Question Video: Finding the 𝑧-Coordinate of a Point That Makes It Collinear with Two Given Points Mathematics • Third Year of Secondary School

Video Transcript

Find 𝑘 so that the points three, nine, negative four; nine, negative three, negative one; negative seven, 29, 𝑘 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, let’s first consider some arbitrary point in space, and we’ll call it 𝑃.

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

Now, let’s imagine that this point 𝑃 and the vector 𝐕 are in some coordinate system with 𝑥- and 𝑦-axes as shown. In this coordinate system, there is exactly one vector with its tail at the origin and its head at the point 𝑃. Let’s call this 𝑃 with a half arrow on top to represent that it is the vector that corresponds to the point 𝑃. Now, remember, the vector 𝐕 has its tail at the point 𝑃, and every point on the line that we’ve drawn can be represented as a scalar multiple of the vector 𝐕.

But as we can clearly see from our drawing, the tail of the vector 𝐕 lines up exactly with the head of the vector 𝐏. 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 𝐕 is exactly the vector 𝐏 plus 𝐕. Likewise, this vector is the vector 𝐏 plus one-half times 𝐕. Continuing in this fashion, every point on the line is the head of some vector that can be written as the vector 𝐏 plus some scalar multiple of the vector 𝐕. If 𝑎 is the scalar, then we can represent each of these vectors as 𝐏 plus 𝑎 times 𝐕 for an appropriately chosen value of 𝑎. It’s important to note that the vectors we’ve just described all have their tails at the origin because the tail of the vector 𝐏 is at the origin. This is important because the information we’re 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 𝑘 is that the points that we’re 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 let’s arbitrarily pick the point 𝑃 to be three, nine, negative four. This means that the vector 𝐏 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, 𝑘 can both be written as the sum of this vector three, nine, negative four plus some multiple of 𝐕.

So, before we can determine 𝑘, we’ll need to determine the vector 𝐕. And we’ll 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 𝐏 plus 𝑎 times the unknown vector 𝐕. To solve for 𝐕, we subtract the vector three, nine, negative four from both sides. Recall that vectors’ 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, we’re only left with 𝑎 times 𝐕.

Now, since six, negative 12, three is clearly not the zero vector, 𝑎 cannot be zero. But if 𝑎 is not zero, then we can divide both sides by 𝑎 to finally solve for 𝐕. At this point, we see that there is an ambiguity in how we proceed. We can pick any value for 𝑎 as long as it isn’t zero. And so, there are infinitely many possible choices for the vector 𝐕. However, all these choices still define the same line because no matter what value for 𝑎 we choose, the vector 𝐕 will always be parallel to six, negative 12, three. So, we can pick whatever value for 𝑎 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, 𝐕 is one-third times six, one-third times negative 12, one-third times three or, equivalently, two, negative four, one.

Alright, now let’s 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, we’re going to replace nine, negative three, negative one with negative seven, 29, 𝑘. We’re also going to replace 𝐕 with the vector that we found parallel to our line. Then, because for two vectors to be equal, corresponding components must also be equal. We’ll be able to solve for the unknown 𝑘. Alright, let’s make these substitutions and clear some space to do the calculations.

And here, we have our new equation, where 𝑏 is the scalar multiple of 𝐕, 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 𝑘 minus negative four, which is 𝑘 plus four. On the right-hand side, again, the first term is zero. And we’re left with 𝑏 times two, negative four, one. Now, remember, by the rules of scalar multiplication, 𝑏 times two, negative four, one is the same thing as two 𝑏, negative four 𝑏, 𝑏.

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

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