Q.Find the distance of the point $(-1, -5, -10)$ from the point of intersection of the line $\vec{r} = 2\hat{i} - \hat{j} + 2\hat{k} + \lambda(3\hat{i} + 4\hat{j} + 2\hat{k})$ and the plane $\vec{r} \cdot (\hat{i} - \hat{j} + \hat{k}) = 5$.

Vector Equation of a Line

A line is fixed by two pieces of information: one point it passes through and the direction it runs in. The vector equation packages both.

Let $\vec{a}$ be the position vector of a known point $A$ on the line, and let $\vec{b}$ be any vector parallel to the line (its direction). For any point $P$ on the line with position vector $\vec{r}$, the displacement $\overrightarrow{AP}$ points along the line, so it is a scalar multiple of $\vec{b}$: $\overrightarrow{AP}=\lambda\vec{b}$. Since $\vec{r}=\vec{a}+\overrightarrow{AP}$,

How to read it

As the parameter $\lambda$ runs through all real numbers, $\vec{r}$ traces every point of the line. At $\lambda=0$ you sit at $A$; positive $\lambda$ moves one way along $\vec{b}$, negative $\lambda$ the other. Think of $\vec{a}$ as "where you start" and $\lambda\vec{b}$ as "how far and which way you walk."

Line through two points

If the line passes through points with position vectors $\vec{a}$ and $\vec{b}$, its direction is $\vec{b}-\vec{a}$, so

Example

The line through $A(1,2,-1)$ parallel to $\vec{b}=2\hat{i}-\hat{j}+3\hat{k}$ is

$\vec{r} = (\hat{i}+2\hat{j}-\hat{k}) + \lambda(2\hat{i}-\hat{j}+3\hat{k}).$ …