Q.Find the distance of the point $(-1, 1)$ from the line $12(x + 6) = 5(y - 2)$.

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

Its magnitude is $\sqrt{25+36+4}=\sqrt{65}$, and $|\vec{b}|=\sqrt{4+1+4}=3$, so

Finding the shortest distance from a point to a line using the cross product is a standard, frequently tested problem in the NCERT Class 12 Three Dimensional Geometry chapter, appearing in CBSE boards, JEE Main and various state CETs. "Distance of a point from a line vector form" is a common search, and this same cross-product technique reappears later when finding the distance between two skew lines.