also saw that the vector
equation of a line is not unique as we can choose any point that lies on the line as
the position vector and any nonzero vector parallel to the line as the direction
vector. Finally, we saw that any three or
more points are collinear if the direction vectors between each pair of points are
equivalent.