### Video Transcript

Find π so that the points three,
nine, negative four; nine, negative three, negative one; negative seven, 29, π are
collinear.

Collinear means that all of the
points lie on the same line, which means that, fundamentally, this is a question
about the connection between points and lines. Mathematically, one of the easiest
and most straightforward ways to connect points to lines is by using vectors. To see this connection, letβs first
consider some arbitrary point in space, and weβll call it π.

Now, letβs consider some vector π
with its tail at point π and pointing in some arbitrary direction. If we imagine extending the arrow
representing π in both directions, we see that the vector π lies exactly on some
line which passes through the point π. If we scale π by multiplying by
any real number, weβll get another vector that also lies on this line. Here, we can see that the two scale
vectors one-half times π and negative one-quarter times π both have their heads on
the line that passes through point π and is in the same direction as π. Indeed, if we draw out every
possible scalar multiple of π, then the heads of those vectors exactly trace out
the entirety of this line.

Now, letβs imagine that this point
π and the vector π are in some coordinate system with π₯- and π¦-axes as
shown. In this coordinate system, there is
exactly one vector with its tail at the origin and its head at the point π. Letβs call this π with a half
arrow on top to represent that it is the vector that corresponds to the point
π. Now, remember, the vector π has
its tail at the point π, and every point on the line that weβve drawn can be
represented as a scalar multiple of the vector π.

But as we can clearly see from our
drawing, the tail of the vector π lines up exactly with the head of the vector
π. But this means, by the rules of
vector addition, that the vector with its tail of the origin and its head at the
head of π is exactly the vector π plus π. Likewise, this vector is the vector
π plus one-half times π. Continuing in this fashion, every
point on the line is the head of some vector that can be written as the vector π
plus some scalar multiple of the vector π. If π is the scalar, then we can
represent each of these vectors as π plus π times π for an appropriately chosen
value of π. Itβs important to note that the
vectors weβve just described all have their tails at the origin because the tail of
the vector π is at the origin. This is important because the
information weβre given to work with is a series of points. And when a vector has its tail at
the origin, the coordinates of the point at its head are exactly the same as the
components of the vector itself.

The last thing we should note
before solving for π is that the points that weβre given are in three dimensions,
but our diagram is in two dimensions. Nevertheless, even though we
derived this representation for a line, that is, the vector to a point plus a scalar
multiple times some vector along the direction of the line from two dimensions, this
form is actually valid for any number of dimensions. In fact, this vector form is the
easiest way to represent a line in any dimension greater than two. Anyway, we know that these three
points are all on some line, so letβs arbitrarily pick the point π to be three,
nine, negative four. This means that the vector π from
the origin to this point will be the vector with components three, nine, and
negative four. Now we know that the two vectors
with components nine, negative three, negative one and negative seven, 29, π can
both be written as the sum of this vector three, nine, negative four plus some
multiple of π.

So, before we can determine π,
weβll need to determine the vector π. And weβll do this from the vector
nine, negative three, negative one. So, we have that the vector nine,
negative three, negative one is equal to the vector three, nine, negative four,
which is exactly the vector π plus π times the unknown vector π. To solve for π, we subtract the
vector three, nine, negative four from both sides. Recall that vectorsβ subtraction is
done component by component. So, on the left-hand side, the
resulting vector would have components nine minus three, negative three minus nine,
and negative one minus negative four. This comes out to six, negative 12,
three. And on the right-hand side, when we
subtract a vector from itself, we just get the zero vector. And so, weβre only left with π
times π.

Now, since six, negative 12, three
is clearly not the zero vector, π cannot be zero. But if π is not zero, then we can
divide both sides by π to finally solve for π. At this point, we see that there is
an ambiguity in how we proceed. We can pick any value for π as
long as it isnβt zero. And so, there are infinitely many
possible choices for the vector π. However, all these choices still
define the same line because no matter what value for π we choose, the vector π
will always be parallel to six, negative 12, three. So, we can pick whatever value for
π would be most convenient, which will usually be the greatest common factor of all
the components of this vector, in this case, three. Now recall that to find the scalar
multiple of some vector, we multiply each component by that scalar. So, π is one-third times six,
one-third times negative 12, one-third times three or, equivalently, two, negative
four, one.

Alright, now letβs describe our
final calculation before we actually go about performing it. In this equation here, which comes
directly from the equation of a line in vector form, weβre going to replace nine,
negative three, negative one with negative seven, 29, π. Weβre also going to replace π with
the vector that we found parallel to our line. Then, because for two vectors to be
equal, corresponding components must also be equal. Weβll be able to solve for the
unknown π. Alright, letβs make these
substitutions and clear some space to do the calculations.

And here, we have our new equation,
where π is the scalar multiple of π, such that this equation holds. Our first step in solving this
equation will be the same as it was before, subtracting three, nine, negative four
from both sides. On the left-hand side, we have
negative seven minus three, which is negative 10, 29 minus nine, which is 20, and π
minus negative four, which is π plus four. On the right-hand side, again, the
first term is zero. And weβre left with π times two,
negative four, one. Now, remember, by the rules of
scalar multiplication, π times two, negative four, one is the same thing as two π,
negative four π, π.

Now, it looks like we have two
unknowns, π and π, but we can actually solve for π because this equality holds
for each component. That is, negative 10 is equal to
two π, and 20 is equal to negative four π. Dividing negative 10 by two, we see
that π is equal to negative five. Just to double-check that weβre
doing this correctly, if π is negative five, then negative four π is negative four
times negative five, which is indeed positive 20. Finally, we have that π plus four
is equal to π, which is negative five. And finally, subtracting four from
both sides, we find that π is negative nine.

Itβs worth making one last note
about our choice for the vector π. If we had chosen a different value
for π, then we would have found a different vector π. However, what weβll find is that by
changing the vector π, that is, changing the right-hand side of this equation, we
always get a corresponding change in the scalar π in exactly such a way that this
equation, π plus four equals negative five, is always true.