# Can a vector equation of a line in 3 dimensions be expressed as $(x,y,z) = t(1,0,1) + (1-t)(4/3,-1/3,5/3)$?

This is a specific equation of course. It is an example. I am not sure if this was a mistake in the lecture series or some exotic way that I have not the smarts to figure out. Usually it is something like

$$r = (x,y,z) + t(x_0,y_0,z_0)$$

The first vector being a position vector and the second vector being the direction vector. Then the equation can be parameterized. No problem with that. But this one has me stumped, why would the professor do this? Can anyone guess the motivation or is it a mistake and I am too stupid to figure it out. It is a lecture on Multivariable calculus on u tube. ( I am trying to learn calculus on my own.) It is lecture # 6 at 27:23. ( Just in case someone is wondering why I am asking such a question:-) )

For sure it can.

For example, take two points in space $P_1$ and $P_2$, if we construct a line that joins them and choose one parameter $t$, for convenience that ranges $[0,1]$. And impose that for $t=0$ we must obtain point $P_1$ and for $t=1$ we must obtain the other, we can write without any loss of generality that any point $x$ in between can be expressed: $$x=t\,P_2 + (1-t)\, P_1 \qquad t\in[0,1]\tag{1}$$

This definition can be extended for any value of $t$, defining now a straight line in space, whose explicit form can be obtained from $(1)$ $$x=t\,P_2 + (1-t)\, P_1 =P_1 + (P_2-P_1)t \qquad t\in\mathbb{R}$$

This latter equation is the one you are familiar with.

• I was just typing this "same" answer. – Chickenmancer May 2 '17 at 13:58
• (I don't think this worth a full answer, so I'll just comment.) Another way to see the usual definition of a line to show up is to write down the three equations ( i.e. eq. (1) in HBR's answer above plus the other two), then eliminate $t$ and solve for $y,z$.What you'll get is $y,z$ as linear functions of $x$ in the familiar way. The reason not to do this is that 1) it's more cumbersome, and 2) you won't be able to represent lines like the $z$ axis in this way, since only one value of $x$ is allowed. The $t$-parametrization doesn't have this restriction. – Semiclassical May 2 '17 at 14:41
• In my first example, yes it is. But this idea can be extended for any value of $t$, I mean that you can use $t=-1$ or $t=10$. The point is that I began to explain what is a segment that links two points to later introduce a more general concept. Anyway.. I will edited the second (last) equation for better clarity. The main point here is the last equation, where I show yo how from your "misterious line" the usual equation is recovered. think the point $P_1$ as your point $(x,y,z)$ and $P_2-P_1$ as yout second point $(x_0,y_0,y_0)$ – HBR May 2 '17 at 19:45
• Forget the first equation, because it is the equation of a segment that begins from one point $P_2$ and ends into another point $P_2$ for $t\in[0,1]$. This first equation is clear because if we have a linear function of $t$, and for $t=0$, $P_1$ results and for $t=1$, $P_2$ results any $t$ in between will result in a point $P_3$ that is between the two points. But... what happens if $t$ is greater that 1? or $t$ is lower than 0? We do have another points that are in the direction of the segment line, $\textit{i.e.}$ we will have a straight line instead if $t\in(-\infty,\infty)$ – HBR May 2 '17 at 20:04
• The first equation is nothing but the preamble to the second one, the equation you asked for. It is like the problem to show how $3×2=6$ without explaining that $3+3=6$. Hope I have explained myself enough. – HBR May 2 '17 at 21:26

There are many different ways to represent 3D lines.

What you have here is a parametric form of the line representing all the points given a parameter $t \in \mathbb{R}$

• The linear interpolation between two points $${\bf r} = (1-t)\, {\bf r}_1 + t\, {\bf r}_2$$
• Or you can look at a point and direction $${\bf r} = {\bf r}_0 + t\, {\bf e}$$

But there are other representations. Without a parameter, but with equation(s) that need to be satisfied

• Point and direction

$$\frac{x-x_0}{e_x} = \frac{y-y_0}{e_y} = \frac{z-z_0}{e_z}$$

• Direction and moment vectors

$$\begin{matrix} L : \begin{Bmatrix} {\bf e} \\ {\bf r_0} \times {\bf e} \end{Bmatrix} & \mbox{or} & L :\begin{Bmatrix} {\bf r}_2 - {\bf r}_1 \\ {\bf r}_2 \times {\bf r}_1 \end{Bmatrix} \end{matrix}$$

• Intersection of two planes

$$\begin{matrix} {\bf r} \cdot {\bf n}_1 = d_1 \\{\bf r} \cdot {\bf n}_2 = d_2 \end{matrix}$$

• I am only interested in r = ( 1-t) + t r2 . What is that ???? can you please Isn't t restricted between 0 and 1? – Sedumjoy May 2 '17 at 19:43
• No $t$ can be anything between $-\infty$ to $+\infty$. If you arrange the line horizontally with point ${\bf r}_1$ to the left and point ${\bf r}_2$ to the right then if $t<0$ the point on the line is to the left of ${\bf r}_1$, when $0<t<1$ then between the points and if $t>1$ to the right of ${\bf r}_2$. – John Alexiou May 2 '17 at 20:05
• The position is a blend of the two fixed points. Like mixiing colors, you are mixing vectors. For example 30% of ${\bf r}_1$ and 70% of ${\bf r}_2$ is expressed as $${\bf r} = (1-0.3)\, {\bf r}_1 + 0.3\, {\bf r}_2$$ – John Alexiou May 3 '17 at 3:00
• I see the trick. they must add up to 1 so they fall on the line segment between the points by vector addition using r1 and r2 as position vectors and then you can extrapolate. I can see the utility now but only if you don't have an explicit direction vector and only have the two points. – Sedumjoy May 3 '17 at 19:49
• You got it now. – John Alexiou May 3 '17 at 19:56