Miscellaneous · Q2: Equation of the line through the origin parallel to the $x$-axis

Basics you need

Solution

1. The direction

   The $x$-axis has direction cosines $\cos 0^\circ, \cos 90^\circ, \cos 90^\circ = 1, 0, 0$⁠. A line parallel to it has the same direction, so we take

   $$\vec b = 1\hat i + 0\hat j + 0\hat k = \hat i,$$

   with direction ratios $1, 0, 0$⁠.

   Parallel lines have proportional direction ratios (Section 11.4, condition (ii)), so the simplest choice is the $x$-axis's own $1, 0, 0$⁠. Any nonzero multiple, such as $2, 0, 0$ or $-1, 0, 0$⁠, gives the same line.
   Direction cosines of the axes (Section 11.2). Lesson: direction cosines and ratios

2. The point

   The line passes through the origin, so $(x_1, y_1, z_1) = (0, 0, 0)$ and its position vector is $\vec a = \vec 0$⁠.

   The position vector of a point is the vector from $O$ to the point. From $O$ to $O$ it is the zero vector.
   Position vector of a point (Section 10.2). Basics: three numbers fix a point

3. The vector equation

   $$\vec r = \vec a + \lambda\vec b = \vec 0 + \lambda\hat i = \lambda\hat i, \qquad \lambda \in \mathbf R.$$
   Start at $O$ and walk $\lambda$ steps of $\hat i$⁠, forwards for $\lambda > 0$ and backwards for $\lambda < 0$⁠. Every point reached has position vector $\lambda\hat i$⁠, that is, coordinates $(\lambda, 0, 0)$⁠.
   Section 11.3.1, equation (1). Lesson: start somewhere, then walk

4. The Cartesian equation, with zero denominators

   The parametric equations are

   $$\begin{aligned} x &= 0 + \lambda(1) = \lambda, \\ y &= 0 + \lambda(0) = 0, \\ z &= 0 + \lambda(0) = 0. \end{aligned}$$

   In the form $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$ this is written

   $$\begin{aligned} &\frac{x - 0}{1} = \frac{y - 0}{0} = \frac{z - 0}{0}, \\ &\text{that is,}\quad \frac{x}{1} = \frac{y}{0} = \frac{z}{0}. \end{aligned}$$
   The Cartesian form comes from solving each parametric equation for $\lambda$⁠, which means dividing by $a$⁠, $b$⁠, $c$⁠. When $b = 0$ we cannot divide: the equation $y = y_1 + \lambda b$ just says $y = y_1$⁠. So a 0 in a denominator is a convention, not a division: it means "this numerator is 0". Here $\frac{y}{0}$ and $\frac{z}{0}$ stand for $y = 0$ and $z = 0$⁠, while $x$ is free.
   Section 11.3.1, equations (2) and (3). Basics: a line is a point plus a direction

5. Check: it is the $x$-axis

   The points of the line are $(\lambda, 0, 0)$⁠, for every real $\lambda$⁠. These are exactly the points of the $x$-axis. For example, $\lambda = 3$ gives $(3, 0, 0)$ and $\lambda = -2$ gives $(-2, 0, 0)$⁠. The point $(0, 1, 0)$ is not on the line, since its $y$ is not 0.

   The only line through $O$ parallel to the $x$-axis is the $x$-axis, because two parallel lines that share a point coincide. So "the equation of a line parallel to the $x$-axis through the origin" is the equation of the $x$-axis. In space a line needs two equations: $y = 0$ alone is the $xz$-plane, and $y = 0$⁠, $z = 0$ together give the line.
   Each point of the line has one value of $\lambda$⁠. Basics: go to A, then walk λ copies of b⃗

Every fact used, and where it comes from