My textbook says, "... an equation of the line through $A$ and $B$ is $x = u + tw = u + t(v — u)$, where $t$ is a real number and $x$ denotes an arbitrary point on the line. An illustration of this is shown below:

See Image

Since $x = u + tw$ is the equation of a line, I presume that it is equivalent to the equation $y = mx + c$, where $m$ is the slope, $x$ is the x-intercept (when $y = 0$), and $c$ is the $y-$intercept (when $x = 0$). Therefore, in the equation $x = u + tw$, $w$ must represent the $x-$intercept $($when $x = 0), u$ must represent the y-intercept (when $w = 0$), and $t$ must represent the slope.

It is clear that there is a connection between the traditional equation of a line, $y = mx + c$, and the vector equation of a line, $x = u + tw = u + t(v — u)$. However, I am struggling to clearly understand the connection between the vectors and the traditional equation:

How can these vectors represent slope ($m$)?

How can these vectors represent x- and y-intercepts?

When I look at the diagram and the vector equations of a line, there does not seem to be a clear indication of how they relate to the cartesian illustration of the traditional equation of a line ($y = mx + c$). For instance, the vectors $w$ and $u$ in $x = u + tw$ would represent the $x$ and $y$ intercepts; But how can a vector represent the $x$ or $y$ intercepts? This makes no sense to me.

To provide further context, take the vector equation of a line $x = (-2,0.1)-t(6,5,2)$. In this case, $(-2,0.1) = u$ and $(6,5,2) = w$; But in what way do the vectors $u$ and $w$ represent $y$ and $x$ intercepts? This is not clear to me.

I would greatly appreciate it if the knowledgeable members of MSE could please take the time to clarify the connection between these two forms of a line.

Thank you.

  • $\begingroup$ Your textbook's picture is pretty poor. It only shows the case where $t=1$. But what's important (and missing) is that $t$ varies, and every value of $t$ gives a different point. $\endgroup$ – bubba Jan 1 '17 at 16:02
  • $\begingroup$ Notation might be the source of your confusion. In $x=u+tw$, $x$, $u$ and $w$ are vectors, but in the equation $y=mx+c$, they are *scalars*—the individual coordinates of a vector, in fact. In particular, the $x$s mean very different things in the two equations. $\endgroup$ – amd Jan 1 '17 at 18:28

The vector equation $\vec{X}(t) = \vec{U} + t\vec{W}$ is really a shorthand for two equations: \begin{align} x &= u_x + t w_x \\ y &= u_y + t w_y \\ \end{align} where $\vec{U}$ has components $(u_x,u_y)$ and $\vec{W}$ has components $(w_x,w_y)$. It's a completely different kind of equation from $y = mx + c$, and it's not helpful to look for correspondences between the two.

However, for 2D lines, the equations are related: from the first equation above, you can get $$ t = \frac{x - u_x}{w_x} $$ Then, if you substitute this in the second equation, you get $$ y = u_y + \frac{x - u_x}{w_x} w_y = \left( \frac{w_y}{w_x} \right)x + \left( u_y - \frac{u_xw_y}{w_x} \right) $$ and this is in the form $y=mx+c$, as you wanted. So, for example, the slope of the line is $\frac{w_y}{w_x}$.

If $w_x = 0$, you get a vertical line (with "infinite" slope) which you can't represent in the form $y=mx+c$. So, the vector form is more general.

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This is perhaps more of a comment than a full answer, but here is one way to think about this.

In vector notation we have $\vec{X}=\vec{u}+t\,\vec{w}$, with $\vec{X}$ pointing to a point on the line parametrised by $t$. In component form we then have, $$ \vec{X} \equiv \begin{pmatrix} x \\ y \end{pmatrix}=\vec{u}+t\,\vec{w} \equiv \begin{pmatrix} u_1 \\ u_2 \end{pmatrix}+t\begin{pmatrix} w_1 \\ w_2 \end{pmatrix}\,. $$ Hence by comparing the two components we see that $y=\frac{w_2}{w_1}x+(u_2-\frac{w_2}{w_1}u_1)\equiv mx+c$. (This would look simpler if we chose to align, say, the $y$-axis with $\vec{u}$.)

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  • $\begingroup$ But is it not an invalid operation to rearrange coordinates as you have? I'm struggling to understand how you got $y=\frac{w_2}{w_1}x+(u_2-\frac{w_2}{w_1}u_1)$ $\endgroup$ – The Pointer Jan 1 '17 at 15:44
  • $\begingroup$ The vector equation is really two equations, given by the two lines. And I've just substituted one into the other. Does that make it clearer? $\endgroup$ – diracula Jan 1 '17 at 15:57
  • $\begingroup$ I'm not sure which equations you're referring to. $\endgroup$ – The Pointer Jan 1 '17 at 16:00
  • $\begingroup$ Sorry, it can be hard to know which notation someone is familiar with. bubba's answer is more explicit, so perhaps that answer will make this point clearer. As bubba says, these vector equations are just shorthand for two equations. We first fix some perpendicular axes, and this defines for every point in the plane an $x$ and a $y$ coordinate. A vector then extends from the origin to one of these points, and so is specified by an $x$ and a $y$. When we have an equation involving vectors, it can be read as one equation relating the $x$ coordinates and one relating the $y$ coordinates. $\endgroup$ – diracula Jan 1 '17 at 16:10
  • $\begingroup$ Yes, I understand now. Thanks for the assistance. $\endgroup$ – The Pointer Jan 1 '17 at 16:11

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