Q.Find the points on the x-axis, whose distances from the line $\dfrac{x}{3} + \dfrac{y}{4} = 1$ are $4$ units.

Distance from a Point to a Line

The distance from a point to a line is the shortest distance — the length of the perpendicular dropped from the point onto the line. In 3D we compute it with vectors and the cross product.

Let the line be $\vec{r}=\vec{a}+\lambda\vec{b}$ (a point $A$ with position vector $\vec{a}$, direction $\vec{b}$), and let $P$ be the given point with position vector $\vec{p}$.

The idea

Look at the triangle formed by $A$, $P$ and the foot of the perpendicular $M$. The segment $\overrightarrow{AP}=\vec{p}-\vec{a}$ is the hypotenuse, and the perpendicular distance $d=PM$ is the side opposite the angle $\theta$ between $\overrightarrow{AP}$ and the line:

But the cross product already contains $\sin\theta$: $|\overrightarrow{AP}\times\vec{b}| = |\overrightarrow{AP}|\,|\vec{b}|\sin\theta$. Dividing by $|\vec{b}|$ isolates the distance.

Example

Distance of $P(1,2,3)$ from the line $\vec{r}=(\hat{i}+\hat{j})+\lambda(2\hat{i}-\hat{j}+2\hat{k})$.

Here $\vec{a}=(1,1,0)$, $\vec{b}=(2,-1,2)$, and $\overrightarrow{AP}=\vec{p}-\vec{a}=(0,1,3)$.

$\overrightarrow{AP}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\0&1&3\\2&-1&2\end{vmatrix} = (2+3)\hat{i} - (0-6)\hat{j} + (0-2)\hat{k} = 5\hat{i}+6\hat{j}-2\hat{k}.$ …