The Straight Line in Space

Through point $(x_0,y_0,z_0)$ with direction ratios $\langle a,b,c\rangle$, the symmetric form is $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}$. Through two points $A, B$, the direction ratios are $\langle x_2-x_1, y_2-y_1, z_2-z_1\rangle$. Setting the chain equal to a parameter $t$ gives the parametric form $(x_0+at,\ y_0+bt,\ z_0+ct)$. Splitting the chain into two equalities expresses the line as the intersection of two planes. Conversely, when a line is GIVEN as the intersection of two planes $P_1, P_2$, its direction ratios are $\vec{n_1} \times \vec{n_2}$ — the cross product of the two normals (the line lies in both planes, so it is perpendicular to both normals).

Points on $(x_0+at,\ y_0+bt,\ z_0+ct)$ are generated by choosing $t$. To test whether $(p,q,r)$ lies on the line, solve $p = x_0+at$ for $t$ and confirm the SAME $t$ reproduces $q$ and $r$. Foot of the perpendicular from a point $P$ to the line, and the distance: let the foot be $F = (x_0+at,\ y_0+bt,\ z_0+ct)$ and impose $\vec{PF}\perp(a,b,c)$: $a(x_0+at-p) + b(y_0+bt-q) + c(z_0+ct-r) = 0$, a linear equation in $t$. Then $d = |PF|$. Closed form: $d = \dfrac{|\vec{AP}\times\vec{u}|}{|\vec{u}|}$ where $A$ is any point on the line and $\vec{u}=(a,b,c)$; equivalently $d^2 = |AP|^2 - \big(\vec{AP}\cdot\hat{u}\big)^2$. The image of $P$ in the line is $P' = 2F - P$.

A line meets the XY-plane where its $z$-coordinate is 0, the YZ-plane where $x=0$, and the ZX-plane where $y=0$. Parametrise the line, set the relevant coordinate to 0, solve for $t$, and back-substitute.

For a line $(x_0+at,\ y_0+bt,\ z_0+ct)$ and plane $Ax+By+Cz = D$, substitute to get a linear equation in $t$. One solution → a single pierce point. No solution (the $t$ terms cancel but the constants disagree) → the line is parallel and misses. Identity (all $t$ work) → the line lies in the plane.

For a line with direction $\langle a,b,c\rangle$ and a plane with normal $\langle A,B,C\rangle$:

• Parallel iff $Aa + Bb + Cc = 0$ (direction ⟂ normal) AND a point of the line is NOT on the plane.
• Lies in the plane iff $Aa+Bb+Cc = 0$ AND a point of the line satisfies the plane equation.