Vector Equation of a Line

Line through point with position vector a, parallel to vector b. λ ∈ ℝ is the parameter. Every point on the line corresponds to a unique λ.

Let the line pass through point $A$ (position vector $\mathbf{a}$) and be parallel to vector $\mathbf{b}$.

For any point $P$ (position vector $\mathbf{r}$) on the line, the vector $\overrightarrow{AP} = \mathbf{r}-\mathbf{a}$ must be parallel to $\mathbf{b}$:

$$\mathbf{r}-\mathbf{a} = \lambda\mathbf{b} \quad \text{for some } \lambda \in \mathbb{R}$$ $$\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$$

Different values of $\lambda$ give different points on the line:

• $\lambda = 0$: point $A$ itself
• $\lambda > 0$: points in the direction of $\mathbf{b}$ from $A$
• $\lambda < 0$: points in the opposite direction

Line through two points $A(\mathbf{a})$ and $B(\mathbf{b})$:

$$\mathbf{r} = \mathbf{a} + \lambda(\mathbf{b}-\mathbf{a}) = (1-\lambda)\mathbf{a}+\lambda\mathbf{b}$$

Converting to Cartesian: If $\mathbf{a} = (x_1,y_1,z_1)$ and $\mathbf{b} = (a,b,c)$, writing $\mathbf{r} = (x,y,z)$:

$$x = x_1+\lambda a, \quad y = y_1+\lambda b, \quad z = z_1+\lambda c$$

Eliminating $\lambda$: $(x-x_1)/a = (y-y_1)/b = (z-z_1)/c$.