Vector form:
$$\mathbf{r} = \mathbf{a} + t\mathbf{d}, \quad t \in \mathbb{R}$$
Parametric form:
$$x = a_1 + td_1, \quad y = a_2 + td_2, \quad z = a_3 + td_3$$
Symmetric form (when all $d_i \neq 0$):
$$\frac{x-a_1}{d_1} = \frac{y-a_2}{d_2} = \frac{z-a_3}{d_3} = t$$
Example 1: Line through $A(1,-1,2)$ parallel to $\mathbf{d} = 3\mathbf{i}-\mathbf{j}+2\mathbf{k}$:
$$\mathbf{r} = (1-1+2) + t(3-1+2): \quad x = 1+3t,\ y=-1-t,\ z=2+2t$$
Or symmetrically: $\dfrac{x-1}{3} = \dfrac{y+1}{-1} = \dfrac{z-2}{2}$.
Two lines $\ell_1: \mathbf{r} = \mathbf{a}+s\mathbf{d}$ and $\ell_2: \mathbf{r} = \mathbf{b}+t\mathbf{e}$:
Testing for intersection: Solve the 3 equations in 2 unknowns ($s, t$). If consistent, lines intersect.
Normal (Cartesian) form:
$$n_1 x + n_2 y + n_3 z = d$$
where $\mathbf{n} = (n_1, n_2, n_3)$ is the normal vector.
Vector form: $\mathbf{n}\cdot\mathbf{r} = d$.
Parametric form: $\mathbf{r} = \mathbf{a} + s\mathbf{b} + t\mathbf{c}$ where $\mathbf{b}, \mathbf{c}$ are non-parallel vectors in the plane.
Example 2: Plane through $(1,0,-1)$, $(2,1,0)$, $(0,1,2)$.
$\mathbf{b} = (1,1,1)$, $\mathbf{c} = (-1,1,3)$.
$\mathbf{n} = \mathbf{b}\times\mathbf{c} = (3-1)\mathbf{i}-(3-(-1))\mathbf{j}+(1-(-1))\mathbf{k} = 2\mathbf{i}-4\mathbf{j}+2\mathbf{k}$.
Simplify: $\mathbf{n} = \mathbf{i}-2\mathbf{j}+\mathbf{k}$.
$d = \mathbf{n}\cdot(1,0,-1) = 1+0-1 = 0$. Plane: $x - 2y + z = 0$.
Substitute parametric equations of the line into the plane equation.
Example 3: Line $\mathbf{r} = (1,2,0)+t(1,-1,2)$; plane $2x+y-z=5$.
Substitute: $2(1+t)+(2-t)-(2t)=5 \Rightarrow 2+2t+2-t-2t=5 \Rightarrow 4-t=5 \Rightarrow t=-1$.
Intersection: $(0,3,-2)$.
Two non-parallel planes intersect in a line. Find it by solving the two plane equations simultaneously.
Example 4: $x+y+z=6$ and $x-y+2z=4$.
Subtract: $2y-z=2 \Rightarrow z = 2y-2$. Let $y=t$: $z=2t-2$, $x=6-y-z=6-t-(2t-2)=8-3t$.
Line: $(8,0,-2)+t(-3,1,2)$.
KEY TAKEAWAY: Lines need a point and a direction; planes need a point and a normal (or equivalently, two in-plane direction vectors). Intersections are found by substitution.
EXAM TIP: To find the equation of a plane containing a given line and point, use the cross product of the direction vector and a connecting vector to find the normal.
COMMON MISTAKE: Forgetting to check whether two 3D lines are skew (no intersection, not parallel) rather than intersecting. Always verify all three equations are consistent.
