### Video Transcript

Find the direction vector of the straight line πΏ for which π₯ minus two over three equals π¦ plus three over two and π§ equals four.

So, here are some equations that define a straight line in 3D space. And weβre looking for the direction vector of this line. That is the vector which points along the line. You might remember that the Cartesian equation of the straight line through the points π₯ one, π¦ one, π§ one with direction vector π, π, π is π₯ minus π₯ one over π equals π¦ minus π¦ one over π equals π§ minus π§ one over π.

And so, given the Cartesian form of the equation of a straight line, it looks very easy to read off the direction vector. π, π, and π are just the denominators in the equation. However, what we have doesnβt exactly have this form. We can read off the denominators of the π₯ and π¦ parts. They are three and two, respectively. But π§ is in a separate equation. And it doesnβt have a denominator. So, what are we to do?

Well, we can assume that we still get the π₯- and π¦-components from the dominators. So, the π₯-component is three, and the π¦-component is two. And that just leaves us with the π§-component to work out. Now what does this π§ equals four tell us? Well, it tells us that every point on the line has π§-coordinate four. π§ is, therefore, not changing at all as you go up or down the line. Itβs staying constant. And so, surely, the π§-component of the direction vector must be zero.

If the π§-component of the direction vector were positive, then going along the line in that direction, we would find that π§ was increasing. But we know that π§ remains constant at four. And similarly, if the π§-component of our direction vector were negative, then going along the line in that direction, π§ would be decreasing. And so, the only possibility is that the π§-component is zero.

The π§-component of the direction vector being zero would explain why our equation doesnβt quite look like the general Cartesian equation that we remember. π is zero. And just substituting it into this equation would mean dividing by zero, which isnβt allowed. So, something has to change. It turns out that this thinking gives the correct answer. The direction vector of our line really is the vector with components three, two, zero. But you may think that the argument we used to find this wasnβt entirely convincing. So, letβs see a better one.

What we do is find the parametric equation. With π₯ one, π¦ one, π§ one, π, π, and π as above, the general form is π₯ equals π₯ one plus ππ‘, π¦ equals π¦ one plus ππ‘, π§ equals π§ one plus ππ‘. This equation, or perhaps more correctly system of equations, involves the parameter π‘. Substituting a value for π‘, gives the coordinates of a point on the line. We set this parameter π‘ equal to π₯ minus two over three, which, of course, is equal to π¦ plus three over two.

We have π‘ in terms of π₯ and π‘ in terms of π¦. Multiplying both sides of the first equation by three, adding two to both sides, and swapping the sides, we get π₯ in terms of π‘. This is very nearly in the form π₯ equals π₯ one plus ππ‘. We just need to swap the terms on the right-hand side.

And we do exactly the same thing for our other equation. We find that π¦ equals negative three plus two π‘, which is in the form required. But we still have this problem with π§. Weβd like an equation of the form π§ equals π§ one plus ππ‘. And what we know about π§ is that it is equal to four. But we can take the equation π§ equals four and write it in the form π§ one plus ππ‘ pretty easily. π§ equals four plus zero π‘. It doesnβt matter what π‘ is. π§ is always four.

These three equations π₯ equals two plus three π‘, π¦ equals negative three plus two π‘, and π§ equals four plus zero π‘ are the parametric equation of our line. And now we really can just read off the values of π, π, and π, which we defined to be the π₯, π¦, and π§ components, respectively, of our direction vector.

So, again, we find that the direction vector has components three, two, zero. The moral of the story here is that this general Cartesian equation only holds if none of the components of the direction vector π, π, or π are zero. If π equals zero, as it does for our line, then the Cartesian form of the equation of the line looks slightly different. And we get something similar if π or π is zero instead, or if some combination of π, π, and π are zero.

Moving to the parametric form of the equation, or indeed the vector form of the equation, doesnβt present such issues. Thereβs no division involved. And so, we donβt have to worry about dividing by zero. And we can just read off the values.